Canonical Forms of Neural Ideals

Abstract

Neural ideals, originally defined in Curto et al., give a way of translating information about the firing pattern of a set of neurons into a pseudomonomial ideal in a polynomial ring. We give a simple criterion for determining whether a neural ideal is in canonical form, along with an improved algorithm for computing the canonical form of a neural ideal.

1 Introduction

Neural rings were first introduced in [4] as an algebraic tool to study receptive field codes (RF codes). In certain regions of the brain, such as the hippocampus, neurons have receptive fields, which are the locations where they fire in response to stimuli. These receptive fields are typically convex subsets of the stimulus space, and the collection of receptive fields covers the stimulus space. The receptive field code consists of a binary codeword for each distinct intersection of receptive fields that appears in the stimulus space. Typically, we consider stimulus spaces as subsets of a subspace X of $\mathbb {R}^d$. One major question in the field is how to identify whether a given binary code can arise from the intersection patterns of some collection of convex sets in some Euclidean space [1,2,3, 5, 8,9,10].

The neural ring serves as a bridge allowing for algebraic methods to be applied to neuroscience and coding theory. Given n neurons, the first step of the construction translates neural firing patterns to a binary code $\mathcal {C} \subseteq \{0,1\}^n$ where each possible combination of neuron firing is translated into a codeword of length n. For a given codeword $c = (c_1,\ldots ,c_n)$, the entry $c_i = 1$ means the ith neuron is firing whereas $c_i = 0$ means the ith neuron is not firing. Given a neural code $\mathcal {C}$, the neural ring is $R/I_{\mathcal {C}}$, where $R=\mathbb {F}_2[x_1,\ldots ,x_n]$ and $I_{\mathcal {C}}$ is the vanishing ideal of $\mathcal {C}$. Using a vanishing ideal enables us to apply tools from algebraic geometry to neural codes. However, this ideal includes the trivial Boolean relations $\{x_i(1 - x_i): 1 \le i \le n\}$; these hold for every neural code and so are not useful in distinguishing neural codes from each other. Instead, [4] defines the neural ideal $J_{\mathcal {C}} \subseteq I_{\mathcal {C}} \subseteq R$ formed from the non-trivial relations of the neural code. The two ideals relate through the equation

$$\begin{aligned} I_{\mathcal {C}} = J_{\mathcal {C}} + \langle x_i(1 - x_i): 1 \le i \le n \rangle \end{aligned}$$

(cf. [4, Lemma 3.2]). As such, one can study the neural ideal in place of the vanishing ideal.

The information encoded in the neural ideal is made more accessible by presenting the ideal in canonical form (cf. [4, Sect. 4.3]). The canonical form of a neural ideal can be used to identify various obstructions to convexity [3]. As such it is useful to know whether or not a neural ideal is being presented in canonical form.

One downside to the neural ideal from an algebraic perspective is that it is generated by pseudomonomials, products $\Pi _{i \in \sigma } x_i \Pi _{i \in \tau } (1-x_i)$, which are neither graded nor local and as a result are not as well studied in commutative algebra. To deal with this issue, Güntürkün, Jeffries, and Sun give a process for polarizing a neural ideal, turning it into a squarefree monomial ideal in the extension ring $\mathbb {F}_2[x_1,\ldots ,x_n,y_1,\ldots ,y_n]$ [7]. In our work, we utilize the polarization technique from [7] to apply monomial ideal results to classify when certain (polarized) neural ideals are presented in canonical form. As a result, our main theorem below establishes sufficient conditions on the generators of a polarized neural ideal to determine when it is in canonical form. Here, saying that two monomials g and h “share an index" means that there is some $1 \le i \le n$ such that $x_i \mid g$ and $y_i \mid h$ or vice versa.

Theorem 1.1

(Theorem 5.5) Let $\mathfrak {a}=(g_1,\ldots ,g_k)$ be a polarized neural ideal such that $g_{j_1} \not \mid g_{j_2}$ for any $1 \le j_1 \ne j_2 \le k$, and $x_iy_i \not \mid g_j$ for any $1 \le i \le n$ and $1 \le j \le k$. If for some pair $g_{j_1},g_{j_2}$ of generators of $\mathfrak {a}$, $g_{j_1}$ and $g_{j_2}$ share exactly 1 index i and no other generator of $\mathfrak {a}$ divides $\frac{\text {lcm}\{g_{j_1},g_{j_2}\}}{x_iy_i}$, then $\mathfrak {a}$ is not in canonical form. Otherwise, $\mathfrak {a}$ is in canonical form.

This result will be useful to researchers trying to prove results about all neural ideals, as they can restrict to the set of canonical forms described by this theorem.

As a consequence of this theorem, we are able to give a shortened algorithm for computing the canonical form of a neural ideal (Algorithm 5.10). This algorithm removes the need to compute the primary decomposition of a neural ideal, which simplifies the process significantly, and appears to speed up the computation in initial tests. It is distinct from the algorithm given in [12], which starts with a binary code and produces a neural ideal in canonical form, whereas ours starts with a neural ideal. As a result, we expect our algorithm to be more useful to mathematicians proving results about families of neural ideals, for example classification results, that can then be applied to neural codes.

In Sect. 6, we prove that it is possible to compute the almost canonical form of a family of neural ideals all at once, by passing to a generic canonical form. We anticipate that algebraic results about the generic canonical form should carry nicely to every member of the corresponding family. This will make generic canonical forms a useful tool for classifying neural ideals by their algebraic properties.

The structure of our paper is as follows. Section 2 covers the necessary background information from [4] and [7]. We also recall some useful observations about squarefree monomial ideals.

Section 3 is dedicated to translating the algorithm from [4] for computing the canonical form of a neural ideal into an algorithm that can be applied to polarized neural ideals (Algorithm 3.2). In Sect. 4, we identify common patterns in the algorithm based off of the generators of our ideal, giving us a number of shortcuts through Algorithm 3.2.

Along with containing our main theorem, Sect. 5 includes a complete classification of the canonical forms of all two-generated neural ideals and a simplified algorithm for computing the canonical form of a neural ideal (Algorithm 5.10). In Sect. 6, we discuss a generic canonical form, a tool that allows us to use the canonical form of a single neural ideal to compute the canonical forms of a number of related neural ideals.

2 Background on Neural Ideals and Polarization

Throughout the paper, we make use of the following notation.

Notation 2.1

• (1) R will denote the ring $\mathbb {F}_2[x_1,\ldots ,x_n]$.

• (2) A pseudomonomial is a product $\prod _{i \in \sigma } x_i \prod _{i \in \tau } (1-x_i)$ in R where $\sigma ,\tau \subseteq \{1,2,\ldots ,n\}$ and $\sigma \cap \tau = \varnothing$.

Definition 2.2

To get the neural ideal, we work with a space $X \subseteq \mathbb {R}^d$ and open sets $U_1,\ldots ,U_n \subseteq X$ where neuron i fires at the points in $U_i$. The neural ideal, as introduced in [4], is an ideal of R that captures the RF-structure of a neural code, specifically the relations

$$\begin{aligned} \bigcap _{i \in \sigma } U_i \subseteq \bigcup _{i \in \tau } U_i, \end{aligned}$$

(1)

where $\sigma ,\tau \subseteq \{1,\ldots ,n\}$ and $\sigma \cap \tau = \varnothing$. More precisely, the neural ideal is generated by the pseudomonomials $\prod _{i \in \sigma } x_i \prod _{i \in \tau } (1-x_i)$ corresponding to the relations above [4, Sect. 4.2].

Note that the Boolean relations $U_i \subseteq U_i$ always hold, but they do not distinguish between distinct neural codes, so they are not included in the neural ideal. Hence we assume that no generator of a neural ideal is divisible by $x_i(1-x_i)$ for any $1 \le i \le n$.

Definition 2.3

The canonical form of the neural ideal captures the minimal relations in the RF-structure of the neural code, in the sense that in Eq. 1 from Definition 2.2 the intersection on the left and the union on the right are irredundant (i.e., if we removed any $U_i$ from either side of the inclusion, the inclusion would no longer hold). The canonical form consists of the pseudomonomials corresponding to the minimal relations and is a generating set for the neural ideal [4, Sect. 4.3].

One downside to these ideals is that they are generated by pseudomonomials, and as a result are neither graded nor local. This is an issue when applying techniques from commutative and homological algebra as most of our tools require rings to be local or graded, for example computing a minimal free resolution and corresponding Betti numbers. In order to resolve this issue, Güntürkün, Jeffries, and Sun [7] developed a method to pass from pseudomonomial ideals in R to squarefree monomial ideals in $S=\mathbb {F}_2[x_1,\ldots ,x_n,y_1,\ldots ,y_n]$ by polarization.

Definition 2.4

([7])

• (1) S will denote the extension ring $\mathbb {F}_2[x_1,\ldots ,x_n,y_1,\ldots ,y_n]$ of R.

• (2) For a monomial m we denote its largest square free divisor as [m]. For example, $[x^3y^2] = xy$. One can quickly check that if $m_1$ and $m_2$ are squarefree monomials, then $[m_1m_2] = \text {lcm}\{m_1,m_2\}$.

• (3) We work with neural ideals, which are generated by pseudomonomials.

• (4) We will refer to a polarization function $\mathcal {P}$ (not a homomorphism) sending pseudomonomials in R to squarefree monomials in S via

  $$\begin{aligned}\prod _{i \in \sigma } x_i \prod _{i \in \tau } (1-x_i) \mapsto \prod _{i \in \sigma } x_i \prod _{i \in \tau } y_i.\end{aligned}$$

• (5) If $\mathcal {A}$ is an ideal of R, then $\mathcal {P}(\mathcal {A})$ is the ideal of S generated by $\mathcal {P}(f)$ for all pseudomonomials $f \in \mathcal {A}$.

• (6) We will refer to a depolarization map $d:S \rightarrow R$, which is the ring homomorphism sending $x_i \mapsto x_i, y_i \mapsto 1-x_i$. The map d induces an isomorphism of S-modules $R \cong S/\left( \sum _{i=1}^n (x_i+y_i-1)\right)$.

We will often work with the polarization of a neural ideal, while explaining how this translates to the depolarized version.

Theorem 2.5

([7, Theorem 3.2]) Let $\mathcal {A}=(g_1,\ldots ,g_k) \subseteq R$ be a neural ideal in canonical form. Then $\mathcal {P}(\mathcal {A})=(\mathcal {P}(g_1),\ldots ,\mathcal {P}(g_k))$.

However, this frequently does not hold when the ideal is not in canonical form.

Example 2.6

For example, $(x_1,x_2(1-x_1))$ contains $x_2$, so $\mathcal {P}(x_1,x_2(1-x_1))$ contains $x_2$. However,

$$\begin{aligned} x_2 \not \in (\mathcal {P}(x_1),\mathcal {P}(x_2(1-x_1)))=(x_1,x_2y_1). \end{aligned}$$

The canonical form of $(x_1,x_2(1-x_1))$ turns out to be $(x_1,x_2)$.

Definition 2.7

([4]) A pseudomonomial prime of R is an ideal p generated by a subset of $\{x_1,\ldots ,x_n,1-x_1,\ldots ,1-x_n\}$ such that for each $1 \le i \le n$, p does not contain both $x_i$ and $1-x_i$.

Theorem 2.8

([4, Theorem 5.4]) Let $\mathcal {A}$ be a pseudomonomial ideal in $\mathbb {F}_2[x_1,\ldots ,x_n]$. Then the primary decomposition of $\mathcal {A}$ consists of pseudomonomial primes, and so $\mathcal {A}$ has no embedded primes. Hence its primary decomposition is unique.

Theorem 2.9

([11, Sect. 4.5]) Let $\mathfrak {a}$ be a squarefree monomial ideal in a polynomial ring $k[x_1,\ldots ,x_n]$. The primary decomposition of $\mathfrak {a}$ consists of primes generated by a subset of $\{x_1,\ldots ,x_n\}$ (monomial primes), and so $\mathfrak {a}$ has no embedded primes. Hence its primary decomposition is unique.

Remark 2.10

Let $g_1,\ldots ,g_k$ be squarefree monomials in a polynomial ring $k[x_1,\ldots , x_n]$. If $g_j=hh'$ for any squarefree monomials $h,h'$ in the ring, then

$$\begin{aligned} (g_1,\ldots ,g_j,\ldots ,g_k)=(g_1,\ldots ,h,\ldots ,g_k) \cap (g_1,\ldots ,h',\ldots ,g_k). \end{aligned}$$

In order to compute the primary decomposition of $(g_1,\ldots ,g_k)$, we may use this rule repeatedly until we get an intersection of monomial primes.

3 Computing the Canonical Form of a Polarized Neural Ideal

In this section we give a polarized version of the algorithm introduced in [4] for computing the canonical form of a neural ideal and prove that it agrees with the original algorithm in [4, Sect. 4.5].

We recall the original algorithm for computing the canonical form of a neural ideal below.

Algorithm 3.1

([4, Sect. 4.5])

• (1) Start with a neural ideal $\mathcal {A}=(g_1,\ldots ,g_k)$ in R.

• (2) Compute the primary decomposition of $\mathcal {A}$. By Theorem 2.8, the ideals $p_1,\ldots ,p_s$ in the primary decomposition will all be generated by a subset of $\{x_1,\ldots ,x_n,1-x_1,\ldots ,1-x_n\}$.

• (3) Compute the set of products $h_1 \cdots h_s$ where $h_\ell$ is a generator of $p_\ell$.

• (4) Set $x_i(1-x_i)=0$ for each $1 \le i \le n$, and as a result remove any product divisible by $x_i(1-x_i)$ and replace any power $x_i^t$ by $x_i$ and $(1-x_i)^t$ by $1-x_i$. All remaining elements are now pseudomonomials.

• (5) Remove any product that is a multiple of a product of lower degree.

The remaining products give the canonical form of $\mathcal {A}$ in R.

We give our polarized version of the algorithm below.

Algorithm 3.2

• (1) Start with a squarefree monomial ideal $\mathfrak {a}=(g_1,\ldots ,g_k) \in S$.

• (2) Compute the primary decomposition of $\mathfrak {a}$. By Theorem 2.9, the ideals $p_1,\ldots ,p_s$ in the primary decomposition will all be generated by a subset of $\{x_1,\ldots ,x_n,y_1, \ldots ,y_n\}$.

• (3) Set $x_i+y_i=1$ for $1 \le i \le n$, and as a result remove any ideal in the primary decomposition containing both $x_i$ and $y_i$ for any $1 \le i \le n$.

• (4) Compute the intersection of the remaining ideals. Since the $p_{\ell }$ are squarefree monomial ideals, this is equivalent to computing $[h_1 \cdots h_s]$ for every set of choices of $h_{\ell }$ a generator of $p_{\ell }$.

• (5) Impose the relations $x_iy_i=0$ for $1 \le i \le n$, and as a result remove any generator divisible by $x_iy_i$ for some i. (These come from imposing the relations $x_i(1-x_i)=0$ in the depolarized ideal.)

• (6) Remove any generator that is a multiple of another generator of lower degree.

The remaining generators give the canonical form of $\mathfrak {a}$ in S.

Lemma 3.3

• (1) Let $f=\prod _{i \in \sigma } x_i \prod _{i \in \tau } (1-x_i)$ be a pseudomonomial in R. Then $d(\mathcal {P}(f))=f$.

• (2) Let $g=\prod _{i \in \sigma } x_i \prod _{i \in \tau } y_i$ be a monomial in S. Then $\mathcal {P}(d(g))=g$.

• (3) Let $\mathfrak {a}=(g_1,\ldots ,g_{k})$ be a squarefree monomial ideal in S. Then $d(\mathfrak {a})~=~(d(g_1), \ldots ,d(g_{k}))R$, though the generators may not be minimal. If $x_i,y_i \in \mathfrak {a}$ for some $1 \le i \le n$, then $d(\mathfrak {a})=(1)$.

• (4) Let p be a pseudomonomial prime of R. Then p is in canonical form.

• (5) Let p be a pseudomonomial prime of R. Then $\mathcal {P}(p)$ is a monomial prime of S.

• (6) Let $\mathfrak {a}$ be a squarefree monomial ideal of S. If for all $1 \le i \le n$ $x_iy_i$ does not divide any generator, then $d(\mathfrak {a})$ is an ideal of R generated by pseudomonomials. In particular, if p is a monomial prime of S such that for all $1 \le i \le n$, $\{x_i,y_i\}$ is not a subset of p, then d(p) is a pseudomonomial prime of R.

• (7) Let $\mathcal {A}$ be a pseudomonomial ideal of R. Then $\mathcal {A}\subseteq d(\mathcal {P}(\mathcal {A}))$. If $\mathcal {A}$ is a pseudomonomial prime, this is an equality.

• (8) Let p be a monomial prime of S such that for $1 \le i \le n$, at most one of $x_i,y_i$ is in p. Then $\mathcal {P}(d(p))=p$.

Proof

Parts (1) and (2) are clear. Part (3) follows from observing that d is an S-module homomorphism.

For (4), we note that p is in canonical form since it is preserved by Algorithm 3.1. Then (5) follows by Theorem 2.5.

Statement (6) follows from (3) and the fact that an ideal generated by some $x_i$ and some $1-x_j$ where there is no index for which we have both $x_i$ and $1-x_i$ as generators is prime.

To prove (7), first let $f \in \mathcal {A}$ be a pseudomonomial. Then by part (1), $f=d(\mathcal {P}(f)) \in d(\mathcal {P}(\mathcal {A}))$. If $\mathcal {A}$ is a pseudomonomial prime, then without loss of generality it has the form $(x_1,\ldots ,x_k,(1-x_{k+1}),\ldots ,(1-x_{k+t}))$. Since this ideal is in canonical form by (4), it follows from Theorem 2.5 that $\mathcal {P}(\mathcal {A})=(x_1,\ldots ,x_k,y_{k+1},\ldots ,y_{k+t})$. Then $d(\mathcal {P}(\mathcal {A}))=\mathcal {A}$.

For (8), assume without loss of generality that

$$\begin{aligned}p=(x_1,\ldots ,x_k,y_{k+1},\ldots ,y_{k+t}).\end{aligned}$$

Then $d(p)=(x_1,\ldots ,x_k,(1-x_{k+1}),\ldots ,(1-x_{k+t}))$ is in canonical form by (4). The result now follows from Theorem 2.5. $\square$

Example 3.4

Consider the ideal $\mathfrak {a}=(x_1,x_2y_1)$ of S. The depolarized ideal is $d(\mathfrak {a})=(x_1,x_2(1-x_1))=(x_1,x_2)$, so we see that there is a “more minimal" way to write $d(\mathfrak {a})$, as mentioned in part (3) of Lemma 3.3.

Lemma 3.5

Let $\mathcal {A}\subseteq R$ be a neural ideal, and let $\mathfrak {a} \subseteq S$ be its polarization. The set of associated primes of $\mathcal {A}$ is equal to the set of primes obtained by depolarizing the set of associated primes of $\mathfrak {a}$, after removing primes containing both $x_i$ and $y_i$ for some $1 \le i \le n$.

Proof

We need to prove two claims:

• (1) If p is a minimal prime of $\mathcal {A}$, then $\mathcal {P}(p)$ is a minimal prime of $\mathfrak {a}$.

• (2) If p is a minimal prime of $\mathfrak {a}$ and there is no $1 \le i \le n$ such that p contains both $x_i$ and $y_i$, then d(p) is a minimal prime of $\mathcal {A}$.

Note that by Theorem 2.9, the irredundant primary decomposition of $\mathfrak {a}$ is unique. Similarly, by Theorem 2.8, the irredundant primary decomposition of $\mathcal {A}$ is also unique. $\square$

Proof of Claim 1

Let p be a minimal prime of $\mathcal {A}$. By Theorem 2.8, p is generated by pseudomonomials $x_i$ and $1-x_i$ (where for each i, at most one of $x_i,1-x_i$ is a generator). By Lemma 3.3, $\mathcal {P}(p)$ is prime, and $\mathfrak {a} \subseteq \mathcal {P}(p)$. Suppose $\mathcal {P}(p)$ is not minimal, i.e. there is a prime q with $\mathfrak {a} \subseteq q \subsetneq \mathcal {P}(p)$. Since $\mathfrak {a}$ is a squarefree monomial ideal, q is generated by a subset of the variables $x_1,\ldots ,x_n,y_1,\ldots ,y_n$. Since $q \subseteq \mathcal {P}(p)$, which does not contain both $x_i$ and $y_i$ for any i, Lemma 3.3 implies d(q) is prime in R. We have

$$\begin{aligned}\mathcal {A}\subseteq d(\mathcal {P}(\mathcal {A}))=d(\mathfrak {a}) \subseteq d(q) \subseteq d(\mathcal {P}(p))=p.\end{aligned}$$

Since p was a minimal prime of $\mathcal {A}$ by assumption, $d(q)=p$. Suppose that $q \subsetneqq \mathcal {P}(p)$. Then there must be some $z \in \{x_1,\ldots ,x_n,y_1,\ldots ,y_n\}$ such that $z \in \mathcal {P}(p)$, but $z \not \in q$. Then $d(z) \in d(\mathcal {P}(p))=p=d(q)$. Applying Lemma 3.3, $z=\mathcal {P}(d(z)) \in \mathcal {P}(d(q))=q$, since q does not contain $x_i$ and $y_i$ for any $1 \le i \le n$. So $q=\mathcal {P}(p)$, which implies that $\mathcal {P}(p)$ is a minimal prime of $\mathfrak {a}$. $\square$

Proof of Claim 2

Let p be a minimal prime of $\mathfrak {a}$ such that for $1 \le i \le n$, p does not contain both $x_i$ and $y_i$. Then by Lemma 3.3, d(p) is a pseudomonomial prime and

$$\begin{aligned}\mathcal {A}\subseteq d(\mathcal {P}(\mathcal {A}))=d(\mathfrak {a}) \subseteq d(p).\end{aligned}$$

Suppose there is some pseudomonomial prime q of R such that $\mathcal {A}\subseteq q \subsetneq d(p)$. Then $\mathcal {P}(q)$ is prime and $\mathfrak {a} \subseteq \mathcal {P}(q) \subseteq \mathcal {P}(d(p))$, which is equal to p by Lemma 3.3. Since p is a minimal prime of $\mathfrak {a}$, $\mathcal {P}(q)=p$. Then by Lemma 3.3 and since d is a homomorphism, $q=d(\mathcal {P}(q))=d(p)$. So d(p) is a minimal prime of $\mathcal {A}$, as desired.

Since neither $\mathcal {A}$ nor $\mathfrak {a}$ has any embedded primes, this proves the result. $\square$

Proposition 3.6

Let $\mathcal {A}\subseteq \mathbb {F}_2[x_1,\ldots ,x_n]$ be a neural ideal, and let $\mathfrak {a}$ be its polarization. The canonical form of $\mathcal {A}$ as computed in Algorithm 3.1 is equal to the depolarization of the canonical form of $\mathfrak {a}$ as computed in Algorithm 3.2.

Proof

First, by Lemma 3.5, at the end of Step 3 of Algorithm 3.2, the remaining set of prime ideals, when depolarized, is equal to the set of prime ideals occurring in the primary decomposition of $\mathcal {A}$.

Next we prove that the set of products left at the end of Step 5 of Algorithm 3.2, depolarized, agree with the products left at the end of Step 4 of Algorithm 3.1. We need to show that if z and w are pseudomonomials in R, if we apply the relations $x_i(1-x_i)=0$ to zw, the result is the same as if we compute $[\mathcal {P}(z)\mathcal {P}(w)]$, impose the relation $x_iy_i=0$, and then depolarize. Suppose

$$\begin{aligned} z=\prod _{i \in \sigma _1} x_i \prod _{i \in \tau _1} (1-x_i), \quad \text {and} \quad w=\prod _{i \in \sigma _2} x_i \prod _{i \in \tau _2} (1-x_i). \end{aligned}$$

Then the result of applying the first process to zw is

$$\begin{aligned} {\left\{ \begin{array}{ll} \prod _{i \in (\sigma _1 \cup \sigma _2)} x_i \prod _{i \in (\tau _1 \cup \tau _2)} (1-x_i) &{} (\sigma _1 \cup \sigma _2) \cap (\tau _1 \cup \tau _2)=\varnothing \\ 0 &{} \text {else}.\\ \end{array}\right. } \end{aligned}$$

The result of applying the second process to $[\mathcal {P}(z)\mathcal {P}(w)]$ is

$$\begin{aligned} {\left\{ \begin{array}{ll} \prod _{i \in (\sigma _1 \cup \sigma _2)} x_i \prod _{i \in (\tau _1 \cup \tau _2)} y_i &{} (\sigma _1 \cup \sigma _2) \cap (\tau _1 \cup \tau _2)=\varnothing \\ 0 &{} \text {else}.\\ \end{array}\right. } \end{aligned}$$

Depolarizing, this agrees with the result of the first process.

Finally, by [7, Lemma 3.1], if z and w are two pseudomonomials in R, $z \mid w$ if and only if $\mathcal {P}(z) \mid \mathcal {P}(w)$. Hence the depolarization of the result of Step 6 of Algorithm 3.2 agrees with the result of Step 5 of Algorithm 3.1. $\square$

4 Deconstructing the Canonical Form Algorithm

In this section we prove a number of lemmas describing in more detail what happens to the generators of a polarized neural ideal under the steps of Algorithm 3.2 and giving us shortcuts through the algorithm.

Definition 4.1

We say two monomials $m_1$ and $m_2$ share an index i if $x_i$ divides one of them and $y_i$ divides the other. If a single monomial is divisible by $x_iy_i$, we do not count this as sharing an index with itself (see Remark 4.8). Moreover, we say that the generators of a squarefree monomial ideal $\mathfrak {a}$ share the index i if there are two generators of the ideal sharing the index i.

Example 4.2

For example, $x_1y_2$ and $x_3y_1$ share the index 1. However, $x_1y_2$ and $x_1y_3$ share no index, even though $x_1$ divides both.

Remark 4.3

We state the results in this section for squarefree monomial ideals rather than for polarized neural ideals so that we can apply them to ideals appearing as intermediate stages in Algorithm 3.2, which may have generators divisible by $x_iy_i$ for some $1 \le i \le n$.

Definition 4.4

Let $\mathfrak {a}$ be a squarefree monomial ideal in S. We say that $\mathfrak {a}$ is recomposed if it is the result of applying Steps 1–4 of Algorithm 3.2 to some squarefree monomial ideal, and call the process of applying Steps 1–4 to a squarefree monomial ideal recomposing it. We say that $\mathfrak {a}$ is in almost canonical form if it is the result of applying Steps 1–5 of Algorithm 3.2 to some squarefree monomial ideal.

Lemma 4.5

Let $\mathfrak {a}$ be a squarefree monomial ideal in S. For some $0 \le t \le n$ and $i_1,\ldots ,i_t \in \{1,\ldots ,n\}$, we can write $\mathfrak {a}=\mathfrak {a}_0 \cap \mathfrak {a}_1 \cap \ldots \cap \mathfrak {a}_t,$ where

• (1) $\mathfrak {a}_0$ is a (possibly empty) intersection of monomial primes, none of which contain both $x_i$ and $y_i$ for any $1 \le i \le n$, and

• (2) for $1 \le j \le t$, $\mathfrak {a}_j$ is an intersection of monomial prime ideals containing both $x_{i_j}$ and $y_{i_j}$.

In particular, $\mathfrak {a}_j$ is either the unit ideal or a squarefree monomial ideal for each $0 \le j \le n$. For any such decomposition, recomposing $\mathfrak {a}$ returns $\mathfrak {a}_0$, and so $\mathfrak {a}$ and $\mathfrak {a}_0$ have the same canonical form. In fact for any $s \le t$, recomposing $\mathfrak {a}'=\mathfrak {a}_0 \cap \mathfrak {a}_s \cap \ldots \cap \mathfrak {a}_t$ yields $\mathfrak {a}_0$, and so $\mathfrak {a}'$ has the same canonical form as $\mathfrak {a}_0$.

Consequently, if $\mathfrak {a}=\mathfrak {b} \cap \mathfrak {c}$ with $\mathfrak {b}$ and $\mathfrak {c}$ squarefree monomial ideals and $(x_i,y_i) \subseteq \mathfrak {b}$ for some $1 \le i \le n$, recomposing $\mathfrak {a}$ returns the same result as recomposing $\mathfrak {c}$.

Remark 4.6

If $\mathfrak {a}_0$ is empty, then the canonical form of $\mathfrak {a}$ has no generators. This corresponds to the code containing all code words.

Proof

We prove this by induction on t, the order of the set

$$\begin{aligned} I=\{i:1 \le i \le n,(x_i,y_i) \subseteq p \text { for some } p \text { in a primary decomposition of } \mathfrak {a}\}. \end{aligned}$$

If $t=0$, $\mathfrak {a}=\mathfrak {a}_0$ so the result is immediate. Suppose $t>0$. Without loss of generality, assume that $I=\{1,\ldots ,t\}$. For $1 \le i \le t$, set $\mathfrak {a}_i$ to be the intersection of the prime ideals p appearing in a primary decomposition of $\mathfrak {a}$ that contain $x_i$ and $y_i$. Set $\mathfrak {a}_0$ to be the intersection of the primes p appearing in the primary decomposition of $\mathfrak {a}$ that do not contain $x_i,y_i$ for any $1 \le i \le n$.

When we perform Step 3 of Algorithm 3.2, we will remove exactly the primes p that contain a pair $x_i,y_i$ for some $1 \le i \le t$. This exactly corresponds to removing $\mathfrak {a}_1,\ldots ,\mathfrak {a}_t$, leaving us with $\mathfrak {a}_0$ when we recompose $\mathfrak {a}$ in Step 4.

The same argument holds for $\mathfrak {a}'$.

For the last statement, if $\mathfrak {a}=\mathfrak {b} \cap \mathfrak {c}$ with $\mathfrak {b}$ and $\mathfrak {c}$ squarefree monomial ideals, where $(x_i,y_i) \subseteq \mathfrak {b}$ for some $1 \le i \le n$, every minimal prime of $\mathfrak {b}$ will be part of $\mathfrak {a}_s$ for some $1 \le s \le t$, so they will all be removed in recomposing $\mathfrak {a}$. Thus it suffices to recompose $\mathfrak {c}$. $\square$

Definition 4.7

We call applying Steps 1–4 of Algorithm 3.2 to a squarefree monomial ideal $\mathfrak {a} \subseteq S$ but only removing primes containing $(x_i,y_i)$ for a particular i recomposing $\mathfrak {a}$ with respect to the index i. We may also refer to recomposing $\mathfrak {a}$ with respect to a subset of $\{1,\ldots ,n\}$, by which we mean we recompose with respect to each index in the set. If no set is specified, we are recomposing with respect to $\{1,\ldots ,n\}$.

Remark 4.8

We observe that a minimal prime of $\mathfrak {a}$ contains $(x_i,y_i)$ if and only if at least two generators of $\mathfrak {a}$ share the index i. The backward direction follows from the process of decomposing a squarefree monomial ideal. For example, $(x_1,x_2y_1)=(x_1,x_2) \cap (x_1,y_1)$. For the forward direction, if some minimal prime of $\mathfrak {a}$ contains $(x_i,y_i)$, then $\mathfrak {a}$ must contain at least one multiple of $x_i$ and at least one (distinct) multiple of $y_i$.

As a result, the indices $i_1,\ldots ,i_t$ from Lemma 4.5 are the indices shared by the generators of $\mathfrak {a}$. Hence it suffices to recompose an ideal with respect to the indices shared by its generators, as recomposing with respect to a non-shared index removes no minimal primes and thus preserves the ideal.

This also explains why, in Definition 4.1, we do not count i as a shared index among the generators of an ideal $\mathfrak {a}$ if $x_iy_i$ divides a single generator, but neither $x_i$ nor $y_i$ divides any other generator of $\mathfrak {a}$ (see Definition 4.1): such a generator does not give us a minimal prime containing $(x_i,y_i)$. For example, $(x_1y_1,x_2)=(x_1,x_2) \cap (y_1,x_2)$.

Remark 4.9

In computing canonical forms, we may remove generators that are a multiple of another generator of lower degree (or duplicate generators) at any stage of the algorithm, rather than waiting until Step 6. Removing such generators preserves the ideal, and hence the primary decomposition as well as the intersection that occurs in Step 4 of Algorithm 3.2.

Lemma 4.10

Let $t \ge 2$. Let

$$\begin{aligned} \mathfrak {a}=\bigcap _{j=1}^t \mathfrak {a}_j, \end{aligned}$$

where $\mathfrak {a},\mathfrak {a}_j$ are squarefree monomial ideals of S. Recomposing $\mathfrak {a}$ with respect to an index i is equivalent to recomposing each of the $\mathfrak {a}_j$ with respect to the index i and then intersecting the results.

Proof

We prove the case $t=2$. The rest follows by induction. We show that the set of primes $\{p_i\}$ appearing in a primary decomposition of $\mathfrak {a}$ is equal to the subset of the primes $\{q_i^{(1)},q_j^{(2)}\}$ appearing in the primary decompositions of $\mathfrak {a}_1$ and $\mathfrak {a}_2$ obtained by removing duplicates and primes that are now non-minimal: we first prove that any prime minimal over $\mathfrak {a}$ is minimal over $\mathfrak {a}_1$ or $\mathfrak {a}_2$. Suppose $\mathfrak {p}$ is minimal over $\mathfrak {a}$ but $\mathfrak {a}_1 \not \subseteq \mathfrak {p}$; we will show $\mathfrak {p}$ is minimal over $\mathfrak {a}_2$. Since $\mathfrak {a}_1 \not \subseteq \mathfrak {p}$, there exists $\alpha \in \mathfrak {a}_1 \backslash \mathfrak {p}$. We observe that $\alpha \mathfrak {a}_2 \subseteq \mathfrak {a}_1 \cap \mathfrak {a}_2 = \mathfrak {a} \subseteq \mathfrak {p}.$ Since $\mathfrak {p}$ is prime and $\alpha \notin \mathfrak {p}$, we must have $\mathfrak {a}_2 \subseteq \mathfrak {p}$.

To see that $\mathfrak {p}$ is minimal over $\mathfrak {a}_2$, we consider another prime $\mathfrak {p}'$ such that $\mathfrak {a}_2 \subseteq \mathfrak {p}' \subseteq \mathfrak {p}$. Since $\mathfrak {a} \subseteq \mathfrak {a}_2$, we have $\mathfrak {a} \subseteq \mathfrak {p}' \subseteq \mathfrak {p}$. Since $\mathfrak {p}'$ is prime and $\mathfrak {p}$ is a minimal prime over $\mathfrak {a}$, we must have $\mathfrak {p}' = \mathfrak {p}$.

Next, we consider a prime $\mathfrak {p}$ that is minimal over $\mathfrak {a}_1$ but not over $\mathfrak {a}$. In that case, there is another prime $\mathfrak {p}'$ minimal over $\mathfrak {a}$ such that $\mathfrak {a} \subseteq \mathfrak {p}' \subsetneq \mathfrak {p}$. By the minimality of $\mathfrak {p}$ over $\mathfrak {a}_1$, we know that $\mathfrak {a}_1 \not \subseteq \mathfrak {p}'$. Since $\mathfrak {p}'$ is minimal over $\mathfrak {a}$ but not over $\mathfrak {a}_1$, we find that $\mathfrak {p}'$ is minimal over $\mathfrak {a}_2$. Consequently, if we take all the primes minimal over $\mathfrak {a}_1$ or $\mathfrak {a}_2$ and remove any primes that are not minimal over $\mathfrak {a}$ or are duplicates, we obtain all the primes minimal over $\mathfrak {a}$.

This remains true after removing any primes containing $(x_i,y_i)$ for some $1 \le i \le n$ from both sets. Computing the intersection of the remaining $p_i$ is equivalent to separately intersecting the remaining $q_i^{(1)}$ and $q_i^{(2)}$, and then intersecting the two resulting ideals. $\square$

Example 4.11

We use Lemmas 4.5 and 4.10 to compute the canonical form of $\mathfrak {a}=(x_1x_2x_3,x_4y_1y_2)$. We begin decomposing $\mathfrak {a}$ as follows:

$$\begin{aligned} (x_1x_2x_3,x_4y_1y_2)&=(x_1,x_4y_1y_2) \cap (x_2,x_4y_1y_2) \cap (x_3,x_4y_1y_2)\\&=(x_1,y_1) \cap (x_1,x_4y_2) \cap (x_2,y_2) \cap (x_2,x_4y_1) \cap (x_3,x_4y_1y_2). \end{aligned}$$

We remove the 1st and 3rd pieces. Notice that it is impossible for any of the other 3 pieces to decompose into primes containing a pair $x_i,y_i$. By Lemma 4.5, these pieces are recomposed, and by Lemma 4.10, their intersection is also recomposed, so we can jump straight to Steps 4–6, performed together:

$$\begin{aligned} (x_1,x_4y_2) \cap (x_2,x_4y_1) \cap (x_3,x_4y_1y_2)&\rightarrow (x_1x_2x_3,x_4y_1y_2). \end{aligned}$$

Notation 4.12

We refer to the generators of an ideal as $g_1,\ldots ,g_k$, frequently pulling out $x_i$ and $y_i$ when i is a shared index between the $g_j$. For example, we write $(x_1g_1,y_1g_2)$. The $g_j$ will always be squarefree monomials.

For ease of notation, we write $x_1,x_2,\ldots$ and $y_1,y_2,\ldots$ instead of $x_{i_1},x_{i_2},\ldots$ and $y_{i_1},y_{i_2},\ldots$. However, all of our results hold if we permute the indices $1,\ldots ,n$ on the variables.

Lemma 4.13

Let $\mathfrak {a}=(g_1,\ldots ,g_k,h_1,\ldots ,h_\ell )$ be a squarefree monomial ideal in S such that for every $1 \le j_1 \le k$ and $1 \le j_2 \le \ell$, $g_{j_1}$ and $h_{j_2}$ do not share any indices (resp. the index i). Then recomposing $\mathfrak {a}$ (resp. with respect to the index i) gives the same result as recomposing $(g_1,\ldots ,g_k)$ and $(h_1,\ldots ,h_\ell )$ (resp. with respect to the index i) and taking the union of their sets of generators.

More generally, recomposing the neural ideal

$$\begin{aligned}\mathfrak {a}=(g_{1,1},\ldots ,g_{1,k_1},g_{2,1},\ldots ,g_{2,k_2},\ldots ,g_{m,k_m})\end{aligned}$$

(resp. with respect to the index i), where $g_{j_1,t_1}$ and $g_{j_2,t_2}$ do not share any indices (resp. the index i) for any $j_1 \ne j_2$, agrees with recomposing each of the $(g_{j,1},\ldots ,g_{j,k_j})$ (resp. with respect to the index i) and taking the union of their sets of generators.

Consequently, the canonical form (resp. almost canonical form) of $\mathfrak {a}$ is the union of the sets of generators of the canonical forms (resp. almost canonical form) of the $(g_{j,1},\ldots ,g_{j,t_j})$, up to removing any generator divisible by another.

Proof

We prove the result in the first paragraph; the second paragraph follows by induction. We first prove the case where the $g_{j_1}$ and $h_{j_2}$ do not share the index i. Suppose the generators of $\mathfrak {a}$ share the index i. Without loss of generality, assume i is a shared index for $(g_1,\ldots ,g_k)$. Then by our hypotheses, $x_i$ and $y_i$ do not divide any $h_j$. Hence no minimal prime of $(h_1,\ldots ,h_\ell )$ contains either $x_i$ or $y_i$.

Consequently, we can write $\mathfrak {a}=(G' \cap G_i)+H=(G'+H)\cap (G_i+H),$ where $H=(h_1,\ldots ,h_{\ell })$ and $G',G_i$ are the components of $G=(g_1,\ldots ,g_k)$ from the statement of Lemma 4.5, where $G_i$ is the squarefree monomial ideal that is the intersection of all minimal primes of G containing $(x_i,y_i)$, and $G'$ is the squarefree monomial ideal that is the intersection of the minimal primes of G not containing $(x_i,y_i)$. In particular, $G'$ is equal to the result of recomposing G with respect to the index i. The minimal primes of $G' + H$ are all of the form $\mathfrak {p}' + \mathfrak {p}$ where $\mathfrak {p}'$ is minimal over $G'$ and $\mathfrak {p}$ is minimal over H [11, Theorem 7.3.4]. Since no minimal prime of H contains $(x_i)$ or $(y_i)$, $G'+H$ is equal to the intersection of the minimal primes of $\mathfrak {a}$ not containing $(x_i,y_i)$, and $G_i+H$ is equal to the intersection of the minimal primes of $\mathfrak {a}$ containing $(x_i,y_i)$.

By Lemma 4.5 applied to $\mathfrak {a}$, recomposing $\mathfrak {a}$ is equal to recomposing $G'+H$ (with respect to any index or set of indices). Further, $G'+H$ has no minimal primes containing $(x_i,y_i)$ so it is already recomposed with respect to the index i. Finally, by Lemma 4.5, recomposing $G'$ agrees with recomposing $(g_1,\ldots ,g_k)$. As a result, if instead we recomposed G and H separately with respect to the index i and add the results, we will get $G'+H$. This proves the result for a single index.

The multiple index result follows by recomposing $\mathfrak {a}$ one index at a time as in Lemma 4.5; at each index i, no $g_{j_1}$ and $h_{j_2}$ share the index i. The final paragraph of the result follows by computing the almost canonical form or canonical form as appropriate. $\square$

Example 4.14

Let $\mathfrak {a}=(x_1x_2,x_2y_1,x_3x_4y_5,x_2x_5y_4)$. We can write $\mathfrak {a}=\mathfrak {b}+\mathfrak {c}$, where $\mathfrak {b}=(x_1x_2,x_2y_1)$ and $\mathfrak {c}=(x_3x_4y_5,x_2x_5y_4)$, since these pieces share no indices with each other. Note that $x_2$ appears in both, but since $y_2$ never appears, 2 is not a shared index. The canonical form of $\mathfrak {b}$ is $(x_2)$ and of $\mathfrak {c}$ is $(x_3x_4y_5,x_2x_5y_4)$, so the almost canonical form of $\mathfrak {a}$ is $(x_2,x_3x_4y_5,x_2x_5y_4)$. Removing $x_2x_5y_4$, which is a multiple of $x_2$, we are left with $(x_2,x_3x_4y_5)$.

Lemma 4.15

Let $\mathfrak {a}=(g_1,\ldots ,g_k)$ be a squarefree monomial ideal of S. If for some $0 \le \ell < k$ and every $\ell +1 \le j \le k$, $g_j$ shares no indices (resp. does not share the index i) with any other generator, then $g_{\ell +1},\ldots ,g_k$ are generators of the recomposition of $\mathfrak {a}$ with respect to every $1 \le i \le n$ (resp. the index i). If in addition $x_iy_i \not \mid g_j$ for any $\ell +1 \le j \le k$, they are generators of the almost canonical form of $\mathfrak {a}$.

In particular, if $\mathfrak {a}$ does not have any shared indices, then the ideal is recomposed. If in addition no generator is divisible by any $x_iy_i$ and no generator is a multiple of another, the ideal is in canonical form.

Proof

This follows from Lemma 4.13. $\square$

Example 4.16

By Lemma 4.15, the neural ideal $(x_1y_2,x_3y_2,x_1x_4)$ is in canonical form.

Lemma 4.17

Let $\mathfrak {a}=(x_1g_{11},\ldots ,x_1g_{1k_1},y_1g_{21},\ldots ,y_1g_{2k_2},g_{31},\ldots ,g_{3k_3}),$ be a squarefree monomial ideal in S, where $k_1,k_2 \ge 0$ and the only factors of $x_1,y_1$ appearing among the generators of $\mathfrak {a}$ are the ones shown. Then recomposing $\mathfrak {a}$ with respect to the index 1 returns

$$\begin{aligned} \mathfrak {a}' =&(x_1g_{11},\ldots ,x_1g_{1k_1}, y_1g_{21},\ldots ,y_1g_{2k_2}, [g_{1j_1}g_{2j_2}]_{1 \le j_1 \le k_1, 1 \le j_2 \le k_2}, g_{31},\ldots ,g_{3k_3}). \end{aligned}$$

If the only indices shared by the generators of $\mathfrak {a}$ are the $x_1,y_1$ shown, $\mathfrak {a}'$ is the almost canonical form of $\mathfrak {a}$. In this case the canonical form of $\mathfrak {a}$ is found by removing generators of $\mathfrak {a}'$ that are a multiple of another generator.

For any pair $g_{1j_1},g_{2j_2}$ that share an index, $[g_{1j_1}g_{2j_2}]$ will be removed in Step 5 of Algorithm 3.2.

Proof

By Lemma 4.15, we may assume $k_3=0$. We prove this by strong induction on $k_1+k_2$. If $k_1+k_2=0$ or 1, then $k_1=0$ or $k_2=0$. In this case, 1 is not a shared index for the generators of $\mathfrak {a}$ and $\mathfrak {a}'=\mathfrak {a}$. The result follows from Lemmas 4.5 and 4.15. We do the base case of $k_1=k_2=1$ so that $k_1+k_2=2$. In this case, we have $\mathfrak {a}=(x_1g_1,y_1g_2)$. We write $\mathfrak {a}=(x_1,y_1) \cap (x_1,g_2) \cap (g_1,y_1g_2).$

Neither piece 2 nor 3 have a minimal prime containing $(x_1,y_1)$, so by Lemma 4.5, we remove the first component and intersect components 2 and 3, leaving us with $(x_1g_1,x_1y_1g_2,[g_1g_2],y_1g_2)$. Since the second generator is divisible by the last, we remove it, leaving us with $(x_1g_1,y_1g_2,[g_1g_2])$.

For the inductive step, suppose that $k_1+k_2 >2$, $k_1,k_2 \ge 1$, and without loss of generality, that $k_1>1$. We have

$$\begin{aligned} \mathfrak {a}&=(x_1,y_1g_{21},\ldots ,y_1g_{2k_2}) \cap (g_{11},x_1g_{12},\ldots ,x_1g_{1k_1},y_1g_{21},\ldots ,y_1g_{2k_2}). \end{aligned}$$

By Lemma 4.10, it suffices to recompose these ideals with respect to the index 1 and intersect the results. By the induction hypothesis, this gives us:

$$\begin{aligned} (x_1,g_{21},\ldots ,g_{2k_2}) \cap&(g_{11},x_1g_{12},\ldots ,x_1g_{1k_1}, y_1g_{21},\ldots ,y_1g_{2k_2},\\&\qquad [g_{1j_1}g_{2j_2}]_{1 \le j_1 \le k_1, 1 \le j_2 \le k_2}). \end{aligned}$$

Intersecting these, we get the desired result. $\square$

Remark 4.18

We can interpret this lemma in terms of the RF-structure of the neural code as well. For each pair of generators $x_1g_1, y_1g_2$, say $g_1=\prod _{i \in \sigma _1} x_i \prod _{i \in \tau _1} y_i$ and $g_2=\prod _{i \in \sigma _2} x_i \prod _{i \in \tau _2} y_i$, where $1 \not \in \sigma _1 \cup \sigma _2 \cup \tau _1 \cup \tau _2$, we have relations

$$\begin{aligned} U_1 \bigcap \left( \bigcap _{i \in \sigma _1} U_i\right) \subseteq \bigcup _{i \in \tau _1} U_i \quad \text {and} \quad \bigcap _{i \in \sigma _2} U_i \subseteq U_1 \bigcup \left( \bigcup _{i \in \tau _2} U_i\right) . \end{aligned}$$

We claim that these two relations together imply the relation

$$\begin{aligned}\left( \bigcap _{i \in \sigma _1} U_i\right) \bigcap \left( \bigcap _{i \in \sigma _2} U_i \right) \subseteq \left( \bigcup _{i \in \tau _1} U_i\right) \bigcup \left( \bigcup _{i \in \tau _2} U_{i}\right) .\end{aligned}$$

To see this, we set $X=U_1$, $A=\bigcap _{i \in \sigma _1} U_i$, $B=\bigcup _{i \in \tau _1} U_i$, $C=\bigcap _{i \in \sigma _2} U_i$, and $D=\bigcup _{i \in \tau _2} U_i$. Then $X \cap A \subseteq B$ and $C \subseteq X \cup D$. Since $A \cap C \subseteq C$, it is also contained in $X \cup D$. Then

$$\begin{aligned}A \cap C \subseteq A \cap (X \cup D)=(A \cap X) \cup (A \cap D) \subseteq B \cup (A \cap D) \subseteq B \cup D,\end{aligned}$$

proving our claim.

As a result, by [4, Sect. 4.3] the neural ideal contains a monomial

$$\begin{aligned} {[}g_1g_2]=\left[ \prod _{i \in \sigma _1 \cup \sigma _2} x_i \prod _{i \in \tau _1 \cup \tau _2} y_i \right] .\end{aligned}$$

If $g_1$ and $g_2$ share no indices, then either this monomial or a monomial dividing it must be in the canonical form. If $g_1$ and $g_2$ share an index, this monomial will be removed in Step 5 of Algorithm 3.2.

Proposition 4.19

Let $\mathfrak {a}=(g_1,\ldots ,g_k) \subseteq S$ be a squarefree monomial ideal. Then recomposing $\mathfrak {a}$ (with respect to any shared index or to all shared indices) returns an ideal whose generators include $g_1,\ldots ,g_k$.

Consequently, for any j such that $x_iy_i \not \mid g_j$ for any $1 \le i \le n$, the generators of the almost canonical form of $\mathfrak {a}$ include $g_j$. Any removal of such a $g_j$ happens in Step 6. Hence for each such j, the canonical form of $\mathfrak {a}$ has a generator dividing $g_j$.

In addition, if $\mathfrak {a}$ has a generator $g_j$ such that $x_iy_i \mid g_j$ for some $1 \le i \le n$, $g_j$ may be removed at any stage of Algorithm 3.2.

Proof

To prove that recomposing $\mathfrak {a}$ does not remove $g_j$ from the ideal, the main thing we need to prove is that removing primes containing an $x_i,y_i$ pair from the primary decomposition of $\mathfrak {a}$ does not remove $g_j$. We prove the result for recomposing with respect to a single index. The result for recomposing with respect to multiple indices follows by induction.

Without loss of generality, suppose that 1 is a shared index for the generators of $\mathfrak {a}$. Rename the generators of $\mathfrak {a}$ as

$$\begin{aligned}\mathfrak {a}=(x_1y_1g_{11},\ldots ,x_1y_1g_{1k_1},x_1g_{21},\ldots ,x_1g_{2k_2},y_1g_{31},\ldots ,y_1g_{3k_3},g_{41},\ldots ,g_{4k_4}),\end{aligned}$$

where $k_j \ge 0$ for $i=1,2,3,4$, and $x_1,y_1 \not \mid g_{ij}$ for any i, j. We work by induction on $k_1$.

If $k_1=0$, $\mathfrak {a}=(x_1g_{21},\ldots ,x_1g_{2k_2},y_1g_{31},\ldots ,y_1g_{3k_3},g_{41},\ldots ,g_{4k_4}).$ By Lemma 4.5, recomposing $\mathfrak {a}$ agrees with recomposing the ideal $\mathfrak {a}'$ we get by removing all primes containing both $x_1$ and $y_1$ from $\mathfrak {a}$. By Lemmas 4.17 and 4.15,

$$\begin{aligned} \mathfrak {a}'=&(x_1g_{21},\ldots ,x_1g_{2k_2}, y_1g_{31},\ldots ,y_1g_{3k_3}, \\&\quad [g_{2j_2}g_{3j_3}]_{1 \le j_2 \le k_2, 1 \le j_3 \le k_3}, g_{41},\ldots ,g_{4k_4}). \end{aligned}$$

This contains all of the original generators of the ideal. Thus when $k_1=0$, the original generators of $\mathfrak {a}$ are generators of the recomposed $\mathfrak {a}'$, and hence of the recomposed $\mathfrak {a}$.

Our induction hypothesis is that for $\ell _1 = k_1-1$, $\ell _2,\ell _3,\ell _4 \ge 0$, and any choices of $h_{ij}$ not divisible by $x_1$ or $y_1$, recomposing the ideal

$$\begin{aligned}\mathfrak {a}=(x_1y_1h_{11},\ldots ,x_1y_1h_{1\ell },x_1h_{21},\ldots ,x_1h_{2\ell _2},y_1h_{31},\ldots ,y_1h_{3\ell _3},h_{41},\ldots ,h_{4\ell _4})\end{aligned}$$

with respect to the index 1 returns

$$\begin{aligned} \mathfrak {a}'=(&x_1y_1h_{11},\ldots ,x_1y_1h_{1\ell },x_1h_{21},\ldots ,x_1h_{2\ell _2},y_1h_{31},\ldots ,y_1h_{3\ell _3},\\&[h_{2j_2}g_{3j_3}]_{1 \le j_2 \le k_2, 1 \le j_3 \le k_3},h_{41},\ldots ,h_{4\ell _4}). \end{aligned}$$

If $k_1>0$, we have:

$$\begin{aligned} \mathfrak {a}=&(x_1,y_1g_{31},\ldots ,y_1g_{3k_3},g_{41},\ldots ) \cap (y_1,x_1g_{21},\ldots ,x_1g_{2k_2},g_{41},\ldots ) \\&\ \cap (g_{11},x_1y_1g_{12},\ldots ,x_1y_1g_{1k_1},x_1g_{21},\ldots ,x_1g_{2k_2},y_1g_{31},\ldots ,y_1g_{3k_3},g_{41},\ldots ). \end{aligned}$$

By Lemma 4.10, recomposing $\mathfrak {a}$ with respect to the index 1 agrees with recomposing (with respect to the index 1) and then intersecting the 3 components. Lemma 4.17 along with the induction hypothesis gives us their recomposed forms:

$$\begin{aligned} (x_1,g_{31},\!\ldots \!,g_{3k_3},g_{41},\!\ldots \!)\!\cap \!(y_1,g_{21},\!\ldots \!,g_{2k_2},g_{41},\!\ldots \!)\!\cap \!(g_{11},x_1y_1g_{12},\!\ldots \!,x_1y_1g_{1k_1}, \\ x_1g_{21},\ldots ,x_1g_{2k_2},y_1g_{31},\ldots ,y_1g_{3k_3}, [g_{2j_2}g_{3j_3}]_{1 \le j_2 \le k_2, 1 \le j_3 \le k_3},g_{41},\ldots ,g_{4k_4}). \end{aligned}$$

Note that in each of the first two pieces, we have already removed a number of pieces divisible by $x_1$ (resp. $y_1$), following Remark 4.9.

Intersecting the first two ideals (while removing monomials divisible by another monomial) gives

$$\begin{aligned} (x_1y_1,x_1g_{21},\!\ldots \!,x_1g_{2k_2}, y_1g_{31},\!\ldots \!,y_1g_{3k_3},[g_{2j_2}g_{3j_3}]_{1 \le j_2 \le k_2, 1 \le j_3 \le k_3},g_{41},\!\ldots \!,g_{4k_4}\!). \end{aligned}$$

Intersecting this with the 3rd piece then gives

$$\begin{aligned} (x_1y_1g_{11},\ldots ,x_1y_1g_{1k_1},x_1g_{21},\ldots ,x_1g_{2k_2},y_1g_{31},[g_{21}g_{31}],\ldots ,[g_{2k_2}g_{31}],\\ \ldots ,y_1g_{3k_3},[g_{21}g_{3k_3}],\ldots ,[g_{2k_2}g_{3k_3}],g_{41},\ldots ,g_{4k_4}), \end{aligned}$$

as desired. The original generators of $\mathfrak {a}$ are still generators of this ideal, proving the first statement.

For the second statement, we see by the first statement that every generator $g_j$ of $\mathfrak {a}$ appears in the recomposed $\mathfrak {a}$, and hence if $x_iy_i \not \mid g_j$ for any $1 \le i \le n$, $g_j$ appears in the almost canonical form of $\mathfrak {a}$. As a result, performing Step 6 of Algorithm 3.2 leaves us with a generator that divides $g_j$ for each j such that $x_iy_i \not \mid g_j$ for any $1 \le i \le n$.

We return to viewing $\mathfrak {a}=(g_1,\ldots ,g_k)$. Now suppose $x_iy_i \mid g_j$ for some $1 \le j \le k$ and $1 \le i \le n$. We see from the proof of the first statement that recomposing $\mathfrak {a}$ with respect to the index i or any other index $i'$ such that $x_{i'}y_{i'} \mid g_j$ preserves $g_j$, but $g_j$ does not contribute to any of the new generators produced by this recomposition. When we perform Step 5 of Algorithm 3.2, $g_j$ will be removed.

We have 3 cases for what happens to $g_j$ when we recompose $\mathfrak {a}$ with respect to an index $i'$ such that $x_{i'}y_{i'} \not \mid g_j$: first, if $x_{i'},y_{i'} \not \mid g_j$, preserves $g_j$, and $g_j$ does not contribute to any of the new generators produced by this recomposition. When we perform Step 5 of Algorithm 3.2, $g_j$ will be removed since it is divisible by $x_iy_i$.

Second, if $x_{i'} \mid g_j$, but $y_{i'} \not \mid g_j$, the recomposition produces $g_j$ and some number of $\left[ \frac{g_j}{x_{i'}}h \right]$. All of these are removed in Step 5 of Algorithm 3.2 as they are divisible by $x_iy_i$. The third case, where $y_{i'} \mid g_j$ and $x_{i'} \not \mid g_j$, is similar.

Since any generator built from $g_j$ will be removed in Step 5 of Algorithm 3.2, removing $g_j$ before recomposing $\mathfrak {a}$ with respect to any index or set of indices does not change the result of Algorithm 3.2. $\square$

Remark 4.20

By Remark 4.9 and Proposition 4.19, we may remove generators divisible by $x_iy_i$ for some $1 \le i \le n$ or by another generator of lower degree at any point of Algorithm 3.2 without altering the result. We use this to shorten computations throughout Sects. 5 and 6. Consequently, we refer to an ideal that only requires Steps 5 and 6 (resp. Step 6) of Algorithm 3.2 to find the canonical form as recomposed (resp. in almost canonical form), even though these ideals may not agree with the results of Step 4 (resp. Step 5) of the algorithm.

5 Classification of Canonical Neural Ideals

In this section, we apply the results of Sect. 4 to determine which polarized neural ideals are in canonical form. Our main result, Theorem 5.5, gives a simple criterion for determining whether a neural ideal is in canonical form based only on the shared indices in its generators. This allows us to describe explicitly the set of squarefree monomial ideals in a particular ring $\mathbb {F}_2[x_1,\ldots ,x_n,y_1,\ldots ,y_n]$ that are neural ideals in canonical form.

As an application of this result, we give a simpler algorithm for computing the canonical form of a neural ideal in Algorithm 5.10. This algorithm starts with a neural ideal, making it useful to mathematicians studying properties of neural ideals with an eye towards applications to neural codes.

Notably, our results do not depend on the total number of neurons, only the indices involved in the generators of the ideal.

Remark 5.1

Any neural ideal with a single generator is in canonical form. This is because the components of its primary decomposition are all principal, and cannot contain $(x_i,y_i)$ for any $1 \le i \le n$. Alternatively, with only one generator, there is no pair of generators that can share an index.

Theorem 5.2

The canonical forms of polarized neural ideals with 2 generators, neither dividing the other, are as follows:

• (1) Any polarized neural ideal of the form $(g_1,g_2)$ where $g_1$ and $g_2$ share no indices or at least two indices, is in canonical form.

• (2) Any polarized neural ideal of the form $(x_1g_1,y_1g_2)$, where $g_1$ and $g_2$ share no indices, is not in canonical form. Recomposing the ideal yields $(x_1g_1,y_1g_2,[g_1g_2])$, and so the canonical form of this neural ideal is:

  • (a) $(x_1g_1,y_1g_2,[g_1g_2])$ if $[g_1g_2] \ne g_1,g_2$,

  • (b) $(x_1g_1,g_2)$ if $[g_1g_2]=g_2 \ne g_1$,

  • (c) $(g_1,y_1g_2)$ if $[g_1g_2]=g_1 \ne g_2$

  • (d) and $(g_1)$ if $g_1=g_2$.

Proof

(1) The case where $g_1$ and $g_2$ share no indices follows from Lemma 4.15.

In the case where $g_1$ and $g_2$ share two or more indices, we can rewrite the ideal as either

$$\begin{aligned} (x_1\cdots x_tg_1,y_1 \cdots y_tg_2), \end{aligned}$$

where $t \ge 2$ and $g_1$ and $g_2$ share no indices, or

$$\begin{aligned} (x_1 \cdots x_ty_{t+1}\cdots y_{t+s}g_1,x_{t+1}\cdots x_{t+s}y_1\cdots y_tg_2), \end{aligned}$$

where $t,s>0$ and $g_1$ and $g_2$ share no indices.

We prove the first case by induction on k. First assume $k=2$. We use Lemma 4.5 to simplify the computation. We have

$$\begin{aligned} (x_1x_2g_1,y_1y_2g_2)= (x_1,y_1) \cap (x_2g_1,y_1) \cap (x_1,y_2g_2) \cap (x_2g_1,y_2g_2). \end{aligned}$$

The first piece is the only piece containing $(x_1,y_1)$, so we remove it. Next we look for components containing $(x_2,y_2)$:

$$\begin{aligned} (x_2g_1,y_1) \cap (x_1,y_2g_2) \cap (x_2g_1,y_2g_2) \hspace{-4cm}&\\&=(x_2g_1,y_1) \cap (x_1,y_2g_2) \cap (x_2,y_2) \cap (x_2,g_2) \cap (g_1,y_2) \cap (g_1,g_2). \end{aligned}$$

Removing $(x_2,y_2)$ and completing Algorithm 3.2 yields $(x_1x_2g_1,y_1y_2g_2)$, as desired. Now suppose $t>2$. We have

$$\begin{aligned}&(x_1 \cdots x_tg_1,y_1 \cdots y_tg_2)=(x_1,y_1) \cap (x_1,y_2 \cdots y_tg_2) \cap (x_2 \cdots x_tg_1,y_1 \cdots y_tg_2). \end{aligned}$$

By Lemma 4.5, the canonical form of the original ideal is equal to the canonical form of the intersection of pieces 2 and 3. By (1) and the induction hypothesis, both pieces are in canonical form. Hence by Lemma 4.10, we may apply Steps 5 and 6 of Algorithm 3.2 to their intersection to get the canonical form of the original ideal. Intersecting the ideals

$$\begin{aligned}(x_1,y_2 \cdots y_tg_1) \cap (x_2 \cdots x_tg_1,y_1 \cdots y_tg_2)\end{aligned}$$

and completing Steps 5 and 6 of Algorithm 3.2, we get the desired result.

We prove the second case by induction on $t+s$, starting with the base case $t=s=1$. In this case, we have

$$\begin{aligned}&(x_1y_2g_1,x_2y_1g_2)=(x_1,y_1) \cap (x_1,x_2g_2) \cap (y_2,x_2)\cap (g_1,x_2) \cap (y_2g_1,y_1g_2). \end{aligned}$$

Removing the pieces $(x_1,y_1)$ and $(y_2,x_2)$ as in Lemma 4.5, we intersect the rest of the ideals following Algorithm 3.2 to get the ideal $(x_1y_2g_1,x_2y_1g_2)$.

Now assume that $t+s>1$, and without loss of generality that $t>1$. We have

$$\begin{aligned}&(x_1 \cdots x_ty_{t+1}\cdots y_{t+s}g_1,x_{t+1}\cdots x_{t+s}y_1 \cdots y_tg_2) \\&\quad =(x_1,x_{t+1}\cdots x_{t+s}y_1 \cdots y_tg_2)\cap (x_2 \cdots x_ty_{t+1} \cdots y_{t+s}g_1,x_{t+1}\cdots x_{t+s}y_1 \cdots y_tg_2). \end{aligned}$$

By the induction hypothesis, the second piece is in canonical form (1 is no longer a shared index). By part (4), the canonical form of the first piece is $(x_1,x_{t+1} \cdots x_{t+s}y_2 \cdots y_tg_2)$. Combining these using Lemma 4.10, we get the desired result.

(2) We have $(x_1g_1,y_1g_2)=(x_1,y_1) \cap (x_1,g_2) \cap (g_1,y_1g_2).$ By Lemma 4.5, the canonical form of the original ideal is equal to the canonical form of the intersection of the second and third ideals. Removing the first piece and combining the other two via Algorithm 3.2, we get the almost canonical form $(x_1g_1,y_1g_2,[g_1g_2])$. The result follows by removing generators that are a multiple of another generator. If $[g_1g_2] \ne g_1,g_2$, we remove no generators. If $[g_1g_2]=g_1\ne g_2$, we remove $x_1g_1$. If $[g_1g_2]=g_2 \ne g_1$, we remove $y_1g_2$. If $[g_1g_2]=g_1=g_2$, we remove $x_1g_1$ and $y_1g_2$. $\square$

Example 5.3

We give examples of each case in Theorem 5.2.

• (1) The neural ideal $(x_1y_2,x_3y_2)$ is in canonical form.

• (2) The neural ideal $(x_1x_2y_3,x_4y_1y_2)$ is in canonical form.

• (3) The neural ideal $(x_1y_2,x_2y_1)$ is in canonical form.

• (4) The neural ideal $(x_1y_2,x_3y_1)$ has canonical form $(x_1y_2,x_3y_1,x_3y_2)$. In contrast, the neural ideal $(x_1,x_3y_1)$ has canonical form $(x_1,x_3)$.

Remark 5.4

Following the lead of Remark 4.18, in the case where the two generators share a single index and either $g_1=1$ or $g_2=1$, we can see that the ideal is not in canonical form more directly from the generators and the conditions for the canonical form given in [4, Theorem 4.3]:

First, we assume that $g_2=1$, so the ideal is $(x_1g_1,y_1)$. Suppose that $g_1 =x_2 \cdots x_ty_{t+1}\cdots y_{t+s}$. Then the generators tell us that

$$\begin{aligned} U_1 \cap U_2 \cap \ldots \cap U_t \subseteq U_{t+1}\cup \ldots \cup U_{t+s}, \, X \subseteq U_1. \end{aligned}$$

(2)

But then $U_1 \cap \ldots \cap U_t=U_2 \cap \ldots \cap U_t,$ so the left hand side in Eq. 2 is non-minimal. Hence $(x_1g_1,y_1)$ is not in canonical form. Further, removing $x_1$ from the first generator fixes the issue, which agrees with the result of Theorem 5.2.

Now suppose that $g_1=1$, so that the ideal becomes $(x_1,y_1g_2)$. Suppose that $g_2=x_2 \cdots x_ty_{t+1}\cdots y_{t+s}$. Then the generators tell us that

$$\begin{aligned} U_1=\varnothing , \, U_2 \cap \ldots \cap U_t \subseteq U_1 \cup U_{t+1} \cup \ldots \cup U_{t+s}. \end{aligned}$$

(3)

Since $U_1=\varnothing$, the right hand side in Eq. 3 is non-minimal. Hence $(x_1,y_1g_2)$ is not in canonical form. In this case, removing $y_1$ from the second generator fixes the issue, which is consistent with the result of Theorem 5.2.

Theorem 5.5

Let $\mathfrak {a}=(g_1,\ldots ,g_k)$ be a polarized neural ideal such that $g_{j_1} \not \mid g_{j_2}$ for any $1 \le j_1 \ne j_2 \le k$, and $x_iy_i \not \mid g_j$ for any $1 \le i \le n$ and $1 \le j \le k$. If for some pair $g_{j_1},g_{j_2}$ of generators of $\mathfrak {a}$, $g_{j_1}$ and $g_{j_2}$ share exactly 1 index i and no other generator of $\mathfrak {a}$ divides $\frac{[g_{j_1}g_{j_2}]}{x_iy_i}$, then $\mathfrak {a}$ is not in canonical form. Otherwise, $\mathfrak {a}$ is in canonical form.

Proof

First assume $\mathfrak {a}$ has at least one pair of generators with exactly one shared index, say the index 1. To distinguish the generators with a factor of $x_1$ or $y_1$, we rewrite $\mathfrak {a}=(x_1g_{11},\ldots ,x_1g_{1k_1},y_1g_{21},\ldots ,y_1g_{2k_2},g_{31},\ldots ,g_{3k_3}),$ where $x_1,y_1 \not \mid g_{j\ell }$ for any $j,\ell$. By Lemma 4.17, recomposing $\mathfrak {a}$ with respect to the index 1 returns

$$\begin{aligned} \mathfrak {a}'=(&x_1g_{11},\!\ldots \!,x_1g_{1k_1},y_1g_{21},\!\ldots \!,y_1g_{2k_2},[g_{1j_1}g_{2j_2}]_{1 \le j_1 \le k_1, 1 \le j_2 \le k_2},g_{31},\!\ldots \!,g_{3k_3}), \end{aligned}$$

which by Lemma 4.5 has the same canonical form of $\mathfrak {a}$.

By our hypothesis, some pair $g_{1j_1}$, $g_{2j_2}$ has no shared indices, so by Proposition 4.19, the canonical form of $\mathfrak {a}$ either has the generator $[g_{1j_1}g_{2j_2}]=\frac{[x_1g_{1j_1}y_1g_{2j_2}]}{x_1y_1}$ or has another generator g that is a proper divisor of $[g_{1j_1}g_{2j_2}]$.

If no original generator of $\mathfrak {a}$ divides $[g_{1j_1}g_{2j_2}]$, then g is not an original generator of $\mathfrak {a}$, so $\mathfrak {a}$ is not in canonical form.

If, however, for every index i and every pair of generators $g_{j_1}$ and $g_{j_2}$ of $\mathfrak {a}$ that share only the index i, some other generator of $\mathfrak {a}$ divides $\frac{[g_{j_1}g_{j_2}]}{x_iy_i}$, then every term of the form $\frac{[g_{j_1}g_{j_2}]}{x_iy_i}$ is removed by Step 6 of Algorithm 3.2, and by Remark 4.9 may be removed now. Hence recomposing $\mathfrak {a}$ returns $\mathfrak {a}$. By the hypotheses that no $x_iy_i$ divides any $g_j$ and no $g_j$ divides another, $\mathfrak {a}$ is in canonical form.

Next assume that no generators of $\mathfrak {a}$ share any indices. Then $\mathfrak {a}$ is in canonical form by Lemma 4.15.

Finally, assume that no generators of $\mathfrak {a}$ share exactly one index, and that at least one pair of generators shares an index. Without loss of generality, assume 1 is a shared index among the generators of $\mathfrak {a}$. Rewrite

$$\begin{aligned}\mathfrak {a}=(x_1g_{11},\ldots ,x_1g_{1k_1},y_1g_{21},\ldots ,y_1g_{2k_2},g_{31},\ldots ,g_{3k_3}),\end{aligned}$$

where $x_1,y_1 \not \mid g_{j\ell }$ for any $j,\ell$.

By Lemma 4.17, recomposing $\mathfrak {a}$ with respect to the index 1 returns

$$\begin{aligned} \mathfrak {a}'=(&x_1g_{11},\!\ldots \!,x_1g_{1k_1},y_1g_{21},\!\ldots \!,y_1g_{2k_2},[g_{1j_1}g_{2j_2}]_{1 \le j_1 \le k_1, 1 \le j_2 \le k_2},g_{31},\!\ldots \!,g_{3k_3}), \end{aligned}$$

which by Lemma 4.5 has the same canonical form as $\mathfrak {a}$.

However, since any two generators that share an index share at least 2 indices, each $[g_{1j_1}g_{2j_2}]$ is divisible by $x_iy_i$ for some $2 \le i \le n$. So they will all vanish when Step 5 of Algorithm 3.2 is applied. By Proposition 4.19, we can remove them now, leaving us with $\mathfrak {a}$ again. We repeat this process for each additional shared index, getting $\mathfrak {a}$ again each time. Then $\mathfrak {a}$ recomposed is equal to $\mathfrak {a}$. By assumption, $x_iy_i$ does not divide any generator of $\mathfrak {a}$, so $\mathfrak {a}$ is then in almost canonical form. Since we assumed no generator of $\mathfrak {a}$ divides another, $\mathfrak {a}$ is in canonical form. $\square$

Note that the divisibility condition in Theorem 5.5 requires the ideal to have at least 3 generators, hence why it is not a part of Theorem 5.2.

Example 5.6

We include an example to illustrate the case where two generators $g_{j_1}$ and $g_{j_2}$ share exactly one index i, but another generator divides $\frac{[g_{j_1}g_{j_2}]}{x_iy_i}$. Let $\mathfrak {a}=(x_1x_2,x_3x_4y_1,x_2x_3)$. The first two generators share only the index 1, but this ideal is nevertheless in canonical form. By Theorem 5.2 along with Lemma 4.15, the almost canonical form of this ideal is

$$\begin{aligned}(x_1x_2,x_3x_4y_1,x_2x_3x_4,x_2x_3).\end{aligned}$$

We remove $x_2x_3x_4$ since it is divisible by $x_2x_3$, returning the original ideal. Hence $\mathfrak {a}$ is in canonical form.

Remark 5.7

We note that any squarefree monomial ideal in $\mathbb {F}_2[x_1,\ldots ,x_n]$ can be realized as a neural ideal in canonical form by passing to the extension ring $\mathbb {F}_2[x_1,\ldots ,x_n,y_1,\ldots ,y_n]$. For example, $(x_1x_2,x_2x_3,x_3x_4,x_4x_5,x_5x_1)$ is the canonical form of a neural ideal in $\mathbb {F}_2[x_1,\ldots ,x_5,y_1,\ldots ,y_5]$.

However, remaining inside of a particular $\mathbb {F}_2[x_1,\ldots ,x_n,y_1,\ldots ,y_n]$, Theorem 5.5 describes the set of squarefree monomial ideals that are canonical forms of neural ideals.

Corollary 5.8

Let $\mathfrak {a}$ be a polarized neural ideal such that any pair of generators that share the index i share at least 2 indices. Then recomposing $\mathfrak {a}$ with respect to the index i returns $\mathfrak {a}$.

If each pair of generators that share any index share at least 2 indices, then recomposing $\mathfrak {a}$ with respect to any set of indices returns $\mathfrak {a}$.

Proof

This follows from the proof of Theorem 5.5. $\square$

Corollary 5.9

Let $\mathfrak {a}$ be a polarized neural ideal. In order to compute the canonical form of $\mathfrak {a}$, it suffices to recompose $\mathfrak {a}$ with respect to the indices $i_1,\ldots ,i_t$ such that for each $1 \le j \le t$, some pair of generators of $\mathfrak {a}$ shares only the index $i_j$.

Proof

By Lemma 4.17, recomposing

$$\begin{aligned} \mathfrak {a}=(x_ig_{11},\ldots ,x_ig_{1k_1},y_ig_{21},\ldots ,y_ig_{2k_2},g_{31},\ldots ,g_{3k_3}) \end{aligned}$$

with respect to the index i returns

$$\begin{aligned} \mathfrak {a}'=(x_ig_{11},\ldots ,x_ig_{1k_1}, y_ig_{21},\ldots ,y_ig_{2k_2}, [g_{1j_1}g_{2j_2}]_{1 \le j_1 \le k_2, 1 \le j_2 \le k_2}, g_{31},\ldots ,g_{3k_3}). \end{aligned}$$

As in the proof of Theorem 5.5, if each pair $x_ig_{1j_1},y_ig_{2j_2}$ sharing the index i also shares an additional index, every $[g_{1j_1}g_{2j_2}]$ can be removed by Proposition 4.19, and we get $\mathfrak {a}$ back. $\square$

We can now give a new algorithm for computing the canonical form, which avoids computing a primary decomposition. Since this algorithm starts from a neural ideal, this algorithm will be more useful to mathematicians studying neural ideals with an eye towards future results classifying neural codes. Those beginning from a neural code will likely prefer the iterative algorithm of [12].

Algorithm 5.10

Begin with a list of generators $L=\{g_1,\ldots ,g_k\}$ for a polarized neural ideal $\mathfrak {a} \subseteq \mathbb {F}_2[x_1,\ldots ,x_n,y_1,\ldots ,y_n]$.

• (1) Set $i=1$.

• (2) For each pair of generators $g_{j_1},g_{j_2}$ that shares only the index i, add a generator $\frac{[g_{j_1}g_{j_2}]}{x_{i_1}y_{i_1}}$ to a new list M.

• (3) Join the lists L and M.

• (4) Increment i and repeat steps 2 and 3 until $i=n+1$.

• (5) Remove any remaining generators divisible by $x_iy_i$ for any $1 \le i \le n$ and any generators divisible by another generator of smaller degree.

Remark 5.11

Note that in Step 4, it is only necessary to remove generators divisible by $x_iy_i$ if ${g_1,\ldots ,g_k}$ included such a generator. This algorithm will not otherwise produce generators divisible by $x_iy_i$ as if $g_{j_1}$ and $g_{j_2}$ share only the index $i_1$, $\frac{[g_{j_1}g_{j_2}]}{x_{i_1}y_{i_1}}$ will not be divisible by $x_iy_i$ for any i.

Remark 5.12

The generators $\frac{[g_{j_1}g_{j_2}]}{x_{i_1}y_{i_1}}$ that we add relate directly to the RF structure of the corresponding neural code: in Remark 4.18 we saw the relations corresponding to the generators $g_{j_1}$ and $g_{j_2}$ imply the relation corresponding to the generator $\frac{[g_{j_1}g_{j_2}]}{x_{i_1}y_{i_1}}$ must also hold.

Proof

By Corollary 5.9, it suffices to recompose $\mathfrak {a}$ with respect to the indices i such that some pair of generators $g_{j_1}$ and $g_{j_2}$ of $\mathfrak {a}$ share only the index i. This is the list made in Step 1. Performing Step 2 recomposes $\mathfrak {a}$ with respect to the index i by Lemma 4.17, with one change: any new generator coming from a pair of generators $g_{j_1},g_{j_2}$ that share more than one index is skipped, as by the proof of Theorem 5.5 they would be removed in Step 5 of Algorithm 3.2 and by Proposition 4.19 they may be removed at any time. Hence after Step 3, $\mathfrak {a}$ is recomposed. Step 4 removes any remaining generators divisible by $x_iy_i$ or by another generator that have not been removed already, completing Steps 5 and 6 of Algorithm 3.2. $\square$

Remark 5.13

We implemented Algorithms 5.10, 3.2, and a polarized version the iterative algorithm from [12] in Macaulay2 [6] to compare how long they took on examples. We include data from a few initial examples here, and plan to include more extensive speed testing data and complexity analysis in an upcoming Macaulay2 package and associated paper.

All times are listed in seconds. If we interrupted a computation before it completed, we list $>t$, where t is the number of seconds at which we interrupted the code.

List of ideals:

• (1) $I_1=(x_1x_2,x_3y_1)$

• (2) $I_2=(x_1x_3,x_2y_1,x_2x_3)$

• (3) $I_3=(x_1x_4,x_2x_3y_1,x_4y_2)$

• (4) $I_4=(x_1y_2,x_2y_3,x_4y_3)$

• (5) $I_5=(x_3y_1,x_1y_2,y_2y_4,x_2x_4y_3,x_3x_4y_1y_2,x_2x_3x_4y_1,x_1x_3x_4y_2)$

• (6) $I_6=(x_1x_2x_4,x_3y_1y_2,x_4y_5,x_4x_5x_6x_7)$

• (7) $I_7=(x_1x_2,x_3y_4,x_6y_7,x_2x_4x_7,x_6y_1y_2,x_2x_3x_5x_6y_7)$

Ideal	Number of neurons	3.2	5.10
$I_1$	3	0.0648534	0.00192403
$I_2$	3	0.061073	0.00157261
$I_3$	4	0.143508	0.00402159
$I_4$	4	0.0800547	0.00117581
$I_5$	4	0.367402	0.0249538
$I_6$	7	>173.618	0.00847768
$I_7$	7	94.8539	0.0130173

Based on these initial results, we are optimistic that Algorithm 5.10 outperforms Algorithm 3.2 in general, making it a better choice for computing the canonical form of a neural ideal. We also find Algorithm 5.10 significantly easier to use when computing small examples by hand than Algorithm 3.2.

List of codes:

• (1) $C_1=\{100,101,001,000,110\}$

• (2) $C_2=\{1001, 0110, 0000, 0111, 0010\}$

• (3) $C_3=\{0000000, 1000000, 0110000, 0100000, 0000011, 0000101\}$

• (4) $C_4=\{0000000, 1000000, 0110000, 0100000, 0000011, 0000101,1000001, 1100010\}$

Code	Number of neurons	3.2	5.10	Iterative
$C_1$	3	0.0539101	0.00636247	0.0323265
$C_2$	4	0.372083	0.0387233	0.0830163
$C_3$	7	38.0215	> 157.579	0.107555
$C_4$	7	>368.461	>189.821	0.097687

However, when going from a neural code on more than 3–4 neurons straight to its canonical form, the iterative algorithm from [12] appears significantly faster than computing the neural ideal as in [4] and then performing Algorithm 5.10 (or Algorithm 3.2, as seen in [12]) to get the canonical form. On two 7-neuron examples we tried, we interrupted our algorithm after some time, but the iterative algorithm finished in under .1 s.

6 Generic Canonical Forms

The results of this section give us a way to compute a single almost canonical form for a set of neural ideals that have identical patterns of shared indices among their generators (e.g. $(x_1x_2,x_3y_1)$ and $(x_1x_2x_3,x_3y_1)$). We do this by passing to an extension ring, computing a “generic" canonical form, and then returning to our original ring. These generic canonical forms create families of canonical forms.

Notation 6.1

Let $g_1,\ldots ,g_k \in S$ be squarefree monomials sharing no indices. We define an extension ring $P=S[z_1,\ldots ,z_k]$ and a map $\pi :P \rightarrow S$ sending $z_j \mapsto g_j$. Then $S \cong P/(z_j-g_j)$.

We compute an analogue of the canonical form for squarefree monomial ideals in P by modifying Algorithm 3.2 as follows:

Algorithm 6.2

Let $P=S[z_1,\ldots ,z_k]$.

• (1) Start with a squarefree monomial ideal $\mathfrak {a}=(f_1,\ldots ,f_m)$ in P, where $f_t=\prod _{i \in \sigma } x_i \prod _{i \in \tau } y_i \prod _{i \in \upsilon } z_i$.

• (2) Compute the primary decomposition of $\mathfrak {a}$. The ideals $p_1,\ldots ,p_s$ in the primary decomposition will all be generated by a subset of $(x_1,\ldots ,x_n,y_1,\ldots ,y_n,z_1,\ldots , z_k)$.

• (3) Impose the relation $x_i+y_i=1$, and as a result remove any ideal in the primary decomposition containing both $x_i$ and $y_i$ for any $1 \le i \le n$.

• (4) Compute $[h_1\cdots h_s]$ for every set of choices of $h_{\ell }$ a generator of $p_{\ell }$. (That is, intersect the remaining prime ideals.)

• (5) Impose the relations $x_iy_i=0$, and remove any monomials with a factor of $x_iy_i$ for some $1 \le i \le n$.

• (6) Remove any remaining products that are multiples of a product of lower degree.

Definition 6.3

We say that an ideal that is the result of Algorithm 6.2 is in generic canonical form. As we did in S, we will refer to an ideal in P that has had Steps 1–4 of Algorithm 6.2 applied to it as generically recomposed, and an ideal that has had Steps 1–5 of Algorithm 6.2 as being in generic almost canonical form.

As with their non-generic version (see Remark 4.20), we refer to an ideal that only requires Steps 5 and 6 (resp. Step 6) of Algorithm 6.2 to find the generic canonical form as generically recomposed (resp. in generic almost canonical form), even though these ideals may not agree with the results of Step 4 (resp. Step 5) of Algorithm 6.2.

Lemma 6.4

Let $b_1,\ldots ,b_k,g_1,\ldots ,g_k \in S$ be squarefree monomials such that the $g_j$ share no indices with the $b_jg_j$. In recomposing $\mathfrak {a}=(b_1g_1,\ldots ,b_kg_k)$, it is not necessary to split the $g_j$ into their factors.

Proof

We proceed by induction on k. We noted in Remark 5.1 that neural ideals with one generator are in canonical form, so we begin with $k=2$. In this case we have $(b_1g_1,b_2g_2)=(b_1,b_2) \cap (b_1,g_2) \cap (g_1,b_2) \cap (g_1,g_2).$ By Lemma 4.15, since no $g_j$ shares any index with any $b_jg_j$, the second, third, and fourth pieces are already recomposed, and by Lemma 4.10, recomposed $\mathfrak {a}$ is equal to the intersection of the recomposed components on the right hand side. This proves the case $k=2$.

Now assume $k>2$. We have

$$\begin{aligned}(b_1g_1,\ldots ,b_kg_k)=(b_1,b_2g_2,\ldots ,b_kg_k) \cap (g_1,b_2g_2,\ldots ,b_kg_k).\end{aligned}$$

By Lemma 4.15, the recomposed second piece is equal to $g_1$ plus the recomposed $(b_2g_2,\ldots ,b_kg_k)$. By the induction hypothesis, we do not need to split the $g_j$ into their factors to compute this. We continue to split the first piece:

$$\begin{aligned}(b_1,b_2g_2,\ldots ,b_kg_k)=(b_1,b_2,b_3g_3,\ldots ,b_kg_k) \cap (b_1,g_2,b_3g_3,\ldots ,b_kg_k).\end{aligned}$$

As before, we do not need to split the $g_j$ into their factors to recompose the second piece. Continuing in this way, we eventually get

$$\begin{aligned}(b_1,b_2,\ldots ,b_{k-1},b_kg_k)=(b_1,\ldots ,b_{k-1},b_k) \cap (b_1,\ldots ,b_{k-1},g_k).\end{aligned}$$

The first piece has no $g_k$, and as before, we do not need to split $g_k$ into its factors to recompose the second piece. $\square$

Notation 6.5

Let $(b_1z_1,\ldots ,b_kz_k)$ be a squarefree monomial ideal of P. When referring to the generically recomposed, almost canonical, or canonical form of $(b_1z_1,\ldots ,b_kz_k)$, we will denote it

$$\begin{aligned}(f(z_1,\ldots ,z_k)):= (f_1(z_1,\ldots ,z_k), \ldots f_{\ell }(z_1,\ldots ,z_k))\end{aligned}$$

for some $\ell$ where each generator $f_j(z_1,\ldots ,z_k) \in P$ is a monomial. This allows us to view the generically recomposed (resp. almost canonical or canonical) form as a function of the $z_j$ that is determined by the $b_j$.

Proposition 6.6

Let $b_1,\ldots ,b_k,g_1,\ldots ,g_k \in S$ be squarefree monomials such that the $g_j$ share no indices with the $b_jg_j$. Let $P=S[z_1,\ldots ,z_k]$ and $\pi :P \rightarrow S$ be the homomorphism sending S to itself and $z_j \mapsto g_j$. If the generically recomposed form of $(b_1 z_1, \ldots , b_k z_k)$ in P is given by $(f(z_1,\ldots ,z_k))$, then the recomposed form of $(b_1g_1, \ldots , b_k g_k)$ in S is given by

$$\begin{aligned} {[}\pi (f(z_1,\ldots ,z_k))]:= ([f_1(g_1,\ldots , g_k)], \ldots , [f_{\ell }(g_1,\ldots ,g_k)]). \end{aligned}$$

Further, if $x_i y_i$ divides $[f_{j}(g_1,\ldots ,g_k)]$ for some $1 \le i \le n$ and $1 \le j \le \ell$, then $x_i y_i$ divides $f_j(z_1,\ldots ,z_k)$. Hence the canonical form of $(b_1g_1,\ldots ,b_kg_k)$ in S may be computed by computing the generic almost canonical form of $(b_1z_1,\ldots ,b_kz_k)$, applying $\pi$ and replacing any $x_i^c$ for $c >1$ with $x_i$, and then applying Step 6 of Algorithm 3.2.

In particular, if we know the canonical form of $(b_1g_1,\ldots ,b_kg_k)$ (where the $g_j$ still share no indices with the $b_jg_j$), and $g \mid g_j$, we can find the canonical form of $(b_1g_1,\ldots ,b_{j-1}g_{j-1},b_jg,b_{j+1}g_{j+1},\ldots ,b_kg_k)$ by computing the canonical form of $(b_1g_1,\ldots ,b_kg_k)$, replacing $g_j$ by g, and then applying Step 6 of Algorithm 3.2 again.

Proof

Say that the generically recomposed $(b_1z_1,\ldots ,b_kz_k)$ in P is given by

$$\begin{aligned}(f(z_1,\ldots ,z_k)):=(f_1(z_1,\ldots ,z_k),\ldots ,f_\ell (z_1,\ldots ,z_k)),\end{aligned}$$

where each $f_s(z_1,\ldots ,z_k) \in P$. Applying $\pi$, we get

$$\begin{aligned}(f_1(g_1,\ldots ,g_k),\ldots ,f_\ell (g_1,\ldots ,g_k)).\end{aligned}$$

We claim that after replacing products with LCMs this is equal to the recomposed $(b_1g_1,\ldots ,b_kg_k)$. By Lemma 6.4, since the $g_j$ contain no $x_i$ or $y_i$ with shared indices, we can leave the $g_j$ alone when decomposing $(b_1g_1,\ldots ,b_kg_k)$. Hence $\pi$ applied to the primary decomposition of $(b_1z_1,\ldots ,b_kz_k)$ gives us a sufficient decomposition of $(b_1g_1,\ldots ,b_kg_k)$ to recompose the latter. Since $\pi (x_i)=x_i$ and $\pi (y_i)=y_i$, the components we remove in Step 3 of Algorithm 3.2 in S are equal to $\pi$ applied to the components we remove in Step 3 of Algorithm 6.2. The only potential difference is in Step 4, and this is resolved by replacing products with LCM’s.

Since the $g_j$ contain no indices shared with the $b_jg_j$ (i.e., if $x_i \mid g_\ell$ for any $\ell$, then $y_i \not \mid \Pi _j b_jg_j$, and the same with $x_i$ and $y_i$ reversed), this will hold after recomposing $(b_1g_1,\ldots ,b_kg_k)$ as well. As a result, any generator of the recomposed $(b_1g_1,\ldots ,b_kg_k)$ divisible by $x_iy_i$ for some $1 \le i \le n$ must be equal to $\pi$ of a generator of the recomposed $(b_1z_1,\ldots ,b_kz_k)$ divisible by the same $x_iy_i$, up to replacing products with LCMs. Applying Step 6 of Algorithm 3.2 to $(f_1(g_1,\ldots ,g_k),\ldots ,f_\ell (g_1,\ldots ,g_k))$ now gives us the canonical form of $(b_1g_1,\ldots ,b_kg_k)$, as desired.

For the final statement, note that

$$\begin{aligned} \mathfrak {a}&=(b_1g_1,\ldots ,b_kg_k) \text { and } \mathfrak {a}'=(b_1g_1,\ldots ,b_jg,\ldots ,b_kg_k) \end{aligned}$$

have the same generic canonical form, $(f(z_1,\ldots ,z_k))$. Applying $\pi _1$ sending $z_j \mapsto g_j$ and $\pi _2$ sending $z_j \mapsto g$, we have effectively replaced $g_j$ by g everywhere it appears. The only possible difference is in Step 6 of Algorithm 3.2. $\square$

Example 6.7

Consider $\mathfrak {a}=(x_1x_2x_4,x_3x_4y_1)$. Set $P=S[z_1,z_2]$, with $\pi (z_1)=x_2x_4$ and $\pi (z_2)=x_3x_4$. Write $\tilde{\mathfrak {a}}=(x_1z_1,y_1z_2) \subseteq P$. Using the proof of Theorem 5.2, this ideal has canonical form $(x_1z_1,y_1z_2,z_1z_2)$ in P. Passing back to S, we get $(x_1x_2x_4,x_3x_4y_1,x_2x_3x_4)$. In this case, no generator is a multiple of another, so we are done. This computation agrees with the result of Theorem 5.2.

Further, by the last statement of the Theorem, the canonical form of $\mathfrak {b}=(x_1x_2,x_3x_4y_1)$ is $(x_1x_2,x_3x_4y_1,x_2x_3x_4)$ and the canonical form of $\mathfrak {c}=(x_1,x_3x_4y_1)$ is $(x_1,x_3x_4)$. All of these agree with the result of Theorem 5.2.

Theorem 6.8

Suppose that the generic almost canonical form of the ideal $(b_1z_1,\ldots ,b_kz_k)$ in $P=S[z_1,\ldots ,z_k]$ is

$$\begin{aligned}(f(z_1,\ldots ,z_k)):=(f_1(z_1,\ldots ,z_k),\ldots ,f_{\ell }(z_1,\ldots ,z_k)).\end{aligned}$$

Let $\mathfrak {a}=(b_1g_{11},\ldots ,b_1g_{1t_1},b_2g_{21},\ldots ,b_2g_{2t_2},\ldots ,b_kg_{k1},\ldots ,b_kg_{kt_k}),$ where for any index i shared by the generators of $\mathfrak {a}$, $x_i,y_i \not \mid \prod g_{j_1j_2}$. Then the almost canonical form of $\mathfrak {a}$ is

$$\begin{aligned} (&[g_{1j_1}g_{2j_2}\cdots g_{kj_k}f(1,\ldots ,1)],\\&[g_{2j_2}\cdots g_{kj_k}f(g_{1t_1},1,\ldots ,1)],\ldots ,[g_{1j_1}\cdots g_{(k-1)j_{k-1}}f(1,\ldots ,1,g_{kt_k})],\\&\vdots \\&[g_{kj_k}f(g_{1t_1},\ldots ,g_{(k-1)t_{k-1}},1)],\ldots ,[g_{1j_1}f(1,g_{2t_2},\ldots ,g_{kt_k})],\\&[f(g_{1t_1},g_{2t_2},\ldots ,g_{kt_k})]), \end{aligned}$$

where $j_i$ ranges over the integers $1,2,\ldots ,t_i-1$ and all products have been replaced with LCM’s.

Remark 6.9

The statement of Theorem 6.8 is asymmetrical in the $g_{ij_i}$, with only the $g_{it_i}$ acting as input to the f’s. This is a result of the particular decomposition used in the proof, and the result works equally well if we fix for each $1 \le i \le k$ a choice of $1 \le s_i \le t_i$, replace every $g_{it_i}$ with $g_{is_i}$, and let $j_i$ range over the integers $1,2,\ldots ,\widehat{s_i},\ldots ,t_i$.

We give an example before proving Theorem 6.8 in order to illustrate the notation.

Example 6.10

We compute the almost canonical form of

$$\begin{aligned}(x_1g_{11},x_1g_{12},y_1g_{21},y_1g_{22},y_1g_{23}),\end{aligned}$$

where the $g_{j_1j_2}$ contain no shared indices. Continuing from Example 6.7, the generic canonical form of $(x_1g_1,y_1g_2)$ is $(x_1z_1,y_1z_2,z_1z_2)$. Here

$$\begin{aligned} f_1(z_1,z_2)&=x_1z_1,&f_2(z_1,z_2)&=y_1z_2,&f_3(z_1,z_2)&=z_1z_2. \end{aligned}$$

In this example, $j_1=1$ and $j_2 \in \{1,2\}$. Applying Theorem 6.8, we have

$$\begin{aligned} g_{1j_1}g_{2j_2}f_1(1,1)=g_{1j_1}g_{2j_2}x_1, \qquad g_{1j_1}g_{2j_2}f_2(1,1)&=g_{1j_1}g_{2j_2}y_1, \\ g_{1j_1}g_{2j_2}f_3(1,1)=g_{1j_1}g_{2j_2}, \end{aligned}$$

so we will only need generators of the third type, which are $[g_{11}g_{21}]$ and $[g_{11}g_{22}]$. Similarly,

$$\begin{aligned} g_{2j_2}f_1(g_{12},1)&=g_{2j_2}x_1g_{12}&g_{2j_2}f_2(g_{12},1)&=g_{2j_2}y_1&g_{2j_2}f_3(g_{12},1)&=g_{2j_2}g_{12}, \end{aligned}$$

from which we get generators $[g_{12}g_{21}],[g_{12}g_{22}],y_1g_{21},$ and $y_1g_{22}$,

$$\begin{aligned} g_{1j_1}f_1(1,g_{23})&=g_{1j_1}x_1&g_{1j_1}f_2(1,g_{23})&=g_{1j_1}y_1g_{23}&g_{1j_1}f_3(1,g_{23})&=g_{1j_1}g_{23}, \end{aligned}$$

from which we get generators $x_1g_{11}$ and $[g_{11}g_{23}]$, and

$$\begin{aligned} f_1(g_{12},g_{23})&=x_1g_{12}&f_2(g_{12},g_{23})&=y_1g_{23}&f_3(g_{12},g_{23})&=g_{12}g_{23}, \end{aligned}$$

from which we get generators $x_1g_{12},y_1g_{23},$ and $[g_{12}g_{23}]$. Hence by Theorem 6.8, the almost canonical form of $\mathfrak {a}=(x_1g_{11},x_1g_{12},y_1g_{21},y_1g_{22},y_1g_{23})$ is

$$\begin{aligned} (&[g_{11}g_{21}],[g_{11}g_{22}],[g_{11}g_{23}],[g_{12}g_{21}],[g_{12}g_{22}],[g_{11}g_{23}],\\&x_1g_{11},x_1g_{12},y_1g_{21},y_1g_{22},y_1g_{23}). \end{aligned}$$

This matches the result of Lemma 4.17. In this situation of only 1 shared index, using Lemma 4.17 might be easier, but as the number of shared indices increases, Theorem 6.8 becomes more useful.

Proof

We prove this by induction on $\sum _{i=1}^k t_i$. If $t_1=\ldots =t_k=1$, we are done by Proposition 6.6. Now suppose $t_i >1$ for at least one $1 \le i \le k$. Without loss of generality, assume $t_1>1$. We have

$$\begin{aligned} (b_1g_{11},\!\ldots \!,b_1g_{1t_1},b_2g_{21},\!\ldots \!)=(b_1,b_2g_{21},\!\ldots \!) \cap (g_{11},b_1g_{12},\!\ldots \!,b_1g_{1t_1},b_2g_{21},\!\ldots \!). \end{aligned}$$

By Lemma 4.10 and Proposition 4.19, the almost canonical form of $\mathfrak {a}$ is equal to the result of applying Step 5 of Algorithm 3.2 to the intersection of the almost canonical forms of the two components. By the induction hypothesis and Proposition 6.6, the almost canonical form of the first piece is

$$\begin{aligned} (&[g_{2j_2}\cdots g_{kj_k}f(1,\ldots ,1)],\\&[g_{3j_3}\cdots g_{kj_k}f(1,g_{2t_2},1,\ldots ,1)],\ldots ,[g_{2j_2}\cdots g_{(k-1)j_{k-1}}f(1,\ldots ,1,g_{kt_k})],\\&\vdots \\&[g_{kj_k}f(1,g_{2t_2},\ldots ,g_{(k-1)t_{k-1}},1)],\ldots ,[g_{2j_2}f(1,1,g_{3t_3},\ldots ,g_{kt_k})],\\&[f(1,g_{2t_2},\ldots ,g_{kt_k})]), \end{aligned}$$

where $j_i$ ranges over the integers $1,2,\ldots ,t_i$ for $2 \le i \le k$. By the induction hypothesis and Lemma 4.15, the almost canonical form of the second piece is

$$\begin{aligned} (&g_{11},[g_{1j_1}g_{2j_2}\cdots g_{kj_k}f(1,\ldots ,1)],\\&[g_{2j_2}\cdots g_{kj_k}f(g_{1t_1},1,\ldots ,1)],\ldots ,[g_{1j_1}\cdots g_{(k-1)j_{k-1}}f(1,\ldots ,1,g_{kt_k})],\\&\vdots \\&[g_{kj_k}f(g_{1t_1},\ldots ,g_{(k-1)t_{k-1}},1)],\ldots ,[g_{1j_1}f(1,g_{2t_2},\ldots ,g_{kt_k})],\\&[f(g_{1t_1},g_{2t_2},\ldots ,g_{kt_k})]), \end{aligned}$$

where $j_i$ ranges over the integers $1,2,\ldots ,t_i-1$ for $2 \le i \le k$ and $j_1$ ranges over the integers $2,\ldots ,t_1-1$.

We claim that

$$\begin{aligned}&[[f(h_1,\ldots ,h_{i-1},1,h_{i+1},\ldots ,h_k)][f(h_1,\ldots ,h_{i-1},h_i,h_{i+1},\ldots ,h_k)]] \\&\quad =[f(h_1,\ldots ,h_{i-1},h_i,h_{i+1},\ldots ,h_k)]. \end{aligned}$$

To see this, note that for any s such that $f_s(z_1,\ldots ,z_k)$ has no factor of $z_i$, the two components agree, and for any s such that $f_s(z_1,\ldots ,z_k)$ has a factor of $z_i$, say $f_s(z_1,\ldots ,z_k)=z_ih$, the first component gives h and the second component gives $h_ih$, so their lcm is $h_ih=f_s(h_1,\ldots ,h_{i-1},h_i,h_{i+1},\ldots ,h_k)$.

Using this claim to intersect the two pieces, the proof is complete. $\square$

Remark 6.11

In computing the canonical form of $\mathfrak {a}$, if $f_s(z_1,\ldots ,z_k)$ contains factors of, without loss of generality, $z_1,\ldots ,z_p$ but not $z_{p+1},\ldots ,z_k$, then

$$\begin{aligned}&f_s(g_{1t_1},\ldots ,g_{pt_p},1,\ldots ,1) = f_s(g_{1t_1},\ldots ,g_{pt_p},g_{(p+1)t_{p+1}},1,\ldots ,1)\\&\quad =\ldots =f_s(g_{1t_1},\ldots ,g_{kt_k}). \end{aligned}$$

(See the proof of Theorem 6.8 for justification of this fact.) So any term with a factor of $f_s(g_{1t_1},\ldots ,g_{pg_p},1,\ldots ,1)$ or any of the intermediate lines is a multiple of $f_s(g_{1t_1},\ldots ,g_{kt_k})$ and will be removed in the canonical form.

We end by computing the canonical forms of a class of polarized neural ideals to illustrate the benefit of the generic approach.

Theorem 6.12

The polarized neural ideal

$$\begin{aligned} \mathfrak {a}=(x_{1}g_1,x_{2}y_{1}g_2,x_{3}y_{2}g_3,\ldots ,x_{{k-1}}y_{{k-2}}g_{k-1},y_{{k-1}}g_k), \end{aligned}$$

where $k > 2$ and the only shared indices are the $x_{i}$ and $y_{i}$ shown, is not in canonical form. The almost canonical form of $\mathfrak {a}$ is

$$\begin{aligned} (&x_{1}g_1,x_{2}[g_1g_2],x_{3}[g_1g_2g_3],\ldots ,x_{{k-1}}[g_1 \cdots g_{k-1}],[g_1 \cdots g_k], \\&\quad x_{2}y_{1}g_2,x_{3}y_{1}[g_2g_3],\ldots ,x_{{k-1}}y_{1}[g_2\cdots g_{k-1}],y_{1}[g_2\cdots g_k],\\&\quad x_{3}y_{2}g_3,\ldots ,x_{{k-1}}y_{2}[g_3 \cdots g_{k-1}],y_{2}[g_3 \cdots g_k],\\&\quad \vdots \\&\quad x_{{k-1}}y_{{k-2}}g_{k-1},y_{{k-2}}[g_{k-1}g_k],\\&\quad y_{{k-1}}g_k). \end{aligned}$$

If for any $1 \le t \le t' < k$, $[g_t \cdots g_k] \ne [g_t \cdots g_{t'}]$, and for any $1 < s' \le s \le k$, $[g_1 \cdots g_s] \ne [g_{s'} \cdots g_s]$, then this is the canonical form of the ideal.

If for some $1 \le s' \le s \le t \le t' \le k$, $[g_1 \cdots g_s]=[g_{s'}\cdots g_s]$ and $[g_t \cdots g_k]=[g_t \cdots g_{t'}]$, where s is minimal, $s'$ is maximal, t is maximal, and $t'$ is minimal with respect to these properties, then the canonical form of $\mathfrak {a}$ is

$$\begin{aligned} (&x_{1}g_1,x_{2}[g_1g_2],\ldots ,x_{s-1}[g_1 \cdots g_{s-1}],x_s[g_{s'} \cdots g_s],\ldots ,x_{t'-1}\\&\quad [g_{s'} \cdots g_{t'-1}],[g_{s'} \cdots g_{t'}], \\&\quad x_{2}y_{1}g_2,x_{3}y_{1}[g_2g_3],\ldots ,x_{s-1}y_1[g_2 \cdots g_{s-1}],\\&\quad \vdots \\&\quad x_{s'}y_{s'-1}g_{s'},\ldots ,x_{s-1}y_{s'-1}[g_{s'}\cdots g_{s-1}],\\&\quad x_{s'+1}y_{s'}g_{s'+1},x_{s'+2}y_{s'}[g_{s'+1}g_{s'+2}],\ldots ,x_{t'-1}y_{s'}[g_{s'+1} \cdots g_{t'-1}],y_{s'}[g_{s'+1} \cdots g_{t'}],\\&\quad \vdots \\&\quad x_ty_{t-1}g_t,\ldots ,x_{t'-1}y_{t-1}[g_t \cdots g_{t'-1}],y_{t-1}[g_t \cdots g_{t'}],\\&\quad x_{t+1}y_tg_{t+1},\ldots ,x_{k-1}y_t[g_{t+1}\cdots g_{k-1}],y_t[g_{t+1}\cdots g_k],\\&\quad \vdots \\&\quad x_{{k-1}}y_{{k-2}}g_{k-1},y_{{k-2}}[g_{k-1}g_k],\\&\quad y_{{k-1}}g_k). \end{aligned}$$

Remark 6.13

If $k=2$, we are in the case of Theorem 5.2, so we skip that case here. If $k=1$, there are no shared indices and so the ideal is in canonical form by Lemma 4.15.

Notice that the canonical form of $\mathfrak {a}$ usually has more generators than $\mathfrak {a}$, often many more. The leftmost column of the canonical form above contains the original generators of the ideal, and all other generators are additional.

Proof

We prove this by induction, beginning with the case $k=3$. We decompose the ideal $(x_1g_1,x_2y_1g_2,y_2g_3)$, removing any component that contains a pair $(x_i,y_i)$ as generators as we go by Lemma 4.5:

$$\begin{aligned}&(x_1g_1,x_2y_1g_2,y_2g_3)=(x_1,x_2y_1g_2,y_2g_3) \cap (g_1,x_2y_1g_2,y_2g_3) \\&=(x_1,y_1,y_2g_3) \cap (x_1,x_2g_2,y_2g_3) \cap (g_1,x_2,y_2g_3) \cap (g_1,y_1g_2,y_2g_3) \\&\quad \rightarrow (x_1,x_2,y_2g_3) \cap (x_1,g_2,y_2g_3) \cap (g_1,x_2,y_2) \cap (g_1,x_2,g_3) \cap (g_1,y_1g_2,y_2g_3) \\&\quad \rightarrow (x_1,x_2,y_2) \cap (x_1,x_2,g_3) \cap (x_1,g_2,y_2g_3) \cap (g_1,x_2,g_3) \cap (g_1,y_1g_2,y_2g_3) \\&\quad \rightarrow (x_1,x_2,g_3) \cap (x_1,g_2,y_2g_3) \cap (g_1,x_2,g_3) \cap (g_1,y_1g_2,y_2g_3). \end{aligned}$$

By Lemma 4.15, since none of the remaining pieces have any shared indices, each piece is recomposed. By Lemma 4.10, to get the recomposed $\mathfrak {a}$, it suffices to intersect these ideals. Intersecting these ideals and removing components divisible by $x_iy_i$ or by another component as in Proposition 4.19, we are left with the almost canonical form$(x_1g_1,x_2g_1g_2,g_1g_2g_3,x_2y_1g_2,y_1g_2g_3,y_2g_3).$

Now assume $k > 3$. We have

$$\begin{aligned}&(x_1g_1,x_2y_1g_2,\ldots ,x_{k-1}y_{k-2}g_{k-1},y_{k-1}g_k) \\&\quad =(x_1,y_1,x_3y_2g_3,\ldots ,x_{k-1}y_{k-2}g_{k-1},y_{k-1}g_k) \\&\qquad \cap (x_1,x_2g_2,x_3y_2g_3,\ldots ,y_{k-1}g_k) \cap (g_1,x_2y_1g_2,x_3y_2g_3,\ldots ,y_{k-1}g_k). \end{aligned}$$

We remove the first piece, which contains the pair $(x_1,y_1)$. By Lemma 4.10, the recomposed $\mathfrak {a}$ is equal to the intersection of the recomposed second piece with the recomposed third piece. By the induction hypothesis, the generically recomposed second piece is

$$\begin{aligned} (&x_1,x_2g_2,x_3g_2g_3,\ldots ,x_{k-1}g_2\cdots g_{k-1},g_2\cdots g_k,\\&\quad x_3y_2g_3,\ldots ,x_{k-1}y_2 g_3 \cdots g_{k-1},y_2 g_3 \cdots g_k,\\&\quad \vdots \\&\quad x_{k-1}y_{k-2}g_{k-1},y_{k-2} g_{k-1}g_k,\\&\quad y_{k-1} g_k). \end{aligned}$$

Viewing $y_1g_2$ as a piece sharing no indices with other generators, the recomposed 3rd piece is

$$\begin{aligned} (&g_1,x_2y_1 g_2,x_3y_1 g_2 g_3,\ldots ,x_{k-1}y_1 g_2\cdots g_{k-1},y_1 g_2\cdots g_k,\\&\quad x_3y_2 g_3,\ldots ,x_{k-1}y_2 g_3 \cdots g_{k-1},y_2 g_3 \cdots g_k,\\&\quad \vdots \\&\quad x_{k-1}y_{k-2} g_{k-1},y_{k-2} g_{k-1} g_k,\\&\quad y_{k-1}g_k). \end{aligned}$$

Intersecting the two pieces and removing generators that are divisible by $x_iy_i$, we get

$$\begin{aligned} (&x_1 g_1,x_2 g_1g_2,\ldots ,x_{k-1} g_1 \cdots g_{k-1}, g_1 \cdots g_k,\\&\quad x_2y_1 g_2,x_3y_1 g_2 g_3,\ldots ,x_{k-1}y_1 g_2 \cdots g_{k-1},y_1 g_2 \cdots g_k,\\&\quad x_3y_2 g_3,\ldots ,x_{k-1}y_2 g_3 \cdots g_{k-1}g,y_2 g_3 \cdots g_k,\\&\quad \vdots \\&\quad x_{k-1}y_{k-2} g_{k-1},y_{k-2} g_{k-1} g_k,\\&\quad y_{k-1} g_k). \end{aligned}$$

Since by assumption, the $g_j$ share no indices, this is the almost canonical form of $\mathfrak {a}$.

If $[g_1\cdots g_s] \ne [g_{s'} \cdots g_s]$ for any $1 < s' \le s \le k$ and $[g_t \cdots g_k] \ne [g_t \cdots g_{t'}]$ for any $1 \le t \le t' < k$, it is not possible for any generator to be a multiple of another generator, so we have found the canonical form. Otherwise, we remove generators that are a multiple of another generator to get the result.

Regardless of the value of $[g_1 \cdots g_k]$, it will always be a generator of the ideal. Since $[g_1 \cdots g_k]$ has no factors of $x_i$ or $y_i$ for $1 \le i \le k-1$, whereas every original generator of $\mathfrak {a}$ has a factor of $x_i$ or of $y_i$, the canonical form is distinct from the original list of generators. $\square$

Remark 6.14

In the case of Theorem 6.12, we note that the canonical form of $\mathfrak {a}$ is never equal to the original ideal. This is best illustrated by noting that $[g_1\cdots g_k]$ is always added as a generator and is not divisible by any of the original generators since it isn’t divisible by any of $x_1,\ldots ,x_{k-1},y_1,\ldots ,y_{k-1}$.

Example 6.15

Consider the polarized neural ideal $\mathfrak {a}=(x_1g_1,x_2y_1g_2,y_2g_3)$, where the $g_j$ contain no shared indices. By Theorem 6.12, the almost canonical form of this ideal is $(x_1g_1,x_2[g_1g_2],[g_1g_2g_3],x_2y_1g_2,y_1[g_2g_3],y_2g_3)$. Consider $g_1=~x_3$, $g_2=x_4$, and $g_3=x_3x_4$, so that $[g_1g_2g_3]=g_3$. Substituting our values of $g_1,g_2,$ and $g_3$ and removing generators that are a multiple of another generator, we are left with $(x_1x_3,x_2y_1x_4,x_3x_4).$ This is a degenerate case like in parts of Theorem 5.2, (4)–instead of adding generators to get the canonical form we change a generator.

If instead we set $g_1=g_2=g_3=x_3$, we find every generator is divisible by $x_3$ thereby making the almost canonical form $(x_3)$. This is similar to the last case of Theorem 5.2.