Lattices and integral quadratic forms

Abstract

In many ways, quaternion algebras are like “noncommutative quadratic field extensions”: this is apparent from their very definition, but also from their description as wannabe $2\times 2$-matrices. Just as the quadratic fields $\mathbb Q (\sqrt{d})$ are wonderously rich, so too are their noncommutative analogues. In this part of the text, we explore these beginnings of noncommutative algebraic number theory.

In many ways, quaternion algebras are like “noncommutative quadratic field extensions”: this is apparent from their very definition, but also from their description as wannabe $2\times 2$-matrices. Just as the quadratic fields $\mathbb Q (\sqrt{d})$ are wonderously rich, so too are their noncommutative analogues. In this part of the text, we explore these beginnings of noncommutative algebraic number theory.

In this chapter, we begin with some prerequisites from commutative algebra, embarking on a study of integral structures and linear algebra over domains.

1 $\triangleright$ Integral structures

Just as we find the integers $\mathbb Z$ inside the rational numbers $\mathbb Q$, more generally we want a robust notion of integrality for possibly noncommutative algebras: this is the theory of orders over a domain.

We first have to understand the linear algebra aspects of this question. Let R be a domain with field of fractions $F:={{\,\mathrm{Frac}\,}}R$, and let V be a finite-dimensional F-vector space. An R-lattice in V is a finitely generated R-submodule $M \subset V$ with $MF=V$. If R is a PID (for example, $R=\mathbb Z$), then M is an R-lattice if and only if $M=R x_1 \oplus \dots \oplus R x_n$ where $x_1,\dots ,x_n$ is a basis for V as an F-vector space.

Between M and V lies intermediate structures, where instead of allowing all denominators (in the field of fractions), we only allow certain denominators; these are the localizations of M. To fix ideas, suppose $R=\mathbb Z$, so $M \simeq \mathbb Z ^n$; we call a $\mathbb Z$-lattice simply alattice. For a prime p, we define thelocalization of $\mathbb Z$ away from p to be

$$\begin{aligned} \mathbb Z _{(p)} :=\{ a/b \in \mathbb Q : p \not \mid b \} \subset \mathbb Q . \end{aligned}$$

In the localization, we can focus on those aspects of the lattice concentrated at the prime p. Extending scalars, $M_{(p)} :=M\mathbb Z _{(p)} \subseteq V$ is a $\mathbb Z _{(p)}$-lattice in V, again called the localization of M at p. These localizations determine the lattice M in the following strong sense (Theorem 9.4.9).

Theorem 9.1.1

(Local-global dictionary for lattices). Let V be a finite-dimensional $\mathbb Q$-vector space, and let $M \subseteq V$ be a lattice. Then the map $N \mapsto (N_{(p)})_p$ establishes a bijection between lattices N and collections of lattices $(N_{(p)})_{p}$ (indexed by primes p) where $M_{(p)}=N_{(p)}$ for all but finitely many primes p.

By this theorem, the choice of “reference” lattice M is arbitrary. Because of the importance of this theorem, a property of a lattice that holds if and only if it holds over every localization is called alocal property.

Finally, often vector spaces come equipped with a measure of length, or more generally a quadratic form; we can restrict these to the lattice $M \subseteq V$ with V a $\mathbb Q$-vector space. More intrinsically, we define aquadratic form $Q :M \rightarrow \mathbb Z$ to be a map satisfying:

• (i) $Q(ax)=a^2 Q(x)$ for all $a \in \mathbb Z$ and $x \in M$, and

• (ii) The associated map $T :M \times M \rightarrow \mathbb Z$ by $T(x,y) =Q(x+y)-Q(x)-Q(y)$ is ($\mathbb Z$-)bilinear.

Condition (i) explains (partly) the ‘quadratic’ nature of the map, and part (ii) is the usual way relating norms (quadratic forms) to bilinear forms. Choosing a basis $e_1,\dots ,e_n$ for $M \simeq \mathbb Z ^n$, we may then write

$$\begin{aligned} Q(x_1e_1+\dots +x_ne_n)=a_{11}x_1^2+a_{12}x_{1}x_2+\cdots +a_{nn}x_n^2 \in \mathbb Z [x_1,\dots ,x_n] \end{aligned}$$

(9.1.2)

as a homogeneous polynomial of degree 2.

2 Bits of commutative algebra

We begin with a brief review of some bits of commutative algebra relevant to our context: we need just enough to do linear algebra over (commutative) domains with good properties. Good general references for the basic facts from algebra we use (Dedekind domains, localization, etc.) are Atiyah–Macdonald [AM69], Matsumura [Mat89, Chapter 8], Curtis–Reiner [CR81, §1, §4], Reiner [Rei2003, Chapter 1], and Bourbaki [Bou98].

Throughout this chapter, let R be a (commutative) noetherian domain with field of fractions $F :={{\,\mathrm{Frac}\,}}R$.

9.2.1

An R-module P is projective  if it is a direct summand of a free module; equivalently, P is projective if and only if every R-module surjection $M \rightarrow P$ of R-modules has a section, i.e., an R-module homomorphism $g :P \rightarrow M$ such that $f \circ g = {{\,\mathrm{id}\,}}_P$.

Accordingly, a free R-module is projective. A projective R-module M is necessarilytorsion free over R, which is to say, if $rx=0$ with $r \in R$ and $x \in M$, then $r=0$ or $x=0$.

9.2.2

Afractional ideal of R is a nonzero finitely generated R-submodule $\mathfrak b \subseteq F$, or equivalently, a subset of the form $\mathfrak b =d\mathfrak a$ where $\mathfrak a \subseteq R$ is a nonzero ideal and $d \in F^\times$. Two fractional ideals $\mathfrak a ,\mathfrak b$ of R are isomorphic (as R-modules) if and only if there exists $c \in F^\times$ such that $\mathfrak b =c\mathfrak a$: indeed, given an isomorphism $\mathfrak a \simeq \mathfrak b$, we may extend scalars to F to obtain an F-linear map $F \simeq F$, which must be given by $c \in F^\times$, and conversely.

ADedekind domain is a noetherian, integrally closed domain such that every nonzero prime ideal is maximal.

Example 9.2.3

A field or a PID is a Dedekind domain; in particular, the rings $\mathbb Z$ and $\mathbb F _p[t]$ are Dedekind domains. If F is a finite extension of $\mathbb Q$ or $\mathbb F _p(t)$, then the integral closure of $\mathbb Z$ or $\mathbb F _p[t]$ in F respectively is a Dedekind domain.

9.2.4

Suppose R is a Dedekind domain. Then a finitely generated R-module is projective if and only if it is torsion free. Moreover, every nonzero ideal $\mathfrak a$ of R can be written uniquely as the product of prime ideals (up to reordering). For every fractional ideal $\mathfrak a$ of R, the set $\mathfrak a ^{-1} :=\{a \in F : a\mathfrak a \subseteq R\}$ is a fractional ideal with $\mathfrak a \mathfrak a ^{-1}=R$. Therefore the set of fractional ideals of R forms a group under multiplication. The set of principal fractional ideals comprises a subgroup, and we define ${{\,\mathrm{Cl}\,}}R$ to be the quotient, or equivalently the group of isomorphism classes of fractional ideals of R.

3 Lattices

Let V be a finite-dimensional F-vector space. ]Vfinite-dimensional vector space

Definition 9.3.1

An R-lattice in V is a finitely generated R-submodule $M \subseteq V$ with $M F=V$. We refer to a $\mathbb Z$-lattice as alattice.

The condition that $MF=V$ is equivalent to the requirement that M contains a basis for V as an F-vector space.

Example 9.3.2

An R-lattice in $V=F$ is the same thing as a fractional ideal of R.

We will be primarily concerned with projective R-lattices; if R is a Dedekind domain, then a finitely generated R-submodule $M \subseteq V$ is torsion free and hence automatically projective (9.2.4).

9.3.3

If there is no ambient vector space around, we will also call a finitely generated torsion free R-module M an R-lattice: in this case, M is a lattice in the F-vector space $M \otimes _R F$ because the map $M \hookrightarrow M \otimes _R F$ is injective (as M is torsion free).

Remark 9 Some authors omit the second condition in the definition of an R-lattice and say that M isfull if $M F=V$. We will not encounter R-lattices that are not full (and when we do, we call them finitely generated R-submodules), so we avoid this added nomenclature.

By definition, an R-lattice can be thought of an R-submodule that “allows bounded denominators”, as follows.

Lemma 9.3.5

Let $M \subseteq V$ be an R-lattice and let $J \subseteq V$ be a finitely generated R-submodule. Then the following statements hold.

• (a) For all $x \in V$, there exists nonzero $r \in R$ such that $rx \in M$.

• (b) There exists nonzero $r \in R$ such that $rJ \subseteq M$.

• (c) J is an R-lattice if and only if there exists nonzero $r \in R$ such that $rM \subseteq J \subseteq r^{-1}M$.

Proof First (a). Since $FM=V$, the R-lattice M contains an F-basis $x_1,\dots ,x_n$ for V, so in particular $M \supseteq Rx_1 \oplus \dots \oplus Rx_n$. For all $x \in V$, writing x in the basis $x_1,\dots ,x_n$ and clearing (finitely many) denominators, we conclude that there exists nonzero $r \in R$ such that $rx \in M$.

For (b), let $y_1,\dots ,y_m$ generate J as an R-module; then for each i, there exist $r_i \in R$ nonzero such that $r_iy_i \in M$ hence $r :=\prod _i r_i \ne 0$ satisfies $rJ \subseteq M$, and therefore $J \subseteq r^{-1} M$. For (c), we repeat (b) with M interchanged with J to find nonzero $s \in R$ such that $sM \subseteq J$, so then

$$ rsM \subseteq sM \subseteq J \subseteq r^{-1} M \subseteq (rs)^{-1} M. $$

For the rest of this section, we suppose that R is a Dedekind domain and treat lattices over R; for further references, see Curtis–Reiner [CR62, §22], O’Meara [O’Me73, §81], or Fröhlich–Taylor [FT91, §II.4]. It turns out that although not every R-lattice has a basis, it can be decomposed as a direct sum, as follows.

Theorem 9.3.6

Let R be a Dedekind domain, let $M \subseteq V$ be an R-lattice and let $y_1,\dots ,y_n$ be an F-basis for V. Then there exist $x_1,\dots ,x_n \in M$ and fractional ideals $\mathfrak a _1,\dots ,\mathfrak a _n$ such that

$$\begin{aligned} M=\mathfrak a _1 x_1 \oplus \dots \oplus \mathfrak a _n x_n \end{aligned}$$

(9.3.7)

and $x_j \in Fy_1+\dots +Fy_j$ for $j=1,\dots ,n$.

Accordingly, we say that every R-lattice M iscompletely decomposable (as a direct sum of fractional ideals), and we call the elements $x_1,\dots ,x_n$ apseudobasis for the lattice M with respect to the coefficient ideals  $\mathfrak a _1,\dots ,\mathfrak a _n$. The matrix with rows $x_i$ in the basis $y_i$ is lower triangular by construction; without loss of generality (rescaling), we may suppose that the diagonal entries are equal to 1, in which case we say that the pseudobasis for M is given inHermite normal form.

More generally, if $M=\mathfrak a _1 x_1 + \dots + \mathfrak a _m x_m$, the sum not necessarily direct, then we say that the elements $x_i$ are apseudogenerating set for M with coefficient ideals $\mathfrak a _i$.

Proof of Theorem 9.3.6

We argue by induction on n, the case $n=1$ corresponding to the case of a single fractional ideal.

Let $W :=F y_1+\dots +Fy_{n-1}$, and let $N=M \cap W$. Then there is a commutative diagram

(9.3.8)

Since $N=W \cap M$, we have $M/N \hookrightarrow V/W$, and $V/W \simeq F$ projecting onto $Fy_n$. Since M/N is nonzero and finitely generated, by 9.2.4 we conclude $M/N \simeq \mathfrak a \subseteq F$ is a fractional ideal, hence projective. Therefore the top exact sequence of R-modules splits (the surjection has a section), so there exists $x \in M$ such that $M = N \oplus \mathfrak a x$ as R-modules. The result then follows by applying the inductive hypothesis to N. $\square$

An argument generalizing that of Theorem 9.3.6 yields the following [O’Me73, 81:11].

Theorem 9.3.9

(Invariant factors). Let R be a Dedekind domain and let $M,N \subseteq V$ be R-lattices. Then there exists a common pseudobasis $x_1,\dots ,x_n$ for M, N; i.e., there exists a basis $x_1,\dots ,x_n$ for V and fractional ideals $\mathfrak a _1,\dots ,\mathfrak a _n$ and $\mathfrak b _1,\dots ,\mathfrak b _n$ such that

$$\begin{aligned} M&= \mathfrak a _1 x_1 \oplus \dots \oplus \mathfrak a _n x_n \\ N&= \mathfrak b _1 x_1 \oplus \dots \oplus \mathfrak b _n x_n \end{aligned}$$

Moreover, letting $\mathfrak d _i :=\mathfrak b _i\mathfrak a _i^{-1}$ we may further take $\mathfrak d _1 \mid \cdots \mid \mathfrak d _n$, and then such $\mathfrak d _i$ are unique.

The unique fractional ideals $\mathfrak d _1,\dots ,\mathfrak d _n$ given by Theorem 9.3.9 are called theinvariant factors of N relative to M.

9.3.10

Let $M \subseteq V$ be an R-lattice with pseudobasis as in (9.3.7). The class $[\mathfrak a _1\cdots \mathfrak a _n] \in {{\,\mathrm{Cl}\,}}R$ is well-defined (Exercise 9.7) and called theSteinitz class.

In fact, if we do not require that $x_j \in Fy_1+ \dots + Fy_i$ for $j=1,\dots ,n$ in Theorem 9.3.6, then we can find a pseudobasis for M with $\mathfrak a _1=\cdots =\mathfrak a _{n-1}=R$, i.e.,

$$\begin{aligned} M=Rx_1\oplus \cdots \oplus R x_{n-1} \oplus \mathfrak a x_n \end{aligned}$$

with $[\mathfrak a ]$ the Steinitz class of M.

4 Localizations

Properties of a domain are governed in an important way by its localizations, and consequently the structure of lattices, orders, and algebras can often be understood by looking at their localizations (and later, completions).

For a prime ideal $\mathfrak p \subseteq R$, we denote by ]$\mathfrak p$prime ideal (of a domain)

$$\begin{aligned} R_{(\mathfrak p )} :=\{r/s \in F : s \not \in \mathfrak p \} \subseteq F \end{aligned}$$

(9.4.1)

the localization of R at $\mathfrak p$. (We reserve the simpler subscript notation for the completion, defined in section 9.5.) p]$M_{(\mathfrak p )}$localization of an R-lattice M at $\mathfrak p$p]$R_{(\mathfrak p )}$localization of a domain R at $\mathfrak p$

Example 9.4.2

If $R=\mathbb Z$ and $\mathfrak p =(2)$, then $R_{(2)}=\{r/s \in \mathbb Q : s \text { is odd}\}$ consists of the subring of rational numbers with odd denominator.

Since R is a domain, the map $R \hookrightarrow R_{(\mathfrak p )}$ is an embedding and we can recover R as an intersection

$$\begin{aligned} R = \bigcap _\mathfrak{p } R_{(\mathfrak p )} = \bigcap _\mathfrak{m } R_{(\mathfrak m )} \subseteq F \end{aligned}$$

(9.4.3)

where the intersections are over all prime ideals of R and all maximal ideals of R, respectively.

Let V be a finite-dimensional F-vector space and let $M \subseteq V$ be an R-lattice. For a prime $\mathfrak p$ of R, let

$$\begin{aligned} M_{(\mathfrak p )} :=MR_{(\mathfrak p )} \subseteq V \end{aligned}$$

be the extension of scalars of M over $R_{(\mathfrak p )}$; identifying $V=MF \simeq M \otimes _R F$ under multiplication, we could similarly define

$$\begin{aligned} M_{(\mathfrak p )} :=M \otimes _R R_{(\mathfrak p )}. \end{aligned}$$

In either lens, $M_{(\mathfrak p )}$ is an $R_{(\mathfrak p )}$-lattice in V. In this way, M determines a collection $(M_{(\mathfrak p )})_\mathfrak{p }$ indexed over the primes $\mathfrak p$ of R.

9.4.4

Returning to 9.2.1, a finitely generated R-module M is projective if and only if it is locally free , i.e., $M_{(\mathfrak p )}$ is free for all prime ideals of R.

The ability to argue locally and then with free objects is very useful, and so very often we will restrict our attention to projective (equivalently, locally free) R-modules.

9.4.5

The localization of a Dedekind domain R is a discrete valuation ring (DVR). A DVR is equivalently a local PID that is not a field. In particular, a DVR is integrally closed and every finitely generated module over a DVR is free.

Consequently, if R is a Dedekind domain, then every fractional ideal of R islocally principal, i.e., if $\mathfrak a \subseteq F$ is a fractional ideal, then for all primes $\mathfrak p$ of R we have $\mathfrak a _{(\mathfrak p )}= a_\mathfrak{p } R_{(\mathfrak p )}$ for some $a_\mathfrak p \in F^\times$.

We now prove a version of the equality (9.4.3) for R-lattices (recall Definition 9.3.1).

Lemma 9.4.6

Let M be an R-lattice in V. Then

$$\begin{aligned} M = \bigcap _\mathfrak p M_{(\mathfrak p )} = \bigcap _\mathfrak m M_{(\mathfrak m )} \subseteq V \end{aligned}$$

where the intersection is over all prime (maximal) ideals $\mathfrak p$.

Proof It suffices to prove the statement for maximal ideals since $M_{(\mathfrak m )} \subseteq M_{(\mathfrak p )}$ whenever $\mathfrak m \supseteq \mathfrak p$. The inclusion $M \subseteq \bigcap _\mathfrak{m } M_{(\mathfrak m )}$ is clear. Conversely, let $x \in V$ satisfy $x \in M_{(\mathfrak m )}$ for all maximal ideals $\mathfrak m$. Let

$$\begin{aligned} (M:x):=\{r \in R : rx \in M \}. \end{aligned}$$

Then (M : x) is an ideal of R. For a maximal ideal $\mathfrak m$ of R, since $x \in M_{(\mathfrak m )}$ there exists $0 \ne r_\mathfrak m \in R \smallsetminus \mathfrak m$ such that $r_\mathfrak m x \in M$. Thus $r_\mathfrak m \in (M:x)$ and (M : x) is not contained in any maximal ideal of R. Therefore $(M:x)=R$ and hence $x \in M$.

Corollary 9.4.7

Let M, N be R-lattices in V. Then the following are equivalent:

• (i) $M \subseteq N$;

• (ii) $M_{(\mathfrak p )} \subseteq N_{(\mathfrak p )}$ for all prime ideals $\mathfrak p$ of R; and

• (iii) $M_{(\mathfrak m )} \subseteq N_{(\mathfrak m )}$ for all maximal ideals $\mathfrak m$ of R.

Proof The implications (i) $\Rightarrow$ (ii) $\Rightarrow$ (iii) are direct; for the implication (iii) $\Rightarrow$ (i), we have $M=\bigcap _\mathfrak p M_{(\mathfrak m )} \subseteq \bigcap _\mathfrak p N_{(\mathfrak m )} = N$ by Lemma 9.4.6.

In particular, it follows from Corollary 9.4.7 that $M=N$ for R-lattices M, N if and only if $M_{(\mathfrak p )}=N_{(\mathfrak p )}$ for all primes $\mathfrak p$ of R.

9.4.8

A property that holds if and only if it holds locally (as in Corollary 9.4.7, for the property that one lattice is contained in another) is called alocal property.

To conclude this section, we characterize in a simple way the conditions under which a collection $(M_{(\mathfrak p )})_\mathfrak p$ of $R_{(\mathfrak p )}$-lattices arise from a global R-lattice. We will see that just as a nonzero ideal of R can be factored uniquely into a product of prime ideals, and hence by the data of these primes and their exponents, so too can a lattice be understood by a finite number of localized lattices, once a “reference” lattice has been chosen (to specify the local behavior of the lattice at other primes).

Theorem 9.4.9

(Local-global dictionary for lattices). Let R be a Dedekind domain, and let $M \subseteq V$ be an R-lattice. Then the map $N \mapsto (N_{(\mathfrak p )})_\mathfrak p$ establishes a bijection between R-lattices $N \subseteq V$ and collections of lattices $(N_{(\mathfrak p )})_\mathfrak{p }$ indexed by the primes $\mathfrak p$ of R satisfying $M_{(\mathfrak p )}=N_{(\mathfrak p )}$ for all but finitely many primes $\mathfrak p$.

In Theorem 9.4.9, the choice of the “reference” lattice M is arbitrary: if $M'$ is another lattice, then by Theorem 9.4.9 $M_{(\mathfrak p )}=M'_{(\mathfrak p )}$ for all but finitely many primes $\mathfrak p$, so we get the same set of lattices replacing M by $M'$. In particular, any lattice $N \subseteq V$ agrees with any other one at all but finitely many localizations.

Remark 21 In Theorem 9.4.9, there is a bit of notational abuse: when we write a collection $(N_{(\mathfrak p )})_\mathfrak{p }$, we do not mean to imply that there is (yet) an R-lattice N such that the localization of N at $\mathfrak p$ is equal to $N_{(\mathfrak p )}$. This conclusion is what is provided by the theorem (the statement of surjectivity), so the notational conflict is only temporary.

Proof of Theorem 9.4.9

Let $N \subseteq V$ be an R-lattice. Then there exists $0 \ne r \in R$ such that $rM \subseteq N \subseteq r^{-1} M$. But r is contained in only finitely many prime (maximal) ideals of R, so for all but finitely many primes $\mathfrak p$, the element r is a unit in $R_{(\mathfrak p )}$ and thus $M_{(\mathfrak p )}=N_{(\mathfrak p )}$.

So consider the set of collections $(N_{(\mathfrak p )})_\mathfrak{p }$ of lattices where $N_{(\mathfrak p )}$ is an $R_{(\mathfrak p )}$-lattice for each prime $\mathfrak p$ with the property that $M_{(\mathfrak p )}=N_{(\mathfrak p )}$ for all but finitely many primes $\mathfrak p$ of R. Given such a collection, we define $N=\bigcap _\mathfrak{p } N_{(\mathfrak p )} \subseteq V$. Then N is an R-submodule of V. We show it is an R-lattice in V. For each $\mathfrak p$ such that $M_{(\mathfrak p )} \ne N_{(\mathfrak p )}$, there exists $r_\mathfrak p \in R$ such that $r_\mathfrak p M_{(\mathfrak p )} \subseteq N_{(\mathfrak p )} \subseteq r_\mathfrak p ^{-1} M_{(\mathfrak p )}$. Therefore, if $r=\prod _\mathfrak p r_\mathfrak p$ is the product of these elements, then $r M_{(\mathfrak p )} \subseteq N \subseteq r^{-1} M_{(\mathfrak p )}$ for all primes $\mathfrak p$ with $M_{(\mathfrak p )} \ne N_{(\mathfrak p )}$. On the other hand, if $M_{(\mathfrak p )}=N_{(\mathfrak p )}$ then already $r M_{(\mathfrak p )} \subseteq M_{(\mathfrak p )} = N_{(\mathfrak p )} \subseteq r^{-1} N_{(\mathfrak p )} = r^{-1} M_{(\mathfrak p )}$. Therefore by Corollary 9.4.7, we have $r M \subseteq N \subseteq r^{-1} M$, and so N is an R-lattice.

By Lemma 9.4.6, the association $(N_{(\mathfrak p )})_\mathfrak{p } \mapsto \bigcap _\mathfrak{p } N_{(\mathfrak p )}$ is an inverse to $N \mapsto (N_{(\mathfrak p )})_\mathfrak{p }$. Conversely, given a collection $(N_{(\mathfrak p )})_\mathfrak{p }$, for a nonzero prime $\mathfrak p$, we have $\bigl (\bigcap _\mathfrak{q } N_\mathfrak q \bigr )_{(\mathfrak p )}=N_{(\mathfrak p )}$ since $(R_\mathfrak q )_{(\mathfrak p )}=F$ so $(N_\mathfrak{q })_{(\mathfrak p )}=V$ whenever $\mathfrak q \ne \mathfrak p$. $\square$

5 Completions

Next, we briefly define the completion and show that the local-global dictionary holds in this context as well. (We will consider completions in the context of local fields more generally starting in chapter , so the reader may wish to return to this section later.) For a general reference on completions (and the induced topology), see e.g. Atiyah–Macdonald [AM69, Chapter 10], Matsumura [Mat89, Chapter 8], Bourbaki [Bou98, Chapter III, §3]. To avoid diving too deeply into commutative algebra we suppose that R is a DVR, with maximal ideal $\mathfrak p$: for example, we might take the localization of a Dedekind domain R at a prime ideal by 9.4.5. There is a natural system of compatible projection maps $R/\mathfrak p ^{n+1} \rightarrow R/\mathfrak p ^n$ indexed by integers $n \ge 1$, and we define thecompletion of R at $\mathfrak p$ to be the inverse (or projective) limit under this system: The completion $R_\mathfrak p$ is again a commutative ring, and we have a natural map $R \rightarrow R_\mathfrak p$ defined by $a \mapsto (a)_n$. Since R has a discrete valuation we have $\bigcap _{n=0}^{\infty } \mathfrak p ^n = \{0\}$, so this map is injective. Moreover, since $\mathfrak p$ is maximal, then in fact this inclusion factors via $R \hookrightarrow R_{(\mathfrak p )} \hookrightarrow R_\mathfrak p$ inducing isomorphisms (Exercise 9.8) for all $e \ge 1$; in particular, the operation of completion is in a sense ‘stronger’ than the operation of localization, and the valuation on R extends naturally to $R_\mathfrak p$, so $R=F \cap R_\mathfrak p \subseteq F_\mathfrak p$. However, once local the completion looks rather similar in the context of lattices, as follows. Let $F_\mathfrak p :=F \otimes _R R_\mathfrak p$ and $V_\mathfrak p :=V \otimes _F F_\mathfrak p$.

Lemma 9.5.3

Let R be a DVR with maximal ideal $\mathfrak p \subseteq R$. Then the maps

(9.5.1)

are mutually inverse bijections between the set of R-lattices in V and the set of $R_\mathfrak p$-lattices in $V_\mathfrak p$.

Proof Let $M \subseteq V$ be an R-lattice. By 9.4.5 we have $M \simeq R^n$ free over R; choose a basis $M=Rx_1 \oplus \dots \oplus Rx_n$. Then $M_\mathfrak p = M \otimes _R R_\mathfrak p \simeq R_\mathfrak p x_1 \oplus \dots \oplus R_\mathfrak p x_n$. Let $M' :=M_\mathfrak{p } \cap V \subseteq V_\mathfrak p$. Then $x' \in M'$ if and only if $x' = a_1x_1 + \dots + a_nx_n$ with $a_i \in R_\mathfrak p \cap F=R$, so indeed $M'=M$. Conversely, let $M_\mathfrak p \subseteq V_\mathfrak p$ and let $M' :=M_\mathfrak p \cap V$. Then $(M')_\mathfrak p \subseteq M_\mathfrak p$, and we prove the opposite inclusion. First, a bit of setup. Let $y_1,\dots ,y_n$ be an F-basis for V, and let $N = R y_1 \oplus \dots \oplus R y_n$. By Lemma 9.3.5, there exists nonzero $r \in R_\mathfrak p$ such that $rN_\mathfrak p \subseteq M_\mathfrak p \subseteq r^{-1}N_\mathfrak p$. Choosing an element $s \in R$ with the same valuation as r, we have $r/s \in R_\mathfrak p ^\times$ so in fact may suppose that $r \in R$. Rescaling the basis vectors $y_i$ and replacing $r^2$ by r we may suppose that $(r N)_\mathfrak p = rN_\mathfrak p \subseteq M_\mathfrak p \subseteq N_\mathfrak p$. From the previous paragraph, we have $rN_\mathfrak p = rN \otimes _R R_\mathfrak p \subseteq M'_\mathfrak p$. Letting $(r)=\mathfrak p ^e$ and taking (9.5.2) on each coordinate, we have an isomorphism $\varphi :N/rN \simeq N_\mathfrak p /rN_\mathfrak p$ induced from the natural inclusion $N \hookrightarrow N_\mathfrak p$. Now to show the inclusion, let $y \in M_\mathfrak p$. Let $x \in N \subseteq V$ be such that $\varphi (x+rN)=y+rN_\mathfrak{p }$; lifting to $N_\mathfrak p$, we find that there exists $z \in rN_\mathfrak{p } \subseteq (M')_\mathfrak p \subseteq M_\mathfrak p$ such that $x=y+z \in M_\mathfrak p \cap V$, so $y=x-z \in (M')_\mathfrak p$.

In particular, Lemma 9.5.3 implies that in the local-global dictionary for lattices over a Dedekind domain R (Theorem 9.4.9), we may also work with collections of $R_\mathfrak p$-lattices $(N_\mathfrak p )_\mathfrak p$ over the completions at primes.

6 Index

Continuing with R a noetherian domain, let $M,N \subseteq V$ be R-lattices.

Definition 9.6.1

The R-index of N in M, written $[M:N]_R$, is the R-submodule of F generated by the set

$$\begin{aligned} \{ \det (\delta ) : \delta \in {{\,\mathrm{End}\,}}_F(V) \text { and } \delta (M) \subseteq N \}. \end{aligned}$$

(9.5.2)

The style of Definition 9.6.1, given by a large generating set (9.6.2), is the replacement for being able to work with given bases; this style will be typical for us in what follows. The determinants $\det (\delta )$ are meant in the intrinsic sense, but can be computed as the determinant of a matrix upon choosing a basis for V.

Lemma 9.6.3

The index $[M:N]_R$ is a nonzero R-module, and if $\alpha \in {{\,\mathrm{Aut}\,}}_F(V)$ then $[\alpha M:N]=\det (\alpha )^{-1}[M:N]$.

Proof Exercise 9.10.

Lemma 9.6.4

If M, N are free (as R-submodules), then $[M:N]_R$ is a free R-module generated by the determinant of any $\delta \in {{\,\mathrm{End}\,}}_F(V)$ giving a change of basis from M to N.

Proof Let $x_1,\dots ,x_n$ be an R-basis for M, thereby an F-basis for V. Let $y_1,\dots ,y_n$ be an R-basis for N; then the map $x_i \mapsto y_i$ first extends to an R-linear isomorphism and thereby to an F-linear map $\delta \in {{\,\mathrm{End}\,}}_F(V)$, and of course $\delta (M) \subseteq N$ by construction, so $\det (\delta ) \in [M:N]_R$. Conversely, let $\delta ' \in {{\,\mathrm{End}\,}}_F(V)$ be such that $\delta '(M) \subseteq N$. The map $\delta '\delta ^{-1} :N \rightarrow N$ is an R-linear map, so $\det (\delta '\delta ^{-1})=\det (\delta ')\det (\delta )^{-1} \in R$, so $\det (\delta ') \in \det (\delta )R$.

Example 9.6.5

If $N=rM$ with $r \in R$, then $[M:N]_R=r^n R$ where $n=\dim _F V$.

Example 9.6.6

If $R=\mathbb Z$ and $N \subseteq M$, then $[M:N]_\mathbb{Z }$ is the ideal generated by $\#(M/N)$, the usual index taken as abelian groups. In this case, for convenience we will often identify $[M:N]_\mathbb{Z }$ with its unique positive generator.

Forming the R-index commutes with localization, as follows.

Lemma 9.6.7

Let $\mathfrak p$ be a prime of R. Then

$$\begin{aligned}{}[M_{(\mathfrak p )}:N_{(\mathfrak p )}]_{R_{(\mathfrak p )}} = ([M:N]_R)_{(\mathfrak p )}. \end{aligned}$$

Proof If $\delta (M) \subseteq N$, then $\delta (M_{(\mathfrak p )}) \subseteq N_{(\mathfrak p )}$ by $R_{(\mathfrak p )}$-linearity, giving the inclusion $(\supseteq )$. For $(\subseteq )$, let $\delta \in {{\,\mathrm{End}\,}}_F(V)$ be such that $\delta (M_{(\mathfrak p )}) \subseteq N_{(\mathfrak p )}$. For any $x \in M$, we have $\delta (x) \in \delta (M) \subseteq \delta (M_{(\mathfrak p )}) \subseteq N_{(\mathfrak p )}$, so there exists $y \in N$ and $s \in R \smallsetminus \mathfrak p$ such that $s\delta (x)=y \in N$. Let $x_1,\dots ,x_m$ generate M as an R-module, and for each i, let $s_i \in R \smallsetminus \mathfrak p$ be such that $s_i\delta (x_i) \in N$. Let $s :=\prod _i s_i$. Then $s\delta (M) \subseteq N$, so $\det (s\delta )=s^n \det (\delta ) \in [M:N]_R$, if $n :=\dim _F V$. Finally, $s \in R_{(\mathfrak p )}^\times$, we conclude that $\det \delta \in ([M:N]_R)_{(\mathfrak p )}$, as desired.

Proposition 9.6.8

Suppose that M, N are projective R-modules. Then $[M:N]_R$ is a projective R-module. Moreover, if $N \subseteq M$ then $[M:N]_R=R$ if and only if $M=N$.

Proof Let $\mathfrak p$ be a prime of R and consider the localization $([M:N]_R)_{(\mathfrak p )}$ at $\mathfrak p$. Since M, N are projective R-modules, they are locally free (9.2.1). By Lemma 9.6.4, the local index $[M_{(\mathfrak p )}:N_{(\mathfrak p )}]_{R_{(\mathfrak p )}}$ is a principal $R_{(\mathfrak p )}$-ideal. By Lemma 9.6.7, we conclude that $[M:N]_R$ is locally principal, therefore projective.

The second statement follows in a similar way: we may suppose that R is local and thus $N \subseteq M$ are free, in which case $M=N$ if and only if a change of basis matrix from N to M has determinant in $R^\times$.

For Dedekind domains, the R-index can be described as follows.

Lemma 9.6.9

If R is a Dedekind domain and $N \subseteq M$, then $[M:N]_R$ is the product of the invariant factors (or elementary divisors) of the torsion R-module M/N.

Proof Exercise 9.12.

7 Quadratic forms

In setting up an integral theory, we will also have need of an extension of the theory of quadratic forms integrally, generalizing those over fields (Section 4.2). For further reading on quadratic forms over rings, we suggest the books by O’Meara [O’Me73], Knus [Knu88], and Scharlau [Scha85].

Definition 9.7.1

Aquadratic map is a map $Q:M \rightarrow N$ between R-modules, satisfying:

• (i) $Q(rx)=r^2Q(x)$ for all $r \in R$ and $x \in M$; and

• (ii) The map $T :M \times M \rightarrow N$ defined by

  $$\begin{aligned} T(x,y) = Q(x+y)-Q(x)-Q(y) \end{aligned}$$

  is R-bilinear.

The map T in (ii) is called theassociated bilinear map.

Remark 32 The bilinearity condition (ii) can be given purely in terms of Q: we require

$$\begin{aligned} Q(x+y+z) = Q(x+y)+Q(x+z)+Q(y+z)-Q(x)-Q(y)-Q(z) \end{aligned}$$

for all $x,y,z\in M$.

Definition 9.7.3

Aquadratic module over R is a quadratic map $Q:M \rightarrow L$ where M is a projective R-module of finite rank and L is a projective R-module of rank 1. Aquadratic form over R is a quadratic module with codomain $L=R$.

A quadratic module $Q:M \rightarrow L$ is free  if M and L are free as R-modules, and a quadratic form $Q:M \rightarrow R$ is free  if M is free as an R-module.

Example 9.7.4

Let $Q:V \rightarrow F$ be a quadratic form. Let $M \subseteq V$ be an R-lattice such that $Q(M) \subseteq L$ where L is an invertible R-module. (When R is a Dedekind domain, we may take $L=Q(M)$, see Exercise 9.13.) Then the restriction $Q|_M :M \rightarrow L$ is a quadratic module over R.

Conversely, if $Q:M \rightarrow L$ is a quadratic module over R, then the extension $Q:M \otimes _R F \rightarrow L \otimes _R F \simeq F$ is a quadratic form over F. Moreover, at the slight cost of some generality (replacing an object by an isomorphic one), by choosing an isomorphism $L \otimes _R F \simeq F$ we may suppose that Q takes values in an invertible fractional ideal $\mathfrak l \subseteq F$.

Example 9.7.5

If $Q:M \rightarrow L$ is a quadratic module and $\mathfrak a \subseteq R$ is a projective R-ideal, then Q extends naturally by property (i) to a quadratic module $\mathfrak a M \rightarrow \mathfrak a ^2 L$.

Definition 9.7.6

Asimilarity between two quadratic modules $Q:M \rightarrow L$ and $Q' :M'\rightarrow L'$ is a pair of R-module isomorphisms and such that $Q'(f(x))=h(Q(x))$ for all $x \in M$, i.e., such that the diagram

(9.5.4)

commutes. Anisometry between quadratic modules is a similarity with $L=L'$ and h the identity map.

Definition 9.7.8

Let $Q:M \rightarrow L$ be a quadratic module over R. Then Q isnondegenerate if the R-linear map is injective; and Q isnonsingular (orregular) if the map (9.7.9) is an isomorphism.

Example 9.7.10

If $R=F$ is a field, then (by linear algebra) Q is nondegenerate if and only if Q is nonsingular.

Example 9.7.11

A quadratic module is nondegenerate if and only if its base extension

$$\begin{aligned} Q_F:M \otimes _R F \rightarrow L \otimes _R F \simeq F \end{aligned}$$

is nondegenerate, since the kernel can be detected over F. Recalling the definition of discriminant (Definition 4.3.3 for ${{\,\mathrm{char}\,}}F \ne 2$ and Definition 6.3.1 in general), we conclude that Q is nondegenerate if and only if ${{\,\mathrm{disc}\,}}Q_F \ne 0$.

The apparent notion of discriminant of a quadratic module needs some care in its definition in this generality; it is delayed until section 15.3, where discriminantal notions are explored in some detail.

Example 9.7.12

Borrowing from the future (see Lemma 15.3.8): if $M \simeq R^n$ is free, then choosing a basis for M and computing (half-)discriminant ${{\,\mathrm{disc}\,}}Q$, we will see that M is nonsingular if and only if ${{\,\mathrm{disc}\,}}Q \in R^\times$.

We now define the notions of genus and classes.

Definition 9.7.13

Let $Q:M \rightarrow L$ be a quadratic module. Thegenus ${{\,\mathrm{Gen}\,}}Q$ is the set of quadratic modules that are locally isometric to Q, i.e., $Q'_{(\mathfrak p )} \sim Q_{(\mathfrak p )}$ for all primes $\mathfrak p \subseteq R$. The class set  ${{\,\mathrm{Cl}\,}}Q$ is the set of isometry classes in the genus. enQ]${{\,\mathrm{Gen}\,}}Q$genus of a quadratic module Q lQ]${{\,\mathrm{Cl}\,}}Q$class set of a quadratic module Q

We conclude with some comments on the codomain of a quadratic map.

Definition 9.7.14

A quadratic module $Q:M \rightarrow L$ isprimitive if Q(M) generates L as an R-module.

9.7.15

If $Q:R^n \rightarrow R$ is a quadratic form, written

$$\begin{aligned} Q(x_1,\dots ,x_n)=\sum _{1 \le i \le j \le n} a_{ij} x_ix_j \in R[x_1,\dots ,x_n], \end{aligned}$$

then Q is primitive if and only if the coefficients $a_{ij}$ generate the unit ideal R.

If R is a Dedekind domain, then $Q(M) \subseteq L$ is again projective (locally at a prime generated by an element of minimal valuation), so one can always replace $Q:M \rightarrow L$ by $Q:M \rightarrow Q(M)$ to get a primitive quadratic module; when R is a PID, up to similarity we may divide through by greatest common divisor of the coefficients $a_{ij}$ in the previous paragraph.

9.7.16

In our admittedly abstract treatment of quadratic modules so far, we have specifically allowed the codomain of the quadratic map to vary at the same time as the domain—in particular, we do not ask that they necessarily take values in R.

Remark 45 In certain lattice contexts with R a Dedekind domain, a quadratic form with values in a fractional ideal $\mathfrak a$ is called an $\mathfrak a$ -modular quadratic form. Given the overloaded meanings of the word modular, we do not employ this terminology. In the geometric context, a quadratic module is called a line-bundle valued quadratic form. Whatever the terminology, we will see in Chapter 22 that it is important to keep track of the codomain of the quadratic map just as much as the domain, and in particular we cannot assume that either is free when R is not a PID.

8 Normalized form

To conclude this chapter, we discuss an explicit normalized form for quadratic forms. Let R be a local PID; then R is either a field or a DVR. In either case, R has valuation $v :R \rightarrow \mathbb Z _{\ge 0} \cup \{\infty \}$ and uniformizer $\pi$; when R is a field, we take a trivial valuation and $\pi =1$.

Let $Q:M \rightarrow R$ be a quadratic form over R. Then since R is a PID, $M \simeq R^n$ is free. We compute a basis for M in which Q has a particularly nice form, diagonalizing Q as far as possible. In cases where $2 \in R^\times$, we can accomplish a full diagonalization; otherwise, we can at least break up the form as much as possible, as follows. For $a,b,c \in R$, the quadratic form $Q(x,y)=ax^2+bxy+cy^2$ on $R^2$ is denoted [a, b, c].

Definition 9.8.1

A quadratic form Q over R isatomic if either:

• (i) $Q \simeq \langle a \rangle$ for some $a \in R^\times$, or

• (ii) $2 \not \in R^\times$ and $Q \simeq [a,b,c]$ with $a,b,c \in R$ satisfying

  $$\begin{aligned} v(b) < v(2a) \le v(2c)\quad \text {and}\quad v(a)v(b)=0. \end{aligned}$$

In case (ii), we necessarily have $v(2)>0$ and $v(b^2-4ac)=2v(b)$.

Example 9.8.2

Suppose $R=\mathbb Z _2$ is the ring of 2-adic integers, so that $v(x)={{\,\mathrm{ord}\,}}_2(x)$ is the largest power of 2 dividing $x \in \mathbb Z _2$. Recall that $\mathbb Z _2^\times /\mathbb Z _2^{\times 2}$ is represented by the elements $\pm 1, \pm 5$, therefore a quadratic form Q over $\mathbb Z _2$ is atomic of type (i) above if and only if $Q(x) \simeq \pm x^2$ or $Q(x) \simeq \pm 5x^2$. For forms of type (ii), the conditions $v(b) < v(2a) = v(a)+1$ and $v(a)v(b)=0$ imply $v(b)=0$, and so a quadratic form Q over $\mathbb Z _2$ is atomic of type (ii) if and only if $Q(x,y) \simeq ax^2 + xy + cy^2$ with ${{\,\mathrm{ord}\,}}_2(a) \le {{\,\mathrm{ord}\,}}_2(c)$. Replacing x by ux and y by $u^{-1} y$ for $u \in \mathbb Z _2^\times$ we may suppose $a=\pm 2^t$ or $a = \pm 5\cdot 2^t$ with $t \ge 0$, and then the atomic representative [a, 1, c] of the isomorphism class of Q is unique.

A quadratic form Q isdecomposable if Q can be written as the orthogonal sum of two quadratic forms ($Q \simeq Q_1 \boxplus Q_2$) and isindecomposable otherwise. It follows by induction on the rank of M that Q is the orthogonal sum of indecomposable forms. We will soon give an algorithmic proof of this fact and write each indecomposable form as a scalar multiple of an atomic form. We begin with the following lemma.

Lemma 9.8.3

An atomic form Q is indecomposable.

Proof If Q is atomic of type (i) then the space underlying Q has rank 1 and is therefore indecomposable. Suppose $Q=[a,b,c]$ is atomic of type (ii) and assume for purposes of contradiction that Q is decomposable. It follows that if $x,y \in M$ then $T(x,y) \in 2R$. Thus we cannot have $v(b)=0$, so $v(a)=0$, and further $v(b) \ge v(2)=v(2a)$; this contradicts the fact that Q is atomic.

Proposition 9.8.4

Let R be a local PID and let $Q:M \rightarrow R$ be a quadratic form. Then there exists a basis of M such that the form Q can be written

$$\begin{aligned} Q \simeq \pi ^{e_1}Q_1 \boxplus \dots \boxplus \pi ^{e_n} Q_n \end{aligned}$$

where the forms $Q_i$ are atomic and $0 \le e_1 \le \dots \le e_n \le \infty$.

In the above proposition, we interpret $\pi ^{\infty } = 0$. A form as presented in Proposition 9.8.4 is callednormalized; this normalized form need not be unique.

Proof When $R=F$ is a field with ${{\,\mathrm{char}\,}}F \ne 2$, we are applying the standard method of Gram–Schmidt orthogonalization to diagonalize the quadratic form. This argument can be adapted to the case where $R=F$ is a field with ${{\,\mathrm{char}\,}}F = 2$, see e.g. Scharlau [Scha85, §9.4]. For the general case, we make further adaptations to this procedure: see Voight [Voi2013, Algorithm 3.12] for a constructive (algorithmic) approach.

9 Exercises

Let R be a noetherian domain with field of fractions $F :={{\,\mathrm{Frac}\,}}R$.

1.:

Let V be a finite-dimensional F-vector space and let $M,N \subseteq V$ be R-lattices. Show that $M+N$ and $M \cap N$ are R-lattices.

$\triangleright$ 2.:

Let B be an F-algebra and let $I \subset B$ be an R-lattice. Show that there exists a nonzero $r \in R \cap I$.

3.:

Give an example of a non-noetherian ring R and modules $N \subset M$ such that M is finitely generated but N is not finitely generated.

4.:

Let k be a field and $R=k[x,y]$. Show that the R-module (x, y) is not projective.

5.:

Let R be a Dedekind domain. Show that every ideal of R is projective, as follows. Let $\mathfrak a \subseteq R$ be a nonzero ideal. (The zero ideal is trivially projective.) Since $\mathfrak a \mathfrak a ^{-1}=R$, we may write $1=\sum _{i=1}^n a_ib_i$ with $a_i \in \mathfrak a$ and $b_i \in \mathfrak a ^{-1}$.

(a):

Define the map $\phi :R^n \rightarrow \mathfrak a$ by $\phi (x_1,\dots ,x_n)=\sum _{i=1}^n a_ix_i$. Observe that $\phi$ is an R-module homomorphism, and construct a right inverse $\psi$ to $\phi$, i.e., $\phi \psi ={{\,\mathrm{id}\,}}_\mathfrak{a }$.

(b):

Using (a), show that $\mathfrak a$ is a direct summand of $R^n$, so $\mathfrak a$ is projective.

6.:

Let $\mathfrak m \subset R$ be a maximal ideal and let M be a finitely generated R-module. Let

be the annihilator  of M. Show that $M_{(\mathfrak m )}=\{0\}$ if and only if $\mathfrak m + {{\,\mathrm{ann}\,}}_R M = R$.

7.:

Suppose R is a Dedekind domain. Let V be a finite-dimensional F-vector space and let $M \subseteq V$ be an R-lattice. Given a pseudobasis $M=\mathfrak a _1 x_1 \oplus \cdots \oplus \mathfrak a _n x_n$ as in (9.3.7), let $[\mathfrak a _1\cdots \mathfrak a _n] \in {{\,\mathrm{Cl}\,}}R$. Show that this class (the Steinitz class, 9.3.10) is well-defined for M independent of the choice of pseudobasis.

8.:

Let R be a DVR with maximal ideal $\mathfrak m$. Show that if $s \not \in \mathfrak m$ then $1/s \in R_\mathfrak m$, so there are natural inclusions

$$\begin{aligned} R \hookrightarrow R_{(\mathfrak m )} \hookrightarrow R_\mathfrak m \end{aligned}$$

from the domain into its localization into the completion, inducing isomorphisms

$$ R/\mathfrak p ^e \xrightarrow {{\sim }}R_{(\mathfrak p )}/\mathfrak p ^e R_{(\mathfrak p )} \xrightarrow {{\sim }}R_\mathfrak p /\mathfrak p ^e R_\mathfrak p $$

for all $e \ge 1$.

$\triangleright$ 9.:

Let V be a finite-dimensional vector space over F and $M \subseteq V$ an R-lattice. Let $\mathfrak p$ be a prime of R. Show that if $M_{(\mathfrak p )} \subseteq V$ is an $R_{(\mathfrak p )}$-lattice then $M_\mathfrak p \cap V = M_{(\mathfrak p )}$. Conclude that Lemma 9.5.3 holds.

10.:

Let V be a finite-dimensional F-vector space and let $M,N \subseteq V$ be R-lattices.

(a):

Show that the index $[M:N]_R$ is a nonzero R-module. [Hint: use Lemma 9.3.5.]

(b):

For $\alpha \in {{\,\mathrm{Aut}\,}}_F(V)$, show $[\alpha M:N]=\det (\alpha )^{-1}[M:N]$.

11.:

Find R-lattices $M,N \subseteq V$ such that $[M:N]_R=R$ but $M \ne N$.

12.:

Prove Lemma 9.6.9, as follows. Suppose R is a Dedekind domain, and let $N \subseteq M \subseteq V$ be R-lattices in a finite-dimensional vector space V over F. Prove that $[M:N]_R$ is the product of the invariant factors (or elementary divisors) of the torsion R-module M/N.

13.:

Suppose R is a Dedekind domain. Let $Q :V \rightarrow F$ be a quadratic form over F, let $M \subseteq V$ be an R-lattice, and let $L :=Q(M) \subseteq F$ be the R-submodule of F generated by the values of Q. Show that L is a fractional R-ideal.

14.:

Consider the ternary quadratic form $Q(x,y,z)=xy+xz$ over $\mathbb Z _2$. Compute a normalized form for Q.

15.:

Consider the following ‘counterexamples’ to Theorem 9.4.9 for more general integral domains as follows. Let $R=\mathbb Q [x,y]$ be the polynomial ring in two variables over $\mathbb Q$, so that $F=\mathbb Q (x,y)$. Let $V=F$ and $I=R$.

(a):

Show that yR has the property that $yR_\mathfrak p \ne R_\mathfrak p$ for infinitely many prime ideals $\mathfrak p$ of R.

(b):

Consider the collection of lattices given by $J_\mathfrak p =f(x)R_\mathfrak p$ if $\mathfrak p = (y, f(x))$ where $f(x) \in \mathbb Q [x]$ is irreducible and $J_\mathfrak p =R_\mathfrak p$ otherwise. Show that $\bigcap _\mathfrak p J_\mathfrak p = (0)$.

[Instead, to conclude that a collection $(J_\mathfrak p )_\mathfrak p$ of $R_\mathfrak p$-lattices arises from a global R-lattice J, one needs that the collection forms a sheaf.]

16.:

In this advanced exercise, we consider generalizations of the notion of lattices to a geometric context; we assume background in algebraic geometry at the level of Hartshorne [Har77, Chapter II].

Let X be a separated, integral scheme—so for each open U, the ring $\mathscr {O}_X(U)$ is a(n integral) domain—and let $\mathscr {O}_X$ be its structure sheaf. Let F be the function field of X (so $F=\mathscr {O}_X(\{\eta \})$ where $\eta$ is the generic point of X). Let V be a finite-dimensional F-vector space.

Define a sheaf of $\mathscr {O}_X$ -lattices in V (also called an $\mathscr {O}_X$ -lattice in V), to be a sheaf $\mathscr {M}$ of $\mathscr {O}_X$-modules such that for each affine open set $U \subseteq X$, the set $\mathscr {M}(U)$ is an $\mathscr {O}_X(U)$-lattice in V. As usual, for $P \in X$ a point, we denote by $\mathscr {M}_{(P)}$ the stalk of $\mathscr {M}$ at P.

(a):

Show that a sheaf of $\mathscr {O}_X$-lattices in V is naturally a subsheaf of the constant sheaf V over X.

(b):

Let $X = \bigcup _i U_i$ be an affine open cover of X, with $U_i={{\,\mathrm{Spec}\,}}R_i$. Since X is separated, each intersection $U_i \cap U_j = {{\,\mathrm{Spec}\,}}R_{ij}$ is affine, so there are natural inclusions $R_i,R_j \hookrightarrow R_{ij} \subseteq F$ of rings for each i, j. Show that a sheaf of $\mathscr {O}_X$-lattices is specified uniquely by $R_i$-lattices $M_i \subseteq V$ for each i, subject to the condition that $M_iR_{ij}=M_jR_{ij}$ for each i, j. [Hint: this is an easy case of gluing, where isomorphism is replaced by equality in V.]

(c):

Now suppose further that X is noetherian, normal, and of dimension $\le 1$ (also called aDedekind scheme). Then the local rings of X at closed points are DVRs with fraction field F, and nonempty affine open subsets of X are the complements of finite subsets of closed points and of the form $U={{\,\mathrm{Spec}\,}}R$ with R an Dedekind domain. (For example, we may take $X={{\,\mathrm{Spec}\,}}R$ for R a Dedekind domain or X a smooth projective integral curve over a field.)

Extend the local-global dictionary for lattices to X, in the following way. Let $U={{\,\mathrm{Spec}\,}}R \subseteq X$ be a nonempty affine open subset, and let $M \subseteq V$ be an R-lattice. Show that the map $\mathscr {N}\rightarrow (\mathscr {N}_{(P)})_P$ establishes a bijection between $\mathscr {O}_X$-lattices $\mathscr {N}$ in V and collections of lattices $(N_{(P)})_P$ indexed by the points $P \in X$, such that for all but finitely many $P \in U$ given by the prime $\mathfrak p \subseteq R$, we have $M_{(\mathfrak p )}=N_{(P)} \subseteq V$.