Skew Category Algebras

Abstract

We study a new (large) class of algebras (that was introduced in Bavula in Math Comput Sci 11(3–4):253–268, 2017)—the skew category algebras. Any such an algebra $\mathcal{C}(\sigma )$ is constructed from a category $\mathcal{C}$ and a functor $\sigma$ from the category $\mathcal{C}$ to the category of algebras. Criteria are given for the algebra $\mathcal{C}(\sigma )$ to be simple or left Noetherian or right Noetherian or semiprime or have 1.

1 Skew Category Algebras, Examples and Constructions

In this paper, K is a commutative ring with 1, algebra means a K-algebra. In general, it is not assumed that a K-algebra has an identity element. Module means a left module. Missed definitions can be found in [1].

Let $\mathcal{C}$ be a category, $\mathrm{Ob} ( \mathcal{C})$ be the set of its objects and $\mathrm{Mor}( \mathcal{C})$ be the set of its morphisms. For each objects $i,j\in \mathrm{Ob} ( \mathcal{C})$, $\mathcal{C}(i,j)$ is the set of morphisms $f:i\rightarrow j$, the objects $i=t(f)$ and $j=h(f)$ are called the tail and head of the morphism f, respectively. For each object $i\in \mathrm{Ob} ( \mathcal{C})$, $e_i$ is the identity morphism $i\rightarrow i$.

Definition 1.1

([2]) Let $\mathcal{C}$ be a category and $\sigma$ be a functor from the category $\mathcal{C}$ to the category of unital K-algebras over a commutative ring K (eg, $K=\mathbb {Z}$ or K is a field). So, for each object $i\in \mathrm{Ob} ( \mathcal{C})$, $D_i:=\sigma (i)$ is a K-algebra and for each morphism

$$\begin{aligned} f: i\mapsto j, \;\; \sigma _f: D_i\rightarrow D_j \end{aligned}$$

is a K-algebra homomorphism, and $\sigma _{fg}= \sigma _f\sigma _g$ for all morphisms f and g such that $t(f) = h(g)$. The direct sum of left K-modules

$$\begin{aligned} \mathcal{C}(\sigma )=\bigoplus _{f\in \mathrm{Mor}( \mathcal{C})} D_{h(f)}f \end{aligned}$$

(1)

where $D_{h(f)}f$ is a free left $D_{h(f)}$-module of rank 1, is a K-algebra with multiplication given by the rule: For all $f,g\in \mathrm{Mor}( \mathcal{C})$, $a\in D_{h(f)}$ and $b\in D_{h(g)}$,

$$\begin{aligned} af\cdot bg={\left\{ \begin{array}{ll} a\sigma _f(b)fg&{} \text {if }t(f) = h(g),\\ 0&{} \text {otherwise}.\\ \end{array}\right. } \end{aligned}$$

(2)

It is a trivial exercise to verify that the multiplication is associative. The K-algebra $\mathcal{C}(\sigma )$ is called a skew category K-algebra. If $K=\mathbb {Z}$, the $\mathbb {Z}$-algebra $\mathcal{C}(\sigma )$ is called a skew category ring.

Definition 1.2

If the direct sum (1) admits an associative product which is given by the rule: For all $f,g\in \mathrm{Mor}( \mathcal{C})$, $a\in D_{h(f)}$ and $b\in D_{h(g)}$,

$$\begin{aligned} af\cdot bg={\left\{ \begin{array}{ll} a\sigma _f(b)c(f,g)fg&{} \text {if }t(f) = h(g),\\ 0&{} \text {otherwise}, \\ \end{array}\right. } \end{aligned}$$

(3)

where

$$\begin{aligned} c(f,g)\in {\left\{ \begin{array}{ll} D_{h(f)}&{} \text {if }t(f) = h(g),\\ \{ 0\}&{} \text {otherwise,} \\ \end{array}\right. } \end{aligned}$$

(4)

then it is called the twisted skew category K-algebra and is denoted by $\mathcal{C}(\sigma , c)$.

The categorical nature of the above classes of rings especially the categorical/explicit nature of their multiplications makes these classes important as far as various computational aspects are concerned.

Let $1_i$ be the identity of the algebra $D_i$. Then $1_ie_i\in D_ie_i\subseteq \mathcal{C}(\sigma )$ where $i\in \mathrm{Ob}( \mathcal{C})$. Abusing the notation, we write $e_i$ for $1_ie_i$. Then $e_i\in \mathcal{C}(\sigma )$.

The$\mathcal{C}$-grading on$\mathcal{C}(\sigma )$. By the very definition, the algebra $\mathcal{C}(\sigma )$ is a $\mathcal{C}$-graded algebra, that is

$$\begin{aligned} D_{h(f)}f\cdot D_{h(g)}g\subseteq D_{h(fg)}fg\;\; \mathrm{for \; all}\;\; f,g\in \mathrm{Mor}( \mathcal{C}). \end{aligned}$$

The algebra $\mathcal{C}(\sigma )$ is a direct sum

$$\begin{aligned} \mathcal{C}(\sigma )= \bigoplus _{i,j\in \mathrm{Ob} ( \mathcal{C})} \mathcal{C}(\sigma )_{ij}\;\; \mathrm{where}\;\; \mathcal{C}(\sigma )_{ij}=\bigoplus _{f\in \mathcal{C}(j,i)}D_if \end{aligned}$$

(5)

and for all $i,j,k,l\in \mathrm{Ob} ( \mathcal{C})$,

$$\begin{aligned} \mathcal{C}(\sigma )_{ij} \mathcal{C}(\sigma )_{kl}\subseteq \delta _{jk} \mathcal{C}(\sigma )_{il} \end{aligned}$$

(6)

where $\delta _{jk}$ is the Kronecker delta. In particular, for each $i\in \mathrm{Ob} ( \mathcal{C})$, $\mathcal{C}(\sigma )_{ii}$ is a K-algebra without 1, in general. For each $i,j\in \mathrm{Ob} ( \mathcal{C})$, $\mathcal{C}(\sigma )_{ij}$ is a $( \mathcal{C}(\sigma )_{ii}, \mathcal{C}(\sigma )_{jj})$-bimodule.

The next two examples show that even for two simplest categories that contain a single object, a single loop or a single invertible loop, the above construction gives apart from a skew polynomial ring or a skew Laurent polynomial ring, new classes of rings.

Example 1

Let $\mathcal{C}$ be a category that contains a single object, say 1, and $\mathrm{Mor}( \mathcal{C})= \{ x^i\, | \, i\in \mathbb {N}\}$ where $e:=x^0$ is the identity morphism. So, $\mathcal{C}(\sigma )= De\oplus Dx\oplus \cdots \oplus Dx^i\oplus \cdots$ where $D= \sigma (1)$ and $ed=\sigma _e(d)e$ and $x^id= \sigma _x^i(d) x^i$ for all $i\ge 1$ where $\sigma _e$ and $\sigma _x$ are ring endomorphisms of D such that $\sigma _e \sigma _x= \sigma _x\sigma _e= \sigma _x$ and $\sigma _e^2= \sigma _e$.

• If $\sigma _e = \mathrm{id}_D$ then $\mathcal{C}(\sigma )= D[x; \sigma _x]$ is a skew polynomial ring.

• If $\sigma _e\ne \mathrm{id}_D$ then $\mathcal{C}(\sigma )$ is not a skew polynomial ring since $ed= \sigma _e(d)e$ and, in general, $\sigma _e(d) e \ne de$ for all $d\in D$ (since $\sigma _e \ne \mathrm{id}_D$). For example, let $D= D_1\times D_2\times D_3$ and $\sigma _e$ and $\sigma _x$ are the projections onto $D_1\times D_2$ and $D_1$, respectively. Then $eD_3=0$.

Example 2

Let $\mathcal{C}$ be a category that contains a single object, say 1, and $\mathrm{Mor}( \mathcal{C})= \{ x^i\, | \, i\in \mathbb {Z}\}$ where $e:=x^0$ is the identity morphism $(xx^{-1}=x^{-1}x=e$). The functor $\sigma$ is determined by the algebra $D=\sigma (1)$ and its algebra endomorphisms $\sigma _e$, $\sigma _x$ and $\sigma _{x^{-1}}$ such that

$$\begin{aligned} \sigma _e^2= \sigma _e, \;\; \sigma _e \sigma _{x^{\pm 1}}= \sigma _{x^{\pm 1}}\sigma _e= \sigma _{x^{\pm 1}}\;\; \mathrm{and}\;\; \sigma _x\sigma _{x^{-1}}=\sigma _{x^{-1}}\sigma _x=\sigma _e. \end{aligned}$$

Then $\mathcal{C}(\sigma )= \oplus _{i\in \mathbb {Z}}Dx^i$.

• If $\sigma _e = \mathrm{id}_D$ then $\sigma _{x^{-1}}=\sigma _x^{-1}$ and $\mathcal{C}(\sigma )= D[x^{\pm 1}; \sigma _x]$ is a skew Laurent polynomial ring.

• If $\sigma _e\ne \mathrm{id}_D$ then $\mathcal{C}(\sigma )$ is not a skew Laurent polynomial ring. For example, let $D= D_1\times D_2$ be a direct product of algebras and $\sigma _e=\sigma _x=\sigma _{x^{-1}}$ be the projection onto $D_1$. Then $eD_2=0$ and $xD_2=x^{-1}D_2=0$.

Example 3

Let $\mathcal{C}$ be a category that contains a single object, say 1, and the monoid $\mathcal{C}(1, 1)$ is generated by elements x and y subject to the defining relation $yx=e$. The functor $\sigma$ is determined by the algebra $D=\sigma (1)$ and its three algebra endomorphisms $\sigma _x$, $\sigma _y$ and $\sigma _e$ such that

$$\begin{aligned} \sigma _y\sigma _x=\sigma _e. \end{aligned}$$

The skew category algebra $\mathcal{C}(\sigma )$ is called the skew semi-Laurent polynomial ring [2]. It is a new class of rings. Suppose, for simplicity, that $\sigma _e=\mathrm{id}_D$. Then the ring $\mathcal{C}(\sigma )$ is generated by a ring D and elements x and y subject to the defining relations:

$$\begin{aligned} yx=1, \;\; xd= \sigma _x(d) x \;\; \mathrm{and }\;\; yd= \sigma _y(d) y \;\; \mathrm{for \; all}\;\; d\in D. \end{aligned}$$

We denote this ring by $D[x,y; \sigma _x, \sigma _y]$. In particular, $D[x,y; \tau , \tau ^{-1}]$ where $\tau$ is an automorphism of D.

Example 4

Let $n\ge 1$ be a natural number and $\mathcal{M}_n$ be the matrix units category:

$$\begin{aligned} \mathrm{Ob}(\mathcal{M}_n)= \{ 1, \ldots , n\}, \;\; \mathcal{M}_n(i,j)=\{ E_{ji}\}\;\; \mathrm{and}\;\; E_{ij}E_{jk}=E_{ik}\;\; \mathrm{for \; all}\;\; i,j,k. \end{aligned}$$

Let D be a ring and $f_1, \ldots , f_n$ be its automorphisms. Define $\sigma$ by the rule $\sigma (i) = D$ and $\sigma (E_{ij})=f_if^{-1}_j$. The skew category algebra

$$\begin{aligned} \mathcal{M}_n(\sigma ) = \oplus _{i,j=1}^n DE_{ij} \end{aligned}$$

is called the skew matrix ring where the multiplication is given by the rule

$$\begin{aligned} dE_{ij}\cdot d'E_{kl}=\delta _{jk} df_if^{-1}_j(d')E_{jl}\;\; \mathrm{for \; all}\;\; d,d'\in D. \end{aligned}$$

The skew graph rings and the skew tree rings.

Definition 1.3

([2]) Let $\Gamma = (\Gamma _0, \Gamma _1)$ be a non-oriented graph without cycles where $\Gamma _0$ is the set of vertices and $\Gamma _1$ is the set of edges. If, in addition, $\Gamma$ is connected then it is called a tree. So, any non-oriented graph without cycles is a disjoint union of its connected components which are trees. Let ${\varvec{\Gamma }}$ be the category groupoid associated with $\Gamma$: $\mathrm{Ob}({\varvec{\Gamma }})=\Gamma _0$, for each $i\in \mathrm{Ob}(\Gamma )$, ${\varvec{\Gamma }}(i,i)=\{ e_{ii} \}$, for distinct $i,j\in \mathrm{Ob}({\varvec{\Gamma }})$ such that $(i,j)\in \Gamma _1$, ${\varvec{\Gamma }} (i,j) = \{ e_{ji}\}$ and ${\varvec{\Gamma }} (j,i) = \{ e_{ij}\}$, $e_{ij}e_{ji}=e_{ii}$ and $e_{ji}e_{ij}=e_{jj}$. Let $\sigma$ be a functor from ${\varvec{\Gamma }}$ to the category of rings. Then ${\varvec{\Gamma }}(\sigma )$ is called the skew graph ring. If $\Gamma$ is a tree then ${\varvec{\Gamma }}(\sigma )$ is called the skew tree ring. We say that the functor $\sigma$ is of isomorphism type if $\sigma (e_{ij}):\sigma (i) \rightarrow \sigma (j)$ is a unital ring isomorphism for all $(i,j) \in \Gamma _1$.

Theorem 1.4

Let $\Gamma$ be a finite tree, $n= |\Gamma _0|$ and the functor $\sigma$ be of isomorphism type. Suppose that for some $i\in \Gamma _0$ the ring $D_i=\sigma (i)$ is a semiprime, left (resp., right) Goldie ring and $Q_l(D_i)$ (resp., $Q_r(D_i))$ is its left (resp., right) quotient ring. Then $\mathbf{\Gamma }(\sigma )$ is a semiprime, left (resp., right) Goldie ring and $Q_l(\mathbf{\Gamma }(\sigma ))\simeq M_n(Q_l(D_i))$ (resp., $Q_r(\mathbf{\Gamma }(\sigma ))\simeq M_n(Q_r(D_i))$) where $M_n(R)$ is a matrix ring over a ring R. In particular, the left (resp., right) uniform dimension of ${\varvec{\Gamma }}(\sigma )$ is $nd_l$ (resp., $nd_r$) where $d_l$ (resp., $d_r$) is a left (resp., right) uniform dimension of $D_i$.

Proof

(Sketch). Let $\mathcal{C}_{D_j}$ be the set of regular elements of the ring $D_j=\sigma (j)$. All the rings $D_j$ are isomorphic. The set of regular elements $S=\oplus _{j=1}^n \mathcal{C}_{D_j}e_{jj}$ is a left Ore set of $\mathbf{\Gamma }(\sigma )$ such that $S^{-1}{} \mathbf{\Gamma }(\sigma )$ is a semisimple Artinian ring. Furthermore, $S^{-1}{} \mathbf{\Gamma }(\sigma )\simeq M_n(Q_l(D_i))$. Hence, $Q_l(\mathbf{\Gamma }(\sigma ))\simeq M_n(Q_l(D_i))$, and so $\mathbf{\Gamma }(\sigma )$ is a semiprime, left Goldie ring. The rest is obvious. $\square$

As a result we have the following corollary.

Corollary 1.5

Let $\Gamma$ be a finite non-orientable graph, i.e., $\Gamma =\coprod _{s=1}^\nu \Gamma ^{(s)}$ is a disjoint union of finite trees $\Gamma ^{(s)}$. Then

• 1. The skew graph ring $\mathbf{\Gamma }(\sigma )$ is a direct product $\prod _{s=1}^\nu \mathbf{\Gamma }^{(s)}(\sigma _s)$ of skew tree rings where $\sigma _s$ is the restriction of the functor $\sigma$ to $\Gamma ^{(s)}(\sigma _s)$.

• 2. If the trees $\Gamma ^{(s)}$ ($s=1, \ldots , \nu$) satisfy the conditions of Theorem 1.4 then $Q_l(\mathbf{\Gamma }(\sigma ))\simeq \prod _{s=1}^\nu Q_l(\mathbf{\Gamma }^{(s)}(\sigma _s))$ (resp., $Q_r(\mathbf{\Gamma }(\sigma ))\simeq \prod _{s=1}^\nu Q_r(\mathbf{\Gamma }^{(s)}(\sigma _s))$) is a direct product of semiprime, left (resp., right) Goldie rings, and so it is a semiprime, left (resp., right) Goldie ring.

2 Properties of Skew Category Algebras

In this section, criteria are given for a skew category algebra $\mathcal{C}(\sigma )$ to be left/right Noetherian or semiprime or simple.

The ideal$\mathfrak {a}$and the algebra$\overline{ \mathcal{C}(\sigma )}$.

Lemma 2.1

Let D be a ring and $\sigma '$ be its ring endomorphism such that $\sigma '^2= \sigma '$. Then $D=\sigma ' (D)\oplus \mathrm{ker } (\sigma ' )$ and the restriction homomorphism $\sigma ' |_{\sigma ' (D)}: \sigma ' (D) \rightarrow \sigma ' (D)$, $d\mapsto d$ is the identity automorphism.

Proof

Straightforward. $\square$

By (5), the formal sum

$$\begin{aligned} e=\sum _{i\in \mathrm{Ob} ( \mathcal{C})}e_i \end{aligned}$$

determines two well-defined maps:

$$\begin{aligned} e\cdot : \mathcal{C}(\sigma )\rightarrow \mathcal{C}(\sigma ),\;\; a\mapsto ea\;\; \mathrm{and}\;\; \cdot e : \mathcal{C}(\sigma )\rightarrow \mathcal{C}(\sigma ), \;\; a\mapsto ae. \end{aligned}$$

Clearly, the map $\cdot e$ is the identity map $\mathrm{id}$ on $\mathcal{C}(\sigma )$ but the kernel $\mathfrak {a}$ of the map $e\cdot$ is equal to

$$\begin{aligned} \mathfrak {a}( \mathcal{C}(\sigma )) := \mathfrak {a}:=\bigoplus _{f\in \mathrm{Mor}( \mathcal{C})} \mathfrak {a}_{h(f)}f \end{aligned}$$

where $\mathfrak {a}_i:= \mathrm{ker } (\sigma _{e_i})$ and $\sigma _i:=\sigma _{e_i}:D_i\rightarrow D_i$ is a K-algebra endomorphism, and $(e\cdot )^2=e\cdot$. Since $\sigma _i^2=\sigma _i$,

$$\begin{aligned} D_i= \sigma _i (D)\oplus \mathfrak {a}_i\;\; \mathrm{for \; all}\;\; i\in \mathrm{Ob}( \mathcal{C}), \end{aligned}$$

(7)

by Lemma 2.1.

$$\begin{aligned} \mathcal{C}(\sigma )=\overline{ \mathcal{C}(\sigma )}\oplus \mathfrak {a}\;\; \mathrm{where} \;\; \overline{ \mathcal{C}(\sigma )}:=\bigoplus _{f\in \mathrm{Mor}( \mathcal{C})} \sigma _{h(f)}(D_{h(f)})f \end{aligned}$$

(8)

is a K-subalgebra of $\mathcal{C}(\sigma )$ such that the maps $(e\cdot )|_{\overline{ \mathcal{C}(\sigma )}}:\overline{ \mathcal{C}(\sigma )}\rightarrow \overline{ \mathcal{C}(\sigma )}$, $c\mapsto c$ and $(\cdot e)|_{\overline{ \mathcal{C}(\sigma )}}:\overline{ \mathcal{C}(\sigma )}\rightarrow \overline{ \mathcal{C}(\sigma )}$, $c\mapsto c$ are the identity map on $\overline{ \mathcal{C}(\sigma )}$.

Lemma 2.2

The set $\mathfrak {a}$ is an ideal of the algebra $\mathcal{C}(\sigma )$ such that $\mathcal{C}(\sigma )\mathfrak {a}=0$, $\mathfrak {a}\, \mathcal{C}(\sigma )=\mathfrak {a}$ and $\mathfrak {a}^2 = 0$.

Proof

$\mathcal{C}(\sigma )\mathfrak {a}= \mathcal{C}(\sigma )\cdot e\cdot \mathfrak {a}=0$, the rest is obvious. $\square$

The next theorem shows that the algebra $\overline{ \mathcal{C}(\sigma )}$ is also a skew category algebra.

Theorem 2.3

• 1. The subalgebra $\overline{ \mathcal{C}(\sigma )}$ of $\mathcal{C}(\sigma )$ is also a skew category algebra $\overline{ \mathcal{C}(\sigma )}= \mathcal{C}(\overline{\sigma } )$ where for each $i\in \mathrm{Ob}( \mathcal{C})$, $\overline{\sigma }(i) := \sigma _i(D_i)$ and for each $f\in \mathcal{C}(i,j)$, $\overline{\sigma }_f:=\sigma _f|_{\sigma _i(D_i)}: \sigma _i (D_i) \rightarrow \sigma _i (D_i)$, $d\mapsto \sigma _f(d)$.

• 2. For all $i\in \mathrm{Ob}( \mathcal{C})$, $\overline{\sigma }_i=\mathrm{id}_{\overline{\sigma }(i)}$.

• 3. $\mathfrak {a}( \mathcal{C}( \overline{\sigma }))=0$.

• 4. The maps $e\cdot$ and $\cdot e$ are the identity maps on $\mathcal{C}(\overline{\sigma })$.

Proof

• 1. Statement 1 follows from (8) and the fact that $\sigma _j\sigma _f=\sigma _f = \sigma _f\sigma _i$ for all elements $f\in \mathcal{C}(i,j)$.

• 2–4. Statement 2 is obvious. Then statements 3 and 4 follow from statement 2. $\square$

The ideal $\mathfrak {a}$ is a $\mathcal{C}$-graded ideal of the algebra $\mathcal{C}(\sigma )$. Furthermore,

$$\begin{aligned} \mathfrak {a}= \bigoplus _{i,j\in \mathrm{Ob} ( \mathcal{C})}\mathfrak {a}_{ij} \end{aligned}$$

where $\mathfrak {a}_{ij}=\bigoplus _{f\in \mathcal{C}(j,i)}\mathfrak {a}_i f\subseteq \mathcal{C}(\sigma )_{ij}$, $0=\mathfrak {a}_{ij}\mathfrak {a}_{kl}\subseteq \delta _{jk}\mathfrak {a}_{il}$ for all $i,j,k,l\in \mathrm{Ob} ( \mathcal{C})$. Since $\overline{ \mathcal{C}(\sigma )}= \mathcal{C}(\overline{\sigma })$ (Theorem 2.3.(1)), the factor algebra

$$\begin{aligned} \overline{ \mathcal{C}(\sigma )}= \mathcal{C}(\sigma )/ \mathfrak {a}= \bigoplus _{f\in \mathrm{Mor}( \mathcal{C})} \overline{D}_{h(f)}f\subseteq \mathcal{C}(\sigma )\end{aligned}$$

is a $\mathcal{C}$-graded algebra where $\overline{D}_i=D_i/\mathfrak {a}_i=\mathrm{im}(\sigma _i)$. Furthermore,

$$\begin{aligned} \overline{ \mathcal{C}(\sigma )}= \bigoplus _{\mathbf{i},j\in \mathrm{Ob} ( \mathcal{C})}\overline{ \mathcal{C}(\sigma )}_{ij}\;\; \mathrm{where}\;\; \overline{ \mathcal{C}(\sigma )}_{ij}= \mathcal{C}(\sigma )_{ij}/ \mathfrak {a}_{ij} \end{aligned}$$

(9)

and $\overline{ \mathcal{C}(\sigma )}_{ij}\overline{ \mathcal{C}(\sigma )}_{kl}\subseteq \delta _{jk}\overline{ \mathcal{C}(\sigma )}_{il}$ for all $i,j,k,l\in \mathrm{Ob} ( \mathcal{C})$.

Theorem 2.4

(Criterion for $\mathcal{C}(\sigma )$ to be a left Noetherian algebra) The algebra $\mathcal{C}(\sigma )$ is a left Noetherian algebra iff the following conditions hold

• 1. the set $\mathrm{Ob} ( \mathcal{C})$ is a finite set,

• 2. the ideal $\mathfrak {a}$ is a finitely generated abelian group,

• 3. for every object $i\in \mathrm{Ob} ( \mathcal{C})$, the K-algebra $\overline{ \mathcal{C}(\sigma )}_{ii}$ is a left Noetherian algebra, and

• 4. for all objects $i,j\in \mathrm{Ob} ( \mathcal{C})$ such that $i\ne j$, the left $\overline{ \mathcal{C}(\sigma )}_{ii}$-module $\overline{ \mathcal{C}(\sigma )}_{ij}$ is finitely generated.

Proof

The algebra $\mathcal{C}(\sigma )= \bigoplus _{j\in \mathrm{Ob} ( \mathcal{C})} \mathcal{C}(\sigma )_{*j}$ is a direct sum of nonzero left ideals where

$$\begin{aligned} \mathcal{C}(\sigma )_{*j}:=\bigoplus _{i\in \mathrm{Ob} ( \mathcal{C})} \mathcal{C}(\sigma )_{ij}. \end{aligned}$$

So, the algebra $\mathcal{C}(\sigma )$ is a left Noetherian algebra iff the set $\mathrm{Ob} ( \mathcal{C})$ is a finite set and all the left ideals $\mathcal{C}(\sigma )_{*j}$ are Noetherian left $\mathcal{C}(\sigma )$-modules iff $|\mathrm{Ob} ( \mathcal{C})|<\infty$, the left $\mathcal{C}(\sigma )$-module $\mathfrak {a}$ is Noetherian and all the left $\overline{ \mathcal{C}(\sigma )}$-modules

$$\begin{aligned} \overline{ \mathcal{C}(\sigma )}_{*j}=\bigoplus _{i\in \mathrm{Ob} ( \mathcal{C})} \overline{ \mathcal{C}(\sigma )}_{ij}\end{aligned}$$

are Noetherian (since $\mathcal{C}(\sigma )=\overline{ \mathcal{C}(\sigma )}\oplus \mathfrak {a}$ is a direct sum of left $\mathcal{C}(\sigma )$-modules) iff conditions 1 and 2 hold (since $\mathcal{C}(\sigma )\mathfrak {a}=0$, Lemma 2.4) and the left $\overline{ \mathcal{C}(\sigma )}_{ii}$-module $\overline{ \mathcal{C}(\sigma )}_{ij}$ is Noetherian for all $i,j\in \mathrm{Ob} ( \mathcal{C})$ (since each left $\overline{ \mathcal{C}(\sigma )}$-submodule M of $\overline{ \mathcal{C}(\sigma )}_{*j}$ is a direct sum

$$\begin{aligned} M=eM=\bigoplus _{i\in \mathrm{Ob} ( \mathcal{C})}e_iM \end{aligned}$$

where $e_iM$ is a left $\overline{ \mathcal{C}(\sigma )}_{ii}$-submodule of $\overline{ \mathcal{C}(\sigma )}_{ij}$ and the functor from the category of all $\overline{ \mathcal{C}(\sigma )}_{ii}$-submodules of $\overline{ \mathcal{C}(\sigma )}_{ij}$ to the category of all $\overline{ \mathcal{C}(\sigma )}$-submodules of $\overline{ \mathcal{C}(\sigma )}_{*j}$,

$$\begin{aligned} N\rightarrow \overline{ \mathcal{C}(\sigma )}N=\bigoplus _{k\in \mathrm{Ob} ( \mathcal{C})} \overline{ \mathcal{C}(\sigma )}_{ki}N \end{aligned}$$

is faithful since $e_i\overline{ \mathcal{C}(\sigma )}N=\overline{ \mathcal{C}(\sigma )}_{ii}N=N$) iff statements 1–4 hold. $\square$

Proposition 2.5

(Criterion for $\mathcal{C}(\sigma )$ to be a right Noetherian algebra) The algebra $\mathcal{C}(\sigma )$ is a right Noetherian algebra iff the following conditions hold

• 1. the set $\mathrm{Ob} ( \mathcal{C})$ is a finite set,

• 2. for every object $i\in \mathrm{Ob} ( \mathcal{C})$, the K-algebra $\mathcal{C}(\sigma )_{ii}$ is a right Noetherian algebra, and

• 3. for all objects $i,j\in \mathrm{Ob} ( \mathcal{C})$ such that $i\ne j$, the right $\mathcal{C}(\sigma )_{jj}$-module $\mathcal{C}(\sigma )_{ij}$ is finitely generated.

Proof

The algebra $\mathcal{C}(\sigma )= \bigoplus _{i\in \mathrm{Ob} ( \mathcal{C})} \mathcal{C}(\sigma )_{i*}$ is a direct sum of nonzero right ideals where

$$\begin{aligned} \mathcal{C}(\sigma )_{i*}=\bigoplus _{j\in \mathrm{Ob} ( \mathcal{C})} \mathcal{C}(\sigma )_{ij}. \end{aligned}$$

So, the algebra $\mathcal{C}(\sigma )$ is a right Noetherian algebra iff the set $\mathrm{Ob} ( \mathcal{C})$ is a finite set and all right ideals $\mathcal{C}(\sigma )_{i*}$ are Noetherian right $\mathcal{C}(\sigma )$-modules iff $|\mathrm{Ob} ( \mathcal{C})|<\infty$ and the right $\mathcal{C}(\sigma )_{jj}$-module $\mathcal{C}(\sigma )_{ij}$ is Noetherian for all $i,j\in \mathrm{Ob} ( \mathcal{C})$ iff $|\mathrm{Ob} ( \mathcal{C})|<\infty$, the rings $\mathcal{C}(\sigma )_{ii}$ are right Noetherian and the right $\mathcal{C}(\sigma )_{jj}$-modules $\mathcal{C}(\sigma )_{ij}$ are finitely generated for all $i\ne j$. $\square$

Example 5

Let $\mathcal{C}$: $1{\mathop {\rightarrow }\limits ^{f}}2$ and the functor $\sigma$ is as follows: $\sigma (1) = \mathbb {Q}$, $\sigma (2) =\mathbb {R}$, $\sigma _{e_1}=\mathrm{id}_{\mathbb {Q}}$, $\sigma _{e_2}=\mathrm{id}_{\mathbb {R}}$ and $\sigma _f:\mathbb {Q}\rightarrow \mathbb {R}$, $q\mapsto q$. Then the algebra $\mathcal{C}(\sigma )$ is isomorphic to the lower triangular matrix algebra $\begin{pmatrix} \mathbb {Q}&{} 0 \\ \mathbb {R}&{}\mathbb {R}\end{pmatrix}$. By Theorem 2.4, the algebra $\mathcal{C}(\sigma )$ is left Noetherian but not right Noetherian, by Proposition 2.5 (since $\mathbb {R}_\mathbb {Q}$ is not a finitely generated right $\mathbb {Q}$-module).

Example 6

Let $\mathcal{C}$: $1{\mathop {\rightarrow }\limits ^{f}}2$ and the functor $\sigma$ is as follows: $\sigma (1) = K[t]$ is a polynomial algebra in the variable t over K, $\sigma (2) =K$, $\sigma _{e_1}:K[t]\rightarrow K[t]$, $t\mapsto 0$; $\sigma _{e_2}=\mathrm{id}_{K}:K\rightarrow K$ and $\sigma _f:K[t]\rightarrow K$, $t\mapsto 0$. Then $\mathfrak {a}= tK[t]e_1$ is not a finitely generated $\mathbb {Z}$-module. So, the algebra $\mathcal{C}(\sigma )$ is not a left Noetherian algebra, by Theorem 2.4. Since the algebra $\mathcal{C}(\sigma )_{11}=K[t]e_1$ is not a right Noetherian algebra, the ring $\mathcal{C}(\sigma )$ is not a right Noetherian ring, by Proposition 2.5.

Lemma 2.6

(Existence of 1 in $\mathcal{C}(\sigma )$) The algebra $\mathcal{C}(\sigma )$ has 1 iff the set $\mathrm{Ob} ( \mathcal{C})$ is a finite set and $\sigma _{e_i}=\mathrm{id}_{D_i}$ for all $i\in \mathrm{Ob} ( \mathcal{C})$. In this case, $e=\sum _{i\in \mathrm{Ob} ( \mathcal{C})}e_i$ is the identity of the algebra $\mathcal{C}(\sigma )$.

Proof

$(\Rightarrow )$ Suppose that 1 is an identity of $\mathcal{C}(\sigma )$. Then necessarily the set $\mathrm{Ob} ( \mathcal{C})$ is a finite set, otherwise $1a=0$ for some nonzero element a of $\mathcal{C}(\sigma )$. The $1=\sum _{i,j} 1_{ij}$ where $1_{ij}\in \mathcal{C}(\sigma )_{ij}$. The equalities $1e_j=e_j=e_j1$ for all $j\in \mathrm{Ob} ( \mathcal{C})$ imply that $1=\sum _{i\in \mathrm{Ob} ( \mathcal{C})} e_i=e$. Then, necessarily $\sigma _{e_i}=\mathrm{id}_{D_i}$ for all $i\in \mathrm{Ob} ( \mathcal{C})$.

$(\Leftarrow )$ Clearly, e is the identity of the algebra $\mathcal{C}(\sigma )$. $\square$

Lemma 2.7

Suppose that $n=|\mathrm{Ob}( \mathcal{C})|<\infty$. If I is an ideal of $\mathcal{C}(\sigma )$ such that $e_iIe_i=0$ for all $i\in \mathrm{Ob}( \mathcal{C})$ then $I^{n+1}=0$.

Proof

By (8), $\mathcal{C}(\sigma )= \overline{ \mathcal{C}(\sigma )}\oplus \mathfrak {a}$. Hence, $I\subseteq \overline{I}\oplus \mathfrak {a}$ where $\overline{I}= (I+\mathfrak {a})/\mathfrak {a}= \sum _{i,j\in \mathrm{Ob}( \mathcal{C})} e_iIe_j\subseteq \overline{ \mathcal{C}(\sigma )}$. Notice that $\overline{I}^n=0$ since $e_iIe_i=0$ for all $i\in \mathrm{Ob}( \mathcal{C})$. Now,

$$\begin{aligned} I^{n+1}\subseteq (\overline{I}+\mathfrak {a})^{n+1} \subseteq \overline{I}^{n+1}+\mathfrak {a}\overline{I}^n =0 \end{aligned}$$

since $\mathfrak {a}^2=0$ and $\mathcal{C}(\sigma )\mathfrak {a}=0$ (Lemma 2.2). $\square$

Recall that a ring is a semiprime ring if the zero ideal is the only nilpotent ideal.

Theorem 2.8

(Criterion for $\mathcal{C}(\sigma )$ to be a semiprime algebra) Suppose that $n:=|\mathrm{Ob}( \mathcal{C})|<\infty$. Then the following statements are equivalent.

• 1. The algebra $\mathcal{C}(\sigma )$ is a semiprime algebra.

• 2. The algebras $\mathcal{C}(\sigma )_{ii}$ are semiprime where $i\in \mathrm{Ob}( \mathcal{C})$ and, for all distinct $i,j\in \mathrm{Ob} ( \mathcal{C})$, $a_{ij} \mathcal{C}(\sigma )_{ji}\ne 0$ and $\mathcal{C}(\sigma )_{ji}a_{ij}\ne 0$ for all nonzero elements $a_{ij}\in \mathcal{C}(\sigma )_{ij}$.

• 3. The algebras $\mathcal{C}(\sigma )_{ii}$ are semiprime where $i\in \mathrm{Ob}( \mathcal{C})$ and each ideal I of $\mathcal{C}(\sigma )$ such that $e_iIe_i=0$ for all $i\in \mathrm{Ob}( \mathcal{C})$ is equal to zero.

Proof

Since $|\mathrm{Ob}( \mathcal{C})|<\infty$, the direct product of algebras $\mathcal{D}:= \prod _{i\in \mathrm{Ob}( \mathcal{C})} \mathcal{C}(\sigma )_{ii}$ is a semiprime algebra iff all the algebras $\mathcal{C}(\sigma )_{ii}$ are semiprime.

$(1\Rightarrow 2)$ If $\mathfrak {b}$ is a nonzero nilpotent ideal of the ring $\mathcal{D}$ and $(\mathfrak {b}) = \mathcal{C}(\sigma )\mathfrak {b} \mathcal{C}(\sigma )$ is the ideal of $\mathcal{C}(\sigma )$ generated by $\mathfrak {b}$ then

$$\begin{aligned} (\mathfrak {b})^k \subseteq (\mathfrak {b}^{\lfloor \frac{k}{n^2}\rfloor })\;\; \mathrm{for\; all}\;\; k\ge 1 \end{aligned}$$

where for a real number r, $\lfloor r\rfloor :=\max \{ z\in \mathbb {Z}\, | \, z\le r\}$, and so the ideal $(\mathfrak {b})$ of the algebra $\mathcal{D}$ is a nilpotent ideal. Therefore, the ring $\mathcal{C}(\sigma )_{ii}$ must be semiprime for all $i\in \mathrm{Ob} ( \mathcal{C})$.

Suppose that there exists a nonzero element $a_{ij}\in \mathcal{C}(\sigma )_{ij}$ for some distinct objects i and j such that either $a_{ij} \mathcal{C}(\sigma )_{ji}=0$ or $\mathcal{C}(\sigma )_{ji}a_{ij}=0$. Then $(a_{ij})^2= (a_{ij} \mathcal{C}(\sigma )_{ji}a_{ij})=0$, a contradiction.

$(2\Rightarrow 1)$ Since all rings $\mathcal{C}(\sigma )_{ii}$ are semiprime, the ideal $\mathfrak {a}$ is equal to zero, by Lemma 2.2. Therefore, if J is a nilpotent ideal of $\mathcal{C}(\sigma )$ then necessarily $J=\bigoplus _{i,j\in \mathrm{Ob} ( \mathcal{C})}J_{ij}$ where $J_{ij}=e_iJe_j$. Furthermore, all $J_{ii}=0$ since the rings $\mathcal{C}(\sigma )_{ii}$ are semiprime (and $J_{ii}^m\subseteq J^m$ for all $m\ge 1$). Suppose that $J\ne 0$. We seek a contradiction. Then $J_{ij}\ne 0$ for some $i\ne j$. Then, by the assumption, either $\mathcal{C}(\sigma )_{ji}J_{ij}$ is a nonzero nilpotent ideal of the algebra $\mathcal{C}(\sigma )_{jj}$ or $J_{ij} \mathcal{C}(\sigma )_{ji}$ is a nonzero nilpotent ideal of the algebra $\mathcal{C}(\sigma )_{ii}$, a contradiction.

$(1\Rightarrow 3)$ The algebras $\mathcal{C}(\sigma )_{ii}$ are semiprime for all $i\in \mathrm{Ob}( \mathcal{C})$, by the implication $(1\Rightarrow 2)$. By Lemma 2.7, each ideal I of $\mathcal{C}(\sigma )$ such that $e_iIe_i=0$ for all $i\in \mathrm{Ob}( \mathcal{C})$ is a nilpotent ideal, so it must be zero (since $\mathcal{C}(\sigma )$ is a semiprime ring).

$(3\Rightarrow 1)$ If I is a nilpotent ideal of $\mathcal{C}(\sigma )$ then for each $i\in \mathrm{Ob}( \mathcal{C})$, $I_{ii}$ is a nilpotent ideals of the semiprime ring $\mathcal{C}(\sigma )_{ii}$, and so $I_{ii}=0$. Then, we must have $I=0$, by the second assumption of statement 3. $\square$

Theorem 2.9

(Simplicity criterion for $\mathcal{C}(\sigma )$) The algebra $\mathcal{C}(\sigma )$ is a simple algebra iff the following conditions hold

• 1. $\mathfrak {a}=0$,

• 2. for every $i\in \mathrm{Ob} ( \mathcal{C})$, the ring $\mathcal{C}(\sigma )_{ii}$ is simple,

• 3. for all distinct $i,j\in \mathrm{Ob} ( \mathcal{C})$, $\mathcal{C}(\sigma )_{ij}$ is a simple $( \mathcal{C}(\sigma )_{ii}, \mathcal{C}(\sigma )_{jj})$-bimodule (in particular, $\mathcal{C}(\sigma )_{ij}\ne 0$), and

• 4. $\mathcal{C}(\sigma )_{ij} \mathcal{C}(\sigma )_{jk}\ne 0$ for all $i,j,k\in \mathrm{Ob} ( \mathcal{C})$.

Proof

$(\Rightarrow )$ Let $\mathcal{C}_{ij}= \mathcal{C}(\sigma )_{ij}$.

• (i) $\mathfrak {a}=0$, by Lemma 2.2.

• (ii) For every$i\in \mathrm{Ob} ( \mathcal{C})$, $\mathcal{C}_{ii}$is a simple ring: Suppose that $\mathfrak {b}$ is a proper ideal of the ring $\mathcal{C}_{ii}$ then $(\mathfrak {b})$ is a proper ideal of $\mathcal{C}(\sigma )$ since $(\mathfrak {b}) \cap \mathcal{C}_{ii}= \mathfrak {b}$, a contradiction.

• (iii) For all distinct objects$i,j\in \mathrm{Ob} ( \mathcal{C})$, $\mathcal{C}_{ij}\ne 0$: Suppose that $\mathcal{C}_{ij}=0$ for some distinct objects i and j. Then the ideal $( \mathcal{C}_{ii})$ of $\mathcal{C}(\sigma )$ is a proper ideal since $( \mathcal{C}_{ii})\cap \mathcal{C}_{jj}= \mathcal{C}_{ji} \mathcal{C}_{ii} \mathcal{C}_{ij}=0$, a contradiction.

• (iv) For all distinct objects$i,j\in \mathrm{Ob} ( \mathcal{C})$, $\mathcal{C}_{ij}$is a simple$( \mathcal{C}_{ii}, \mathcal{C}_{jj})$-bimodule: Suppose that $\mathfrak {b}$ is a proper $( \mathcal{C}_{ii}, \mathcal{C}_{jj})$-sub-bimodule of $\mathcal{C}_{ij}$ then $(\mathfrak {b})$ is a proper ideal of the algebra $\mathcal{C}(\sigma )$ since $(\mathfrak {b}) \cap \mathcal{C}_{ij}=\mathfrak {b}$, a contradiction.

• (v) $\mathcal{C}_{ij} \mathcal{C}_{jk}\ne 0$for all objects$i,j,k\in \mathrm{Ob} ( \mathcal{C})$: The statement (v) holds in the following cases $i=j=k$ (by (ii)), $i=j$ or $j=k$ (by (iii)). Suppose that $i=k$ and $\mathcal{C}_{ij} \mathcal{C}_{ji}=0$, we seek a contradiction. Then the ideal $( \mathcal{C}_{ij})$ of $\mathcal{C}(\sigma )$ is a proper ideal since $( \mathcal{C}_{ij})\cap \mathcal{C}_{ii}= \mathcal{C}_{ij} \mathcal{C}_{ji}=0$, a contradiction. Suppose that $\mathcal{C}_{ij} \mathcal{C}_{jk}=0$ for some distinct i, j and k. Then the ideal $( \mathcal{C}_{ij})$ of $\mathcal{C}(\sigma )$ is a proper ideal since $( \mathcal{C}_{ij})\cap \mathcal{C}_{kk}= \mathcal{C}_{ki} \mathcal{C}_{ij} \mathcal{C}_{jk}=0$, a contradiction.

$(\Leftarrow )$ Suppose that conditions 1–4 hold. By conditions 1–3, condition 4 can be replaced by condition $4'$: $\mathcal{C}_{ij} \mathcal{C}_{jk}= \mathcal{C}_{ik}$ for all $i,j,k\in \mathrm{Ob} ( \mathcal{C})$. Let J be a nonzero ideal of $\mathcal{C}(\sigma )$. We have to show that $J= \mathcal{C}(\sigma )$. By condition 1, $e_iJe_j\ne 0$ for some i and j. By conditions 2 and 3, $J_{ij}=J\cap \mathcal{C}_{ij}= \mathcal{C}_{ij}$. By condition $4'$, $\mathcal{C}_{st}= \mathcal{C}_{si} \mathcal{C}_{ij} \mathcal{C}_{jt}\subseteq J$ for all s, t. This means that $J= \mathcal{C}(\sigma )$, as required. $\square$