Proof. We prove the direction which is used later on and leave the other one to the interested reader. So assume the poset is Noetherian. First observe that for every infinite sequence $(x_i)_{i\in \mathbb N}$ of elements, there exists $i\in \mathbb N$ such that $x_j\le x_i$ for infinitely many $j\in \mathbb N$ . This follows by looking at the ascending chain of ideals generated by the elements $x_0,\ldots ,x_n$ , which must stabilize. Now define $\nu \colon \mathbb N\to \mathbb N$ recursively by taking for $\nu (0)$ the smallest $i\in \mathbb N$ such that $x_j\le x_i$ for infinitely many $j\in \mathbb N$ . For $n> 0,$ set

$$\begin{align*}\nu(n)=\mathrm{min}\{i> \nu(n-1)\mid x_j\le x_i\le x_{\nu(n-1)} \text{ for infinitely many }j\in\mathbb N\}.\\[-34pt] \end{align*}$$

Proof. (1) $\Rightarrow $ (2): Let $\phi \in Q(x)$ . The ideal of $Q(x)$ that is generated by all paths $\alpha \phi $ with $\alpha $ an arrow starting at $t(\phi )$ is finitely generated. It follows that there are only finitely many arrows starting at $t(\phi )$ . Suppose $\phi _1,\phi _2,\ldots $ is an infinite sequence of paths such that there is more than one arrow starting at each $t(\phi _i)$ . Because $Q(x)$ is Noetherian, we may assume $\phi _{i+1}\le \phi _{i}$ for all i. For each $i,$ we may choose an arrow $\alpha _i$ such that $\phi _{i+1}\not \le \alpha _i\phi _i $ . Then the ideal of $Q(x)$ generated by $\{ \alpha _i\phi _i\mid i\in \mathbb N\}$ cannot be generated by finitely many elements. This is a contradiction.

(2) $\Rightarrow $ (3): Let n be the number of paths $\phi \in Q(x)$ such that there is more than one arrow starting at $t(\phi )$ , and let d be the maximal number of such arrows. Then there are at most $d^n$ maximal paths starting at x.

(3) $\Rightarrow $ (1): Let $I\subseteq Q(x)$ be an ideal, so a left ideal with respect to the composition of paths in Q. For each maximal path starting at x and given by a (finite or infinite) sequence of arrows $\ldots \alpha _2\alpha _1$ , we check if there exists $n\in \mathbb N$ such that $\alpha _n\ldots \alpha _2\alpha _1$ belongs to I. In that case, we choose n minimal, and it is easily checked that I is generated by all paths of this form.

Proof. Suppose the poset $Q(x)$ is Noetherian. Then Proposition 2.3 implies that there are only finitely many maximal paths starting at x. From this finite set of paths, the decomposition (2.1) is constructed as follows. Set $Q^0=Q_x$ when $Q_x$ is finite. Otherwise, there exists a maximal path $\ldots \beta _2\beta _1$ in $Q(x)$ and $r\ge 1$ such that $Q_x\smallsetminus \{\beta _r\}=Q'\amalg Q^1$ with $Q^1$ of type $A_\infty $ and set of arrows $\{\beta _i\mid i>r\}$ . Set $Q^0=Q'$ if this quiver is finite. Otherwise, there exists another maximal path in $Q'$ and we continue as before by removing finitely many arrows. It is straightforward to check that such a decomposition of $Q_x$ is equivalent to the condition on $Q(x)$ to have only finitely many maximal paths starting at x. So the assertion follows from Proposition 2.3.

Proof. We apply Proposition 2.4 and note that, in the decomposition (2.1), we have $Q_x=Q^0$ , since for $i\neq 0,$ the quiver $Q^i$ has infinitely many vertices.

Proof. Suppose that $\mathcal A$ is locally Noetherian. Then we may take the Noetherian objects as the collection of generators, because for a morphism $\phi \colon X\to Y$ with $X=\sum _i X_i$ written as the directed union of its Noetherian subobjects, we have

$$\begin{align*}\mathrm{Hom}(X,Y)\cong\lim_i \mathrm{Hom}(X_i,Y).\end{align*}$$

If $\phi \neq 0$ , then one of the composites $X_i\hookrightarrow X\to Y$ needs to be non-zero.

Now suppose there is a collection of Noetherian generators $(G_i)_{i\in I}$ . Let X be an object and $X'\subseteq X$ the directed union of its Noetherian subobjects. It is clear that every composite $G_i\to X\twoheadrightarrow X/X'$ is zero. Thus, $X'=X$ .

Proof. Combine (3.1) with the fact that $\mathrm {Hom}(-,-)$ preserves colimits in the first argument.

Proof. Let $X=\sum _i X_i$ and $Y=\sum _j Y_j$ be objects in $\mathcal A$ , written as directed union of their Noetherian subobjects. Using (3.1), we have

$$\begin{align*}h_X\cong\underset{i}{\mathrm{colim}}\ \mathrm{Hom}(-,X_i)\end{align*}$$

and compute

$$ \begin{align*} \mathrm{Hom}(h_X,h_Y)&\cong\mathrm{Hom}(\underset{i}{\mathrm{colim}}\ \mathrm{Hom}(-,X_i),\underset{j}{\mathrm{colim}}\ \mathrm{Hom}(-,Y_j))\\ &\cong\lim_i\mathrm{Hom}(\mathrm{Hom}(-,X_i),\underset{j}{\mathrm{colim}}\ \mathrm{Hom}(-,Y_j))\\ &\cong \lim_i\ \underset{j}{\mathrm{colim}}\ \mathrm{Hom}(X_i,Y_j)\\ &\cong\mathrm{Hom}(X,Y). \end{align*} $$

The second isomorphism uses that $\mathrm {Hom}(-,-)$ preserves colimits in the first argument, the third follows from Yoneda’s lemma, and the last is (3.2). Thus, $\mathcal A\to \bar {\mathcal A}$ is fully faithful. For the rest of the assertion, see [Reference Gabriel6, Section II.4].

Proof. This follows from the adjunction isomorphism (4.1).

Proof. Let $M[Q(x,-)]=\sum _{i\in I}N_i$ be a directed union of subobjects $N_i$ . Then the adjunction isomorphism (4.1) yields a morphism $M\to \sum _{i\in I}N_i(x)$ , which factors through $N_{i_0}(x)\to \sum _{i\in I}N_i(x)$ for some $i_0\in I$ since M is finitely generated. Thus, the identity $M[Q(x,-)]\to \sum _{i\in I}N_i$ factors through $N_{i_0}\to \sum _{i\in I}N_i$ , so $M[Q(x,-)]=N_{i_0}$ .

Proof. Let $U\subseteq M[Q(x,-)]$ be a subrepresentation and $\phi \in Q(x,y)$ a path. We set

$$\begin{align*}Q(x,y)_\phi:=\{\psi\in Q(x,y)\mid \phi\preceq \psi\}\end{align*}$$

and write $U_\phi $ for the image of the composite

$$\begin{align*}U(y)\cap M[Q(x,y)_\phi]\hookrightarrow M[Q(x,y)_\phi]\twoheadrightarrow M,\end{align*}$$

where the second map denotes the projection corresponding to $\phi $ . For $\psi \in Q(x,y)$ , we have

(4.2) $$ \begin{align} \phi \leq \psi\quad\implies\quad U_\psi\subseteq U_\phi. \end{align} $$

More precisely, if there is a path $\chi $ such that $\phi =\chi \psi $ , then composition with $\chi $ induces a map

$$\begin{align*}Q(x,y)_\phi\longrightarrow Q(x,y)_\psi\end{align*}$$

such that $\phi $ is the only element that is mapped to $\psi $ . Thus, $U(\chi )$ maps $U_\psi $ into $U_\phi $ .

Given subrepresentations $U\subseteq V\subseteq M[Q(x,-)]$ , we have $U_\phi \subseteq V_\phi $ for all $\phi \in Q(x)$ . If $U\neq V$ , we claim that there exists $\phi \in Q(x)$ such that $U_\phi \neq V_\phi $ . To see this, choose $y\in Q_0$ such that $U(y)\neq V(y)$ . Observe that the union

(4.3) $$ \begin{align} Q(x,y)=\bigcup_{\phi\in Q(x,y)}Q(x,y)_\phi \end{align} $$

is directed, and therefore condition (AB5) implies

$$\begin{align*}U(y)=\sum_{\phi\in Q(x,y)}U(y)\cap M[Q(x,y)_\phi].\end{align*}$$

Thus, there exists $\phi \in Q(x,y)$ with

$$\begin{align*}U(y)\cap M[Q(x,y)_\phi]\neq V(y)\cap M[Q(x,y)_\phi].\end{align*}$$

Using that $Q(x)$ is Noetherian, we may choose $\phi $ maximal with respect to $\preceq $ . Next observe that

(4.4) $$ \begin{align} Q(x,y)_\phi\smallsetminus\{\phi\}=\bigcup_{\phi\prec\psi}Q(x,y)_\psi \end{align} $$

is a directed union. This yields the following commutative diagram with exact rows, using again the condition (AB5):

For $\phi \prec \psi ,$ the maximality of $\phi $ implies

$$\begin{align*}U(y)\cap M[Q(x,y)_\psi]= V(y)\cap M[Q(x,y)_\psi]\end{align*}$$

and therefore $U_\phi \neq V_\phi $ .

Now suppose there is a proper ascending chain

$$\begin{align*}U^1\subseteq U^2 \subseteq U^3\subseteq \cdots\end{align*}$$

of subrepresentations of $M[Q(x,-)]$ . Pick for each $i\in \mathbb N$ a path $\phi _i$ such that $U^i_{\phi _i}\subsetneq U^{i+1}_{\phi _i}$ . The set $\{\phi _i\mid i\in \mathbb N\}$ is infinite since M is Noetherian. So we may assume $\phi _i\neq \phi _j $ for all $i\neq j$ in $\mathbb N$ . Because $Q(x)$ is Noetherian, there exists a map $\nu \colon \mathbb N\to \mathbb N$ such that $i<j$ implies $\nu (i)<\nu (j)$ and $\phi _{\nu (j)}\leq \phi _{\nu (i)}$ (see Lemma 2.1). Thus, for all $i\in \mathbb N,$ we have

$$\begin{align*}U^{\nu(i)}_{\phi_{\nu(i)}}\subsetneq U^{\nu(i)+1}_{\phi_{\nu(i)}} \subseteq U^{\nu(i+1)}_{\phi_{\nu(i)}}\subseteq U^{\nu(i+1)}_{\phi_{\nu(i+1)}},\end{align*}$$

where the last inclusion follows from (4.2). This infinite chain is impossible since M is Noetherian, and therefore the representation $M[Q(x,-)]$ is Noetherian.

Proof. (1) Any inclusion of sets $X\subseteq Y$ induces an inclusion $A[X]\subseteq A[Y]$ of A-modules, with equality $A[X]=A[Y]$ only if $X=Y$ .

(2) Any inclusion of A-modules $M\subseteq N$ induces an inclusion $M[Q(x,-)]\subseteq N[Q(x,-)]$ , with equality only if $M=N$ .

Proof of Theorem 1.1.

(1) $\Rightarrow $ (2): Let $M\in \mathrm {Rep}(Q,\mathcal A)$ . We need to show that M is the directed union of its Noetherian subobjects. We may assume that $\mathcal A$ is essentially small, because it suffices to consider the full subcategory of $\mathcal A$ that is obtained by closing the set of objects $M(x)$ with $x\in Q_0$ under finite coproducts, subobjects, and quotients. Let $\bar {\mathcal A}$ denote the completion of $\mathcal A$ , as in Proposition 3.4. The representations $N[Q(x,-)]$ with $N\in \mathrm {Noeth}\bar {\mathcal A}$ and $x\in Q_0$ are Noetherian and generate $\mathrm {Rep}(Q,\bar {\mathcal A})$ by Lemmas 4.1 and 4.3. Thus, $\mathrm {Rep}(Q,\bar {\mathcal A})$ is locally Noetherian (see Lemma 3.2). It follows that M is the directed union of its Noetherian subobjects, because we can view $\mathrm {Rep}(Q,\mathcal A)$ as a full subcategory which is closed under subobjects.

(2) $\Rightarrow $ (3): This is clear since $\mathrm {Mod} A$ is locally Noetherian for every left Noetherian ring A (see Example 3.5).

(3) $\Rightarrow $ (1): Fix $x\in Q_0$ . The regular module A is a finitely generated object in $\mathrm {Mod} A$ , and therefore the representation $A[Q(x,-)]$ is finitely generated in $\mathrm {Rep}(Q,\mathrm {Mod} A)$ by Lemma 4.2. It follows that $A[Q(x,-)]$ is Noetherian since $\mathrm {Rep}(Q,\mathrm {Mod} A)$ is locally Noetherian, and we conclude from Lemma 4.4 that the poset $Q(x)$ is Noetherian.

Proof. We give the argument for the implication that is not covered by Theorem 1.1. So suppose that $\mathrm {Rep}(Q,\mathrm {Mod} A)$ is locally Noetherian. As in the above proof of (3) $\Rightarrow $ (1), we deduce from the first part of Lemma 4.4 that Q is left Noetherian. Using the second part of that lemma, it follows that the ring A is left Noetherian.

Proof. The assertion follows from Theorem 1.1 and Corollary 2.5, because of the equivalence (5.1).

Proof. Recall that a ring is left artinian if and only if its category of left modules is locally finite [Reference Stenström12, Section VIII.1]. On the other hand, the quiver Q is left finite if and only if both chain conditions hold for the ideals of $Q(x)$ for all vertices x. Thus, we can adapt the proof of Theorem 1.1.

Proof. The special case that $\mathcal C$ is the path category of a quiver is Theorem 1.1. In fact, the Gröbner enrichment for a path category is specified right before Lemma 4.3. The proof of the general case is essentially the same. The crucial ingredient is Lemma 4.3 where the Gröbner enrichment is used; its definition is extracted from the proof of that lemma. For instance, condition (G1) is needed for (4.2), and condition (G2) ensures that in (4.3) and (4.4) the union is directed.