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For a finite acyclic quiver, the arrow ideal of a path algebra $R$ is nilpotent, and $A/R\cong {\prod }_{i\in Q_0}k$. If $S$ is a simple module, its submodule $RS$ is either zero or $S$. The latter would imply $R^{d}S=S$ for every $d$, contradicting nilpotence. Thus $RS=0$, and a simple module over the product of fields is supported at one coordinate. Therefore the simples are exactly $S(i)$, with $k$ at $i$, zero elsewhere, and zero arrows; the vertex $i$ is unique.

A finite-dimensional semisimple module is consequently ${⨁}_{i}S(i{)}^{\oplus n_i}$, where $n_i=\dim X_i$. Its dimension vector of a quiver representation determines its isomorphism class.

For an arbitrary finite quiver, cycles allowed, the vertex projective module of a path algebra is $P(i)=A{e}_i$. Its space at vertex $j$ has basis all paths from $i$ to $j$, and an arrow acts by adjoining that arrow at the end of the path. Its endomorphism ring iswhere $e_{i}A{e}_i$ is spanned by the closed paths based at $i$. The opposite multiplication appears because endomorphisms act by right multiplication.

The evaluation isomorphism for a vertex projective isFor $x\in X_i$, its inverse sends a path $p$ starting at $i$ to $px$. This proves both injectivity and surjectivity, and is natural in $X$. Vertex evaluation is exact, so $P(i)$ is a projective module; alternatively it is a direct summand of the free module $A$.

The closed-path corner of a path algebra is a domain: in a product of two nonzero linear combinations, choose their longest path lengths. Concatenation in that top degree has a unique cut at those lengths, so a product of two nonzero top coefficients cannot cancel. Hence its only idempotents are zero and one. The same is true of the opposite ring, proving $P(i)$ is an indecomposable module, even when cycles make it infinite-dimensional.

For $1\to 2\leftarrow 3$, the paths starting at vertex $1$ are $e_1$ and the arrow $1\to 2$. Thus the displayed $k\stackrel{1}{\to }k\leftarrow 0$ is $P(1)$ and is projective.

A nonzero finite-dimensional representation is a brick module when its endomorphism ring is a division algebra. Over the algebraically closed field $k$, this means ${\mathrm{End}}_Q(X)=k$: for any endomorphism $f$, an eigenvalue $\lambda$ makes $f-\lambda I$ noninvertible, hence zero in a division algebra.

For a counterexample to the converse of “brick implies indecomposable”, take the one-loop representation $k^2$ with nilpotent Jordan block $N=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$. Its endomorphism ring is $k[N]\cong k[t]/(t^2)$, a local endomorphism ring of dimension two. It has no nontrivial idempotents, so the module is indecomposable, but it is not a brick.

For the one-arrow quiver, splitting the kernel, image and target complement decomposes any representation into copies of $k\to 0$, $0\to k$ and $k\stackrel{1}{\to }k$. Each has endomorphism ring $k$. Therefore every indecomposable of the one-arrow quiver is a brick.

Rows in the following matrix index the source, columns the target, in the order $X,Y,Z$. Solving the arrow-square equation for a quiver representation morphism givesIn particular, the nonzero morphisms between distinct representations are $X\to Z$ and $Z\to Y$, each a one-dimensional family of scalar multiples of the vertexwise inclusion or projection. A map $Y\to Z$ is forced to vanish at the source by the identity arrow of $Z$; similarly a map $Z\to X$ is forced to vanish at the target. Maps between $X$ and $Y$ are zero. Each endomorphism ring is $k$.

The extension complex of quiver representations for the one-arrow quiver isIts cokernel is ${\mathrm{Ext}}_Q^1(U,V)$. Substitution of the three representations givesHere the first argument is the quotient endpoint of a short exact sequence. Thus the only possible nonsplit endpoint pair is subobject $X$, quotient $Y$. The sequenceis nonsplit, since $Z$ is indecomposable. More explicitly, all extensions with these endpoints have a middle arrow $k\stackrel{\lambda }{\to }k$: $\lambda =0$ gives the split short exact sequence, while each $\lambda \ne 0$ gives a middle representation isomorphic to $Z$. With endpoint identifications fixed the extension classes form $k$; up to endpoint automorphisms all nonzero classes give this same nonsplit sequence. There are no other nonsplit sequences with the listed endpoints, including equal endpoints.

First split off the kernel of each source-to-sink arrow. Each kernel contributes copies of the simple representation supported at that source. The remaining arrows are injective, so identify the three source spaces with subspaces $U_1,U_2,U_3$ of the sink space $V$. We now give an elementary three-subspace decomposition.

Split off the common intersection $U_1\cap U_2\cap U_3$: any projection onto it preserves all three subspaces. Next split off each pairwise intersection. For example, once the triple intersection has gone, $W=U_1\cap U_2$ is disjoint from $U_3$; a projection onto $W$ that kills $U_3$ preserves $U_1,U_2,U_3$. This extracts blocks whose membership is exactly $\{1,2\}$. Repeating for the other pairs leaves pairwise disjoint subspaces.

For each $i$, choose a complement $W_i$ to $U_i\cap (U_j+U_{\ell })$ in $U_i$. A projection onto $W_i$ killing $U_j+U_{\ell }$ preserves the triple, so these complements split off as blocks of membership exactly $\{i\}$. In the remaining triple, every $U_i$ lies in the sum of the other two and all pairwise intersections are zero. Thus the sum is $U_1\oplus U_2$, and both projections of $U_3$ onto $U_1,U_2$ are isomorphisms. Hence $U_3$ is the graph of an isomorphism $T:U_1\to U_2$. Choose a basis $u_a$ of $U_1$ and the basis $T{u}_a$ of $U_2$; this graph decomposes into two-dimensional blocks with three distinct lines. A complement to $U_1+U_2+U_3$ in $V$ contributes sink-only simple blocks.

The complete list of indecomposable blocks is therefore:

Each of the first eleven blocks has endomorphism ring $k$. For the exceptional indecomposable of the three-subspace quiver, an endomorphism of $k^2$ preserving the first two lines is diagonal; preserving the third makes its two diagonal entries equal. Its endomorphism ring is also $k$, so all twelve are indecomposable. Their dimension vectors of quiver representations distinguish them. The decomposition argument proves completeness, and hence there are exactly $12$ isomorphism classes. No general classification theorem is needed.

A nonzero finite-dimensional quiver representation is a brick module when ${\mathrm{End}}_Q(X)=k{1}_X$. If $X=U\oplus V$ with both summands nonzero, projection onto $U$ is an idempotent endomorphism different from zero and the identity. That is impossible in $k{1}_X$. Hence every brick is indecomposable.

For the requested converse counterexample, take the dual-number algebra $A=k[\epsilon ]/({\epsilon }^2)$ and its left regular module $A$. The algebra is local with maximal ideal $(\epsilon )$, and it is not simple because this is a nonzero proper ideal. The module is indecomposable: its endomorphism ring is $A^{\mathrm{op}}\cong A$, which has no nontrivial idempotents. But this ring has dimension two over $k$, so the regular module is not a brick. Thus indecomposable need not mean brick without the positive-definiteness hypothesis used below.

If $P$ is a prime ideal of the commutative Artinian ring $R$, then $R/P$ is an Artinian domain. For $0\ne x\in R/P$, the descending chain $(x)\supseteq (x^2)\supseteq \cdots$ stabilizes, so $x^n=x^{n+1}a$ for some $n$; cancellation gives $xa=1$. Thus $R/P$ is a field and $P$ is maximal.

There are only finitely many maximal ideals. Otherwise, distinct maximal ideals $P_i$ are pairwise comaximal and their finite products give the strictly descending chaincontradicting the Artinian condition. This proves the statement about the prime ideals of a commutative Artinian ring.

The nilradical $N$ is the ideal of all nilpotent elements, equivalently the intersection of all prime ideals. Its powers stabilize, say $N^r=N^{r+1}=I$. If $I\ne 0$, choose an ideal $J$ minimal subject to $IJ\ne 0$, then choose $x\in J$ with $Ix\ne 0$. Since $I^2=I$, minimality gives $Rx=Ix$, so $x=ax$ for some $a\in I$. But $a$ is nilpotent, hence $1-a$ is a unit, contradicting $(1-a)x=0$. Therefore the nilradical of a commutative Artinian ring is nilpotent.

If the maximal ideals are $P_1,\dots ,P_s$, the Chinese remainder theorem givesa finite product of fields. Each layer $N^j/N^{j+1}$ is an Artinian module over this product and therefore finite-dimensional. Since $N$ is nilpotent, these finitely many layers show that every ideal of $R$ is finitely generated. This proves the Artinian commutative ring is Noetherian theorem.

Now assume $R$ is local with maximal ideal $P$ and ${\dim }_{R/P}P/P^2=1$. The Nakayama lemma gives $P=(x)$. Since $P$ is nilpotent, its powers form a finite chain ending in zero. For a nonzero ideal $J$, choose the largest $r$ with $J\subseteq (x^r)$ and take $y\in J\setminus (x^{r+1})$. Write $y=x^{r}u$; then $u\notin P$, so $u$ is a unit. Henceand every ideal of $R$ is principal.

Finally, the Artin–Wedderburn theorem says that a finite-dimensional semisimple $k$-algebra has the formwhere each $D_i$ is a finite-dimensional division ring over $k$; the factors are unique up to permutation and isomorphism. By the assumed complete reducibility, write the right regular module aswith pairwise nonisomorphic simple right modules $S_i$. The Schur lemma says that $D_i={\mathrm{End}}_T(S_i)$ is a division ring and that homomorphisms between distinct $S_i$ vanish. Left multiplication and the endomorphism ring of the displayed direct sum therefore giveConversely, the regular module of $M_n(D)$ is a direct sum of $n$ simple column modules, proving that every algebra on the right is semisimple. Uniqueness follows from uniqueness of the simple summands and their multiplicities.

No. Let $Q$ be the quiver $1\to 2$ over a field $k$, and take the representation of a quiverAn endomorphism is a pair of scalar maps $(a,b)$ satisfying $b=a$, so its endomorphism ring is $k$. Every nonzero endomorphism is therefore an isomorphism, making this representation a brick module. It nevertheless has the proper nonzero subrepresentation $0\to k$, so it is not an irreducible module.

Equivalently, this is a nonsimple module over the path algebra $kQ$ whose endomorphism ring is a division ring.

For a nonzero finitely generated indecomposable module $P$ that is a projective module over a right Artinian ring $A$, its top $P/PJ(A)$ is a simple module. The Jacobson radical $J(A)$ is a nilpotent ideal, so the top is nonzero; the quotient is a semisimple module over the semisimple ring $A/J(A)$. If it decomposed, a nontrivial idempotent of its endomorphism ring would lift by projectivity to an endomorphism of $P$. The Hopkins-Levitzki theorem gives finite composition length, and the Fitting lemma makes the lift either invertible or nilpotent. Its induced map on the top would have the same property, impossible for that nontrivial idempotent.

The vertex projective module of a path algebra $P(i)=kQ{e}_i$ has at vertex $j$ the span of paths from $i$ to $j$. Each arrow appends itself to such paths. It is a projective module because it is a summand of the free module $kQ$. Its endomorphism ring is $(e_{i}kQ{e}_i{)}^{\mathrm{op}}$, acting by right multiplication. The closed-path corner of a path algebra is a domain, so this ring has no nontrivial idempotents and $P(i)$ is indecomposable. Directed cycles may make these projectives infinite-dimensional.