Classification of indecomposable modules over a given ring

Question

Let $K$ be a field, $x$ an indeterminate and $n$ a positive integer.

How can one classify all the indecomposable modules up to isomorphism over the ring $R=K[x]/(x^n)$ ?

Answer 1

As pointed out by user26857, we know by the structure theorem for finitely generated modules over a principal ideal domain that any finitely generated indecomposable module over $R$ is isomorphic to $K[x]/(x^i)$ for some $i\in \{1,\ldots, n\}$.

By using the axiom of choice, we can prove that any indecomposable $R$-module is finitely generated, thus proving that the above classify all indecomposable modules.

Let $M$ be an indecomposable $R$-module. It is useful to note that all elements of $R$ have the form $\lambda x^i$, with $\lambda$ invertible. The annihilator of $M$ in $R$ has the form $(x^m)$ for some $m\in\{1, \ldots, n\}$. Let $z\in M$ be such that $x^{m-1}z \neq 0$. Let us prove that $z$ generates $M$.

Let $X$ be the set of submodules of $M$ which have trivial intersection with $Rz$. Then any chain in $X$ has an upper bound, namely the union of the submodules of that chain. By Zorn's lemma, $X$ admits a maximal element $V$.

Let us now prove that $M=Rz\oplus V$. Assume the contrary, and let $w\in M\setminus (Rz\oplus V)$.

If $Rw\cap(Rz\oplus V)$ is trivial, then $(Rw+V)\cap Rz$ is trivial as well, contradicting the maximality of $V$.

Thus $Rw\cap(Rz\oplus V)$ is non-trivial; let $p$ be the smallest integer such that there exists a non-zero element $x^pw = \lambda x^qz+v$, with $\lambda\in R$ invertible and $v\in V$. Note that $p\leq q$, since $0=x^{m-p}x^pw=\lambda x^{m-p+q}z + x^{m-p}v = 0$ implies that $x^{m-p+q}z=0$, which in turn implies that $m-p+q \geq m$.

Consider the element $w-\lambda x^{q-p}z$. The claim is that $R(w-\lambda x^{q-p}z)+V$ has trivial intersection with $Rz$. Indeed, let $x^t(w-\lambda x^{q-p}z) + u = \mu x^s z$ be an element of this intersection, with $u\in V$ and $\mu\in R$ invertible. Then $x^tw \in Rz\oplus V$, so $t\geq p$ by minimality of $p$. Thus $x^t(w-\lambda x^{q-p}z) + u = \mu x^s z$ is both an element of $V$ and of $Rz$, and so it is zero.

Therefore $R(w-\lambda x^{q-p}z)+V$ has trivial intersection with $Rz$, and by maximality of $V$, we have that $w-\lambda x^{q-p}z \in V$. Hence $w\in Rz\oplus V$, contradicting the assumption that $w\in M \setminus (Rz\oplus V)$.

Hence $M= Rz\oplus V$, and since $M$ is indecomposable, $V=0$. We have proved that any indecomposable module over $R$ is generated by one element, and thus that the list of indecomposable modules at the top of this post is the complete list of indecomposable $R$-modules.

Answer 2

$R=k[x]/(x^n)$ is an example of what is called a Nakayama $k$-algebra. These algebras are characterized by the fact that indecomposable projective modules and indecomposable injective modules are uniserial modules (a module $M$ is called uniserial if for any two submodules $A$ and $B$ of $M$ either $A\subseteq B$ or $B\subseteq A$; that is, submodules form a chain).

Nakayama algebras are artinian rings of finite representation type (i.e. there are up to isomorphism only finitely many indecomposable modules), so every indecomposable module is automatically finitely generated.

In our case, $R$ is a commutative local artinian ring with unique maximal ideal $J=(x)/(x^n)$. Hence, $R_R$ is indecomposable. Therefore, since $R_R$ is always a projective module, it must be a uniserial module by the characterization above for Nakayama algebras. We say in this case that $R$ is a commutative artinian uniserial ring.

Over commutative artinian uniserial rings every indecomposable module is uniserial (indeed, the same is true over any non-necessarily commutative Nakayama algebra). We notice now that every uniserial module $M$ must be cyclic because otherwise we can find $a,b\in M$ s.t. $aR\not\subset bR$ and $bR\not\subset aR$, which contradicts the uniseriality of $M$.

Over any commutative ring, any cyclic module is isomorphic to $R_R/I$ for some ideal $I$ of $R$. Thus, in order to classify all the indecomposable modules over $R$ we need first to find all the ideals of $R$. Indeed, $R$ is an ideal and the rest are of the form $I_i=(x^i)/(x^n)$ for some $1\leq i\leq n$. By uniserialy of $R_R$ each $R_R/I_i$ is indecomposable, and $R_R/I_i\cong R_R/I_j\Leftrightarrow i=j$. So all the indecomposable modules over $R$ are classified up to isomorphism. They are $R_R$ and the ones of the form $R_R/I_i$.

An example of a non-commutative Nakayama $k$-algebra is the upper triangular matrix ring $R=\big(\begin{smallmatrix} k & k\\ 0 & k \end{smallmatrix}\big)$. $R$ is the path algebra associated to the quiver $A_2: 1\rightarrow 2$, so every module over $R$ can be identified with a tuple $(V,W,f)$ where $V$ and $W$ are $k$-vector spaces $V$ and $f$ is a $k$-linear map from $V$ to $W$. In this case, the only indecomposables are $(0,k,0),(k,0,0)$ and $(k,k,Id)$.