A module has only finite many submodule up to isomorphism

Question

Let $A$ be a commutative Artin ring, $M$ be a finitely generated $A$-module. Then are there only finitely many submodules of $M$ up to isomorphism? For example, when $A$ is a field, it is true. But I don't know the general case.

Answer 1

It is possible for a finitely generated $A$-module to have infinitely many submodules up to isomorphism.

Let $k$ be an infinite field, and let $A = k\langle x,y \rangle/(x^2, y^2, xy, yx)$. Then $A$ is a $k$-algebra and is finite-dimensional over $k$, so it is an Artin ring. Consider the (unique up to isomorphism) indecomposable injective $A$-module $I$: it is a $3$-dimensional $k$-vector space on which the actions of $x$ and $y$ are defined by the matrices $$ I_x = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{bmatrix}, I_y = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}. $$ Next, for each $\lambda\in k$, define an $A$-module $M_\lambda$ as a $2$-dimensional $k$-vector space with action of $x$ and $y$ defined by $$ M_{\lambda, x} = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}, M_{\lambda,y} = \begin{bmatrix} 0 & 0 \\ \lambda & 0 \end{bmatrix}. $$ Then the $M_\lambda$ are pairwise non-isomorphic; to see this, simply note that if $X = PM_{\lambda,x}P^{-1}$ and $Y=PM_{\lambda,y}P^{-1}$ are simultaneous conjugates of the matrices $M_{\lambda,x}$ and $M_{\lambda,y}$, then they satisfy $\lambda X - Y = 0$ and $\mu X - Y \neq 0$ for any $\mu \neq \lambda$. Also, each $M_\lambda$ is isomorphic to a submodule of $I$, as the embedding of $k^2$ into $k^3$ given by the matrix $$ \iota_\lambda:\begin{bmatrix} 1 & 0 \\ \lambda & 0 \\ 0 & 1 \end{bmatrix} $$ is such that $\iota_\lambda \circ M_{\lambda,x} = I_x \circ \iota_{\lambda}$ and $\iota_\lambda \circ M_{\lambda,y} = I_y \circ \iota_{\lambda}$, so it defines a monomorphism of $A$-modules. Since $k$ is infinite, this proves the assertion.