arXiv2021
A set of polynomials $M$ is called a {\it submodule} of $\mathbb{C} [x_1, \dots, x_n ]$ if $M$ is a translation invariant linear subspace of $\mathbb{C} [x_1, \dots, x_n ]$. We present a description of the submodules of $\mathbb{C} [x,y]$ in terms of a special type of submodules. We say that the submodule $M$ of $\mathbb{C} [x,y]$ is an {\it L-module of order} $s$ if, whenever $F(x,y)=\sum_{n=0}^N f_n (x) \cdot y^n \in M$ is such that $f_0 =\ldots = f_{s-1}=0$, then $F=0$. We show that the proper submodules of $\mathbb{C} [x,y]$ are the sums $M_d +M$, where $M_d =\{ F\in \mathbb{C} [x,y] \colon \textit{deg}_x F <d\}$, and $M$ is an L-module. We give a construction of L-modules parametrized by sequences of complex numbers. A submodule $M\subseteq \mathbb{C} [x_1, \dots, x_n ]$ is {\it decomposable} if it is the sum of finitely many proper submodules of $M$. Otherwise $M$ is {\it indecomposable}. It is easy to see that every submodule of $\mathbb{C} [x_1, \dots, x_n]$ is the sum of finitely many indecomposable submodules. In $\mathbb{C} [x,y]$ every indecomposable submodule is either an L-module or equals $M_d$ for some $d$. In the other direction we show that $M_d$ is indecomposable for every $d$, and so is every L-module of order $1$. Finally, we prove that there exists a submodule of $\mathbb{C} [x,y]$ (in fact, an L-module of order $1$) which is not relatively closed in $\mathbb{C} [x,y]$. This answers a problem posed by L. Székelyhidi in 2011.