# Graded torsion module is zero in high enough degree

If $R = k[x,y]$ with the usual grading for some field $k$, we say that a graded $R$-module $M$ is graded torsion if $M$ is annihilated by some power of $\mathfrak m = (x,y)$.

Is it true that there exists $N$ such that $M_n = 0$ for all $n \geq N$ if and only if $M$ is graded torsion?

One direction is obvious: if $M_n = 0$ for all $n\geq N$ then $\mathfrak m^N M = 0$.

If $M$ is finitely generated then I think we can argue as follows:

Let $m_1, \dots, m_k$ be a set of homogeneous generators for $M$ and let $d = \text{max}_i \ \text{deg} \ m_i$, and suppose $\mathfrak m^N M = 0$. Then let $n\in M$ be homogeneous of degree at least $Nd$. Then $n = a_1m_1 + \dots + a_km_k$. Then since each $a_im_i$ has degree at least $Nd$, each $a_i$ must be homogeneous of degree at least $N$, meaning $a_i\in \mathfrak m^N$ and so $a_im_i = 0$ for all $i$ and so $n= 0$. Thus $M_j = 0$ for all $j \geq Nd$.

Is this correct? And is the claim true if $M$ is not finitely generated?

• $\mathfrak m^N M = 0$ shows that $M$ is an $R/\mathfrak m^N$-module. But $R/\mathfrak m^N$ is artinian, and whenever $M$ is finitely generated $M$ itself is artinian. Then $M_n=0$ from some $N$ on. On the other side, if $M$ is not finitely generated one can't expect this. Consider $M$ a graded infinitely dimensional $K$-vector space. Then it is a graded $R$-module and $\mathfrak mM=0$. – user26857 Sep 20 '17 at 18:24