Let $R = k[x_1, \ldots, x_n$] be a graded ring with $deg(x_i) = 1 \ \forall i = 1, \ldots, n$ and let $\mathfrak{m} = (x_1, \ldots, x_n)$ be the graded maximal ideal. Let $M$ be a finitely generated $R$-module. Assume that $\mathfrak{m}$ contains a nonzerodivisor $x$ on M, i.e. $x$ is $M$-regular.
Can we always chose $x$ to be linear, i.e. $x \in \mathfrak{m}_1$?
What I was thinking, is that if by contradiction all linear elements are zerodivsors on $M$, then $x_1, \ldots, x_n$ are zerodivisors on $M$. Wouldn't that mean that all monomials are zerodiviors and therefore all elements in $\mathfrak{m}$ are zerodivisors on $M$?
I know of a proof in Bruns and Herzog's Cohen-Macaulay rings (Prop. 1.5.12), but there they need the assumption that $k$ is infinite. So I was wondering, where my proof fails, i.e. why this would not hold for a finite field $k$?