Let $R$ be a $\mathbf N^r$-graded ring, for instance a polynomial ring in $r$-variables. A prime ideal $\mathfrak p\subseteq R$ is associated to a graded $R$-module $M$ if there is a (not necessarily homogeneous) $m\in M$ such that $\mathfrak p=\operatorname{Ann}(m)$.
Question: Is it true that such a $\mathfrak p$ is necessarily homogeneous?
I came across this claim and haven't found an error in the proof presented to me. However, the following makes me doubt this holds true.
Attempted Counterexample: Consider $R=\mathbf F_2[x,y]/(x^2, y^2)$ as $\mathbf N^2$-graded ring and $R$ as a module over itself. Then $\operatorname{Ann}(x+y)=(x+y)$, which is a prime ideal, but obviously not $\mathbf N^2$-homogeneous.
Question:
- What's wrong about that example?
- Does the claim hold true if one only considers polynomial rings (and not quotients thereof)?
$(x+y)$ isn't a $\mathbf{N}^2$-graded submodule of $R$. It has no homogeneous elements, and so as an abelian group, it cannot be the direct sum over its graded pieces.