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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)?
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  • $\begingroup$ $(x+y)$ is not an $\Bbb N^2$ graded submodule of $R$, $\endgroup$ – Lord Shark the Unknown Apr 25 '18 at 15:43
  • $\begingroup$ @LordSharktheUnknown see comment to Hurkyl s answer $\endgroup$ – Bubaya Apr 25 '18 at 15:57
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    $\begingroup$ See also Eisenbud, "Commutative algebra with a view toward algebraic geometry", Exercise 3.5 ("General Graded Primary Decomposition"). $\endgroup$ – Minseon Shin Apr 25 '18 at 17:01
  • $\begingroup$ @MinseonShin Your hint was rather enlightening. For other readers: The mistake in my attempt was that in fact $\operatorname{Ann}(m)=(x+y, xy)$, which is not prime. But primeness was one of the premises of the claim. $\endgroup$ – Bubaya Apr 25 '18 at 17:30
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$(x+y)$ 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.

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  • $\begingroup$ Why is this a problem? The assertion certainly would hold if in the definition of a associated prime ideal, I required $m$ to be homogeneous. However, the text I am working through does not make this requirement. $\endgroup$ – Bubaya Apr 25 '18 at 15:57

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