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.


  • What's wrong about that example?
  • Does the claim hold true if one only considers polynomial rings (and not quotients thereof)?
  • $\begingroup$ $(x+y)$ is not an $\Bbb N^2$ graded submodule of $R$, $\endgroup$ – Angina Seng Apr 25 '18 at 15:43
  • $\begingroup$ @LordSharktheUnknown see comment to Hurkyl s answer $\endgroup$ – Bubaya Apr 25 '18 at 15:57
  • 1
    $\begingroup$ 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

$(x+y)$ 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.

| cite | improve this answer | |
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.