Let $S$ be an $\mathbb{N}$-graded ring. Define $\operatorname{Proj}S$ to be the set of graded prime ideals of $S$ not containing $S_+=\sum_{d>0}S_d$. I thought the following standard result would be true:

Every graded ideal $J\subset S$ not containing $S_+$ is contained in a maximal such ideal, which is prime.

After a few hours of futile attempt to prove it, now I'm beginning to suspect that it might not be true. Is it?

  • $\begingroup$ Isn't this the projective version of the Nullstellensatz? $\endgroup$ – Zhen Lin Nov 17 '13 at 9:24
  • $\begingroup$ I thought Nullstelensatz is for $\mathbb{P}^n$ and polynomial rings over a field. $\endgroup$ – ashpool Nov 17 '13 at 12:18
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    $\begingroup$ You must suppose $\sqrt{J}$ does not contain $S_+$. Then the result can be proven using Krull's intersection theorem. $\endgroup$ – Cantlog Nov 18 '13 at 22:38
  • $\begingroup$ @Cantlog Thanks for pointing that out. I guess I can refine the question as: if $\operatorname{Proj}S\neq\emptyset$, then does it have a maximal element? $\endgroup$ – ashpool Nov 24 '13 at 14:14

Both are not true:

  1. The ideal $(x^2)\subset\mathbb{C}[x]$ is a maximal graded ideal not containing $S_+=(x)$, but it is not prime.
  2. $(x_1^2, x_2^2, \ldots)\subset\mathbb{C}[x_1,x_2,\ldots]$ is a graded ideal not containing $S_+=(x_1, x_2, \ldots)$, but it is not contained in any maximal such ideal.

As Cantlog pointed out, if we replace the condition $S_+\not\subset J$ by $S_+\not\subset\sqrt{J}$, then the second part (that the maximal element is a prime ideal) can be proved using the standard argument. The first part (existence of a maximal element) can also be proved if $S_+$ is finitely genereated, using Zorn's Lemma. Zorn's Lemma doesn't seem to work if $S_+$ is not finitely generated, though.


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