MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Consider a nonsingular projective variety $X$ over an algebraically closed field $k$ and let $Y \subseteq X$ be a nonsingular closed subvariety. Let $\mathcal I \subseteq \mathcal O_X$ be the ideal sheaf of $Y$ and let $\mathcal L$ be a very ample invertible sheaf on $X$. If it makes the setup any easier we can assume that $X$ is projectively normal under the associated embedding to some projective space. Set $$ R(\mathcal L) := \bigoplus_{n \geq 0} \Gamma( X, \mathcal L^n ), $$ the section ring associated to $\mathcal L$. Then for each $m \geq 0$ we have the homogeneous ideal $$ R_m(\mathcal L) := \bigoplus_{n \geq 0} \Gamma( X, \mathcal I^m \otimes_{\mathcal O_X} \mathcal L^n ) $$ of the ring $R(\mathcal L)$. These ideals define a decreasing multiplicative filtration on $R(\mathcal L)$ so we can consider the associated (bi-)graded ring $gr R(\mathcal L)$.

Here is my question: Under what conditions can we conclude that this graded ring is Noetherian? Is this always true? Unfortunately I do not have a deep knowledge of commutative algebra so I have gotten stuck on this point. If it makes the answer easier we can even assume that $Y$ is a point.

share|cite|improve this question
I think all that's happening here is that you're filtering your $R$ by powers of an ideal $I$. No? In which case $R$ is Noetherian so $gr_I\ R$ is too. – Allen Knutson May 24 '13 at 3:40
That makes sense, although it's not clear to me why the multiplication map $ R_1(\mathcal L)^{\otimes m} \to R_m(\mathcal L)$ should be surjective (but I might be missing something easy here). – Chuck Hague May 24 '13 at 14:28

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.