Let $R$ be a polynomial ring, $I$ a homogeneous ideal, and let $\operatorname{pd}(R/I)$ denote its projective dimension. Is there a characterization of homogeneous elements $a\in R\setminus I$ for which we have strict inequality: $$(\dagger) \ \ \operatorname{pd}(R/I)<\operatorname{pd}(R/(I+(a)))?$$
For example I think that if $a$ is not a zero divisor for $R/I$ then $(\dagger)$ holds by Corollary 4.3.14 in Weibel's book. I'm wondering if there are other examples of $a$'s where $(\dagger)$ holds? Or is it an if and only if, i.e. $(\dagger)$ holds if and only if $a$ is a nonzero divisor for $R/I$? Any examples or references or comments would be greatly appreciated.

  • 1
    $\begingroup$ Take $R=k[x,y]$, $I=(x^2,xy)$ and $a=x$. Then projective dimension of $R/I=2$, projective dimension of $R/I+(a)=1$. $\endgroup$ – Mohan Nov 26 '17 at 20:41
  • $\begingroup$ Thanks for the example! I guess I'm particularly interested in examples where $(\dagger)$ does hold, especially if its a zero divisor if $R/I$. Still, thanks :-) $\endgroup$ – Chris McDaniel Nov 26 '17 at 20:59
  • 1
    $\begingroup$ The same example works. Take $I=(x^2), a=xy$ (or the other way works too). Then projective dimension of $R/I=1$, $xy$ is a zero divisor in $R/I$ and projective dimension of $R/(x^2,xy)=2$. $\endgroup$ – Mohan Nov 26 '17 at 22:26
  • $\begingroup$ I'm voting to close this question because it's already been answered in the comments. $\endgroup$ – David White Nov 27 '17 at 15:40
  • $\begingroup$ Hmmm...well I guess I got one example, but the question whether there is a characterization of such a's has certainly not been settled... $\endgroup$ – Chris McDaniel Nov 27 '17 at 15:57

Your Answer

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

Browse other questions tagged or ask your own question.