Consider the polynomial ring $R = k[x_1, \ldots, x_n]$ in $n$ indeterminates over an algebraically closed field $k$ (my relevant case is the complex numbers). Furthermore, consider an algebraic variety V in the affine space $k^n$, whose associated polynomial ideal we label as $I$ [we note that I is radical]. Finally, define $R_V = R/I$, that is the quotient ring of $R$ by $I$ -- the coordinate ring of the variety $V$.

In this question, it is stated that if $I$ is generated by m elements and $\mathrm{codim}(I) = m$, (i.e. $I$ is generated by a regular sequence), then (for a Cohen-Macaulay ring such as $R$) $R/I$ is also Cohen-Macaulay.

I can easily find a reference for this fact if $R$ is a local Cohen-Macaulay ring. However, I am unable to find a reference in the case that $R$ is only Cohen-Macaulay (such as the polynomial ring in question). If my understanding is correct, can anyone provide a reference for this fact (potentially alongside a proof)?

  • 1
    $\begingroup$ Isn't a Noetherian ring Cohen-Macauley if and only if all its local rings are Cohen-Macauley? So the local case implies the global case. e.g. stacks.math.columbia.edu/tag/00NC $\endgroup$
    – Will Sawin
    Aug 9 at 11:29
  • $\begingroup$ To my understanding, this is the definition of globally Cohen-Macaulay, yes. However, one then needs to show (e.g.) that it is true for all of the localizations at maximal ideals. It is not clear to me how to derive this from the statement that I is radical and generated by a regular sequence. (Indeed it's not clear to me that it necessarily follows.) $\endgroup$
    – V.S.
    Aug 9 at 12:58
  • $\begingroup$ So To show $(R/I)_x. = R_x/I_x$ is Cohen-Macauley, you can use the fact that $R_x$ is a local Cohen-Macauley ring and $I_x$ is generated by a regular sequence, plus the reference you already found. $\endgroup$
    – Will Sawin
    Aug 9 at 13:37

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.