Let $(R, \mathfrak{m})$ be a Cohen-Macaulay ring, let $f_1, \dotsc, f_d \in \mathfrak{m}$ be a regular sequence, and let $n_1, \dotsc, n_d > 0$ be weights (feel free to assume that $n_1 = 1$ if it helps). I would like to consider the graded $R$-subalgebra $R' \subset R[T]$ generated by the $f_iT^{n_i}$ and form the "weighted blowup" $X = Proj(R')$. Is $X$ Cohen-Macaulay? If $n_1 = 1$ (the only case I care about), is at least the locus of $X$ where ``$f_1T$'' does not vanish, i.e., the last (first?) affine coordinate patch, Cohen-Macaulay?

I suspect that the answer may be contained in the commutative algebra literature and that maybe even $R'$ itself is Cohen-Macaulay--this would be even better--but I do not know where. For instance, in the case $n_1 = ... = n_d = 1$, when we are talking about the usual blowup / Rees algebra, the Cohen-Macaulayness of $R'$ was proved by Barshay in 1973.

I would be very grateful if someone could point me to the relevant literature, I know that Cohen-Macaulayness of various graded rings has been studied extensively, but I don't know where to look for this particular question. I tried using Hyry's paper on Cohen-Macaulayness of multigraded rings implying that of the diagonal subring, but its results did not seem to help directly (because of generation in degree $1$ assumptions).


1 Answer 1


Yes. Here's a proof when all the weights are 1 (so we are looking at a usual blowup. For further details (and some further references) you might be interested in Prop. 5.5 of https://arxiv.org/abs/1703.02269.

Let $Y = \mathrm{Spec}R$, let $I = (f_1, \dots, f_d)$, let $Z = V(I) \subset Y$, let $\pi : X \to Y$ be the projection and let $E = \pi^{-1}(Z)$. By the universal property of blowing up, $E$ is a Cartier divisor, hence if $E$ is Cohen-Macaulay so is $X$. Since $f_1, \dots, f_d$ is a regular sequence, the co-normal bundle $I/I^2$ of $Z$ is locally free, and so $E$ is a $\mathbb{P}^{d-1}$-bundle over $Z$. So if $Z$ is Cohen-Macaulay, so is $E$. Finally, $Z$ is Cohen-Macaulay since $R$ is Cohen-Macaulay and $f_1, \dots, f_d$ is a regular sequence.

Perhaps the same strategy works for weighted blowups — one would have to verify:

  • $E$ is a Cartier divisor,
  • $E \to Z$ is a flat family of weighted projective spaces (It would then follow that since $Z$ is Cohen-Macaulay, so is $E$ - the key point here is that weighted projective spaces are Cohen-Macaulay, since finite quotient singularities are Cohen-Macaulay).
  • $\begingroup$ Thank you for your response. I know the case when all the weights are $1$; some other references for it are Lemma A.6.1 in Fulton's Intersection theory or, for a stronger statement, 1.3--1.4 in expose VII of SGA6. Could you elaborate on the argument in the general weighted case? For one thing, is $E$ still a Cartier divisor then? $\endgroup$ Sep 3, 2020 at 9:03
  • $\begingroup$ Ah, good point. I'm not sure (edited accordingly). Sorry I can't elaborate further off the top of my head, since I don't have much working experience with weighted-blowups. $\endgroup$
    – cgodfrey
    Sep 4, 2020 at 19:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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