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

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