# Weighted blowup of a Cohen-Macaulay ring along a regular sequence is Cohen-Macaulay?

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).

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.
• $$E$$ 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).
• 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? Sep 3 '20 at 9:03