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Karl Schwede
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There is some literature in the commutative algebra world that might be relevant, although I don't know it super-well. I can point you in the right direction at least. I

I don't think anything follows immediately from Grothendieck duality, especially for blowing up height-1 primes.

Say $R$ is a 2-dimensional and Cohen-Macaulay ring (equivalently S2). Suppose that $p$ is an arbitrary height-1-prime. Since $R$ is 2-dimensional, the analytic spread of $p$ is at most $2$. (See the book of Swanson-Huneke on integral closure, there are some subtleties with analytic spread in the case of a finite residue field -- be careful). We have two cases.

The analytic spread of $p$ is 2

Probably this is the more interesting case. Suppose there happens to be *only one* ideal $J$, with two generators such that $J p = p^2$ (this isn't as strong as it might sound, it's a bit stronger than requiring that $J$ and $p$ have the same normalized blowup). Additionally suppose that $R_p$ is regular (maybe this is too strong). Then [Theorem 3.1 in this paper by Santiago Zarzuela][2] proves that the Rees Algebra of the ideal is Cohen-Macaulay, and hence so is the blowup. There is also a lot of potentially relevant stuff in [this paper of Huckaba and Huneke][3]. (You can look at the papers which cite it on mathscinet to find even more).

The analytic spread of $p$ is 1

In particular, then the blowup of $p$ is some finite integral extension of $R$ (in particular, the blowup is an affine scheme dominated by the normalization of $R$). Since $R$ was S2, this implies that $p$ is a prime defining an irreducible component of the non-normal locus. You want to keep the blowup S2... I don't know in general if this is possible but I recall that some conditions for such blowups being normalizations appeared towards the end of [this paper by Greco and Traverso][4]. Does your surface happen to be seminormal?

There is some literature in the commutative algebra world that might be relevant, although I don't know it super-well. I can point you in the right direction at least. I don't think anything follows immediately from Grothendieck duality, especially for blowing up height-1 primes.

Say $R$ is a 2-dimensional and Cohen-Macaulay ring (equivalently S2). Suppose that $p$ is an arbitrary height-1-prime. Since $R$ is 2-dimensional, the analytic spread of $p$ is at most $2$. (See the book of Swanson-Huneke on integral closure, there are some subtleties with analytic spread in the case of a finite residue field -- be careful). We have two cases.

The analytic spread of $p$ is 2

Probably this is the more interesting case. Suppose there happens to be *only one* ideal $J$, with two generators such that $J p = p^2$ (this isn't as strong as it might sound, it's a bit stronger than requiring that $J$ and $p$ have the same normalized blowup). Additionally suppose that $R_p$ is regular (maybe this is too strong). Then [Theorem 3.1 in this paper by Santiago Zarzuela][2] proves that the Rees Algebra of the ideal is Cohen-Macaulay, and hence so is the blowup. There is also a lot of potentially relevant stuff in [this paper of Huckaba and Huneke][3]. (You can look at the papers which cite it on mathscinet to find even more).

The analytic spread of $p$ is 1

In particular, then the blowup of $p$ is some finite integral extension of $R$ (in particular, the blowup is an affine scheme dominated by the normalization of $R$). Since $R$ was S2, this implies that $p$ is a prime defining an irreducible component of the non-normal locus. You want to keep the blowup S2... I don't know in general if this is possible but I recall that some conditions for such blowups being normalizations appeared towards the end of [this paper by Greco and Traverso][4]. Does your surface happen to be seminormal?

There is some literature in the commutative algebra world that might be relevant, although I don't know it super-well.

I don't think anything follows immediately from Grothendieck duality, especially for blowing up height-1 primes.

Say $R$ is a 2-dimensional and Cohen-Macaulay ring (equivalently S2). Suppose that $p$ is an arbitrary height-1-prime. Since $R$ is 2-dimensional, the analytic spread of $p$ is at most $2$. (See the book of Swanson-Huneke on integral closure, there are some subtleties with analytic spread in the case of a finite residue field -- be careful). We have two cases.

The analytic spread of $p$ is 2

Probably this is the more interesting case. Suppose there happens to be *only one* ideal $J$, with two generators such that $J p = p^2$ (this isn't as strong as it might sound, it's a bit stronger than requiring that $J$ and $p$ have the same normalized blowup). Additionally suppose that $R_p$ is regular (maybe this is too strong). Then [Theorem 3.1 in this paper by Santiago Zarzuela][2] proves that the Rees Algebra of the ideal is Cohen-Macaulay, and hence so is the blowup. There is also a lot of potentially relevant stuff in [this paper of Huckaba and Huneke][3]. (You can look at the papers which cite it on mathscinet to find even more).

The analytic spread of $p$ is 1

In particular, then the blowup of $p$ is some finite integral extension of $R$ (in particular, the blowup is an affine scheme dominated by the normalization of $R$). Since $R$ was S2, this implies that $p$ is a prime defining an irreducible component of the non-normal locus. You want to keep the blowup S2... I don't know in general if this is possible but I recall that some conditions for such blowups being normalizations appeared towards the end of [this paper by Greco and Traverso][4]. Does your surface happen to be seminormal?
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Karl Schwede
  • 20.5k
  • 3
  • 53
  • 98

There is some literature in the commutative algebra world that might be relevant, although I don't know it super-well. I can point you in the right direction at least. I don't think anything follows immediately from Grothendieck duality, especially for blowing up height-1 primes.

Say $R$ is a 2-dimensional and Cohen-Macaulay ring (equivalently S2). Suppose that $p$ is an arbitrary height-1-prime. Since $R$ is 2-dimensional, the analytic spread of $p$ is at most $2$. (See the book of Swanson-Huneke on integral closure, there are some subtleties with analytic spread in the case of a finite residue field -- be careful). We have two cases.

The analytic spread of $p$ is 2

Probably this is the more interesting case. Suppose there happens to be *only one* ideal $J$, with two generators such that $J p = p^2$ (this isn't as strong as it might sound, it's a bit stronger than requiring that $J$ and $p$ have the same normalized blowup). Additionally suppose that $R_p$ is regular (maybe this is too strong). Then [Theorem 3.1 in this paper by Santiago Zarzuela][2] proves that the Rees Algebra of the ideal is Cohen-Macaulay, and hence so is the blowup. There is also a lot of potentially relevant stuff in [this paper of Huckaba and Huneke][3]. (You can look at the papers which cite it on mathscinet to find even more).

The analytic spread of $p$ is 1

In particular, then the blowup of $p$ is some finite integral extension of $R$ (in particular, the blowup is an affine scheme dominated by the normalization of $R$). Since $R$ was S2, this implies that $p$ is a prime defining an irreducible component of the non-normal locus. You want to keep the blowup S2... I don't know in general if this is possible but I recall that some conditions for such blowups being normalizations appeared towards the end of [this paper by Greco and Traverso][4]. Does your surface happen to be seminormal?