Perfectoid multiplier/test ideals in regular rings and bounds on symbolic powers

Abstract

Using perfectoid algebras we introduce a mixed characteristic analog of the multiplier ideal, respectively test ideal, from characteristic zero, respectively $p > 0$, in the case of a regular ambient ring. We prove several properties about this ideal such as subadditivity. We then use these techniques to derive a uniform bound on the growth of symbolic powers of radical ideals in all excellent regular rings. The analogous result was shown in equal characteristic by Ein–Lazarsfeld–Smith and Hochster–Huneke.

Appendix A Blowups

In this appendix we briefly recall (and in some cases prove) facts about blowups of ideals. These are well known but we record them here for ease of the reader. Note, we are working with potentially non-Noetherian rings in most cases.

Setting A.1

Throughout this section, R will be a reduced ring and $J \subseteq R$ will be a finitely generated ideal. We let $X = {{\mathrm{{Spec}}}}R$ and let $Y \xrightarrow {\ \ }X$ be the blowup of J in X. In particular, set $S = R \oplus JT \oplus JT^2 \oplus \dots$ where the T serve as a dummy variable to help distinguish degree, and thus $Y = {{\mathrm{{Proj}}}}S$.

Lemma A.2

If $J = ( z_1, \ldots , z_m )$, then the complements $U_i$ of $V(z_i T) \subseteq Y$ form an affine cover of Y with $U = {{\mathrm{{Spec}}}}R[z_1/z_i, \ldots , z_m/z_i]$.

In the above $R[z_1/z_i, \ldots , z_m/z_i]$ is viewed as the subring of elements of $S[ (z_i T)^{-1} ]$ of the form $g T^n / (z_i T)^n$ as in [40, Tag 052P].

Proof

Note any homogeneous prime of S does not contain some $z_i$ and so this follows from for instance [40, Tag 0804]. $\square$

Lemma A.3

Suppose that $f \in R$ is integral over J. Define $J' = J + (f )$ and let $Y' \xrightarrow {\ \ }X$ be the blowup of $J'$. Then $Y' \xrightarrow {\ \ }X$ factors through Y and $Y'$ is a partial normalization of Y generated locally by adding a single integral element to the rings defining the affine charts $U_i$.

Proof

Write $f^n + a_1 f^{n-1} + \dots + a_n = 0$ with $a_i \in J^i$. Now write $J = (z_1, \ldots , z_m )$ and form the Rees algebra S as above. Let $S' = R \oplus J' T \oplus J'^2 T \oplus \dots \supseteq S$. We will first prove that the $U_i' = Y' \setminus V(z_i T)$ form an open cover of $Y'$ (in particular, we do not need V(fT)). Suppose that $Q \subseteq S'$ is a homogeneous prime ideal containing all of the $z_i T$ but not fT. Obviously Q contains $0 = f^n T^n + a_1 f^{n-1} T^n + \dots + a_n T^n$ also note that Q contains $a_n T^n$ since $a_n T^n \in ( z_1, \dots , z_n )^n T^n$. But then since Q does not contain fT, Q must contain

$$\begin{aligned} f^{n-1} T^{n-1} + a_1 f^{n-2} T^{n-1} + \dots + a_{n-1} T^{n-1}. \end{aligned}$$

But Q also contains $a_{n-1} T^{n-1}$ as before and so continuing in this way, we eventually deduce that $f T \in Q$, a contradiction. Thus we have shown that $\{U_i\}$ form an open cover of ${{\mathrm{{Proj}}}}S' = Y'$.

On the other hand, each $U_i' = {{\mathrm{{Spec}}}}R[z_1/z_i, \dots , z_m/z_i, f/z_i]$ and $y = f/z_i$ satisfies the monic polynomial equation

$$\begin{aligned} (f/z_i)^n + (a_1/z_i) (f/z_i)^{n-1} + \dots + (a_n/z_i^n) = 0 \end{aligned}$$

where each $a_j/z_i^j \in R[z_1/z_i, \dots , z_m/z_i]$ by construction. The lemma follows. $\square$

Next we recall a partial converse to the previous Lemma.

Lemma A.4

Suppose additionally to Setting A.1 that R is normal, and that the normalization $\mu : Y' \xrightarrow {\ \ }Y$ is finite over Y. Then $\pi : Y' \xrightarrow {\ \ }X$ is the blowup of $\overline{J^n}$ for some $n > 0$ where ${\overline{\bullet }}$ denotes the integral closure of the ideal.

Proof

Write $J = (z_1, \dots , z_m)$ and consider the ring $R_i\! :=\! R[z_1/z_i, \dots , z_m/z_i]$ defining an affine chart $U_i$ on Y. Suppose that $x \in \mathcal {O}_{Y'}(\mu ^{-1} U_i)$, and hence x is integral over $R_i$. It follows that x satisfies some integral equation

$$\begin{aligned} x^l + f_{1} x^{l-1} + \dots + f_{l-1} x^1 + f_l = 0 \end{aligned}$$

with $f_j = f_j(z_1/z_i, \dots , z_m/z_i) \in R_i$. Note that we can pick a sufficiently large h such that $f_jz_i^h\in J^h$ for all j (i.e., clearing all the denominators of $f_j$). It follows that $f_j z_i^{hj} \in J^{hj} \subseteq R$ for all j. Multiplying by $z^{hl}$ we get

$$\begin{aligned} (xz_i^h)^l + f_{1} z_i^h (xz_i^h)^{l-1} + \dots + f_{l-1} z_i^{h(l-1)} (x z_i^h)^1 + f_l z_i^{hl} = 0. \end{aligned}$$

Now, $x z_i^h$ is in R since it is integral over R and R is normal. Since $f_j z_i^{hj} \in J^{hj}$ for all j, we also have $xz_i^h \in \overline{J^h}$ and thus $x\in R[\frac{\overline{J^h}}{z_i^h}]$. We can do this for the finitely many generators of each chart, and pick $h\gg 0$ that works for all these generators. It follows that there exists $h\gg 0$ such that $\mathcal {O}_{Y'}(\mu ^{-1} U_i)\subseteq R[\frac{\overline{J^h}}{z_i^h}]$ for every i. But then $\mathcal {O}_{Y'}(\mu ^{-1} U_i)= R[\frac{\overline{J^h}}{z_i^h}]$ because the latter is integral over $R_i$ and $\mathcal {O}_{Y'}(\mu ^{-1} U_i)$ is the integral closure of $R_i$. Therefore $Y'$ is the blow up of $\overline{J^h}$ as desired. $\square$

Remark A.5

Another way to prove this when R is normal, Noetherian and excellent is to consider the Rees algebra S, and observe that the normalization $S'$ of S is

$$\begin{aligned} S' = R \oplus {\overline{J}} T \oplus \overline{J^2} T^2 \oplus \dots , \end{aligned}$$

see for instance [32, Proposition 5.2.1]. It easily follows that ${{\mathrm{{Proj}}}}S'$ is the normalization of ${{\mathrm{{Proj}}}}S$ [37, 6.C.9 Exercise]. Since S is excellent, $S'$ is finite over S and hence Noetherian. We thus see that $S'^{n}$, the nth Veronese of $S'$, is generated in degree 1 for n sufficiently divisible [11, Chapter III, §1.3, Proposition 3]. But ${{\mathrm{{Proj}}}}S'^{n} \cong {{\mathrm{{Proj}}}}S'$ is the blowup of $\overline{J^n}$.

Finally, we now move to blowups in Noetherian regular local rings. First we recall some notation, suppose that $\pi : Y \xrightarrow {\ \ }X = {{\mathrm{{Spec}}}}A$ is a finite type birational map between normal Noetherian integral schemes where X is regular (or at least Gorenstein). We also fix a choice of a dualizing complex on A. Since A is Gorenstein and integral, this complex has cohomology only in a single degree (which we select to be $-\dim X$), and that cohomology is a line bundle which is denoted by $\omega _X$. We then define the dualizing complex on Y to be where we have sheafified our dualizing complex on A. We also set and observe that this is not necessarily a line bundle.

By a canonical divisor on X we mean any Weil divisor $K_X$ on X such that $\mathcal {O}_X(K_X) \cong \omega _X$. Since X is Gorenstein, $\mathcal {O}_X(K_X)$ is a line bundle and hence $K_X$ is Cartier. Likewise a canonical divisor on Y is any Weil divisor $K_Y$ so that $\mathcal {O}_Y(K_Y) \cong \omega _Y$.

Lemma A.6

There exist canonical divisors $K_Y$ and $K_X$ that agree where $\pi$ is an isomorphism. Furthermore, for any choice of $K_X$, there is such a compatible choice of $K_Y$.

Our proof also holds if X is not necessarily Gorenstein but only normal with a dualizing complex.

Proof

First notice that even though $\omega _Y$ is not a line bundle, $\omega _Y$ is still a reflexive rank-1 sheaf, and so there exists a $K_Y$ with $\mathcal {O}_Y(K_Y) = \omega _Y$. Consider the divisor $\pi _* K_Y$ on X obtained by throwing away any irreducible component of $K_Y$ that is mapped to a subscheme of codimension $\ge 2$. This divisor agrees with $K_Y$ wherever $\pi$ is an isomorphism, which is a set U whose complement has codimension $\ge 2$ on X. In particular, $\mathcal {O}_U(\pi _* K_Y) \cong \omega _X|_U$. Thus $\mathcal {O}_X(\pi _* K_Y)$ is a reflexive sheaf that agrees with $\omega _X$ outside a set of codimension $\ge 2$, and so $\mathcal {O}_X(\pi _* K_Y) \cong \omega _X$, [24]. Setting $K_X = \pi _* K_Y$ proves the first part of the lemma.

Now suppose that $K_X'$ is another choice of canonical divisor. Since $\mathcal {O}_X(K_X') \cong \omega _X \cong \mathcal {O}_X(K_X)$, we see that $K_X' \sim K_X$ and so there exists some element f of the fraction field K(A) so that $K_X' = K_X + {{\mathrm{{div}}}}_X(f)$. We then set $K_Y' = K_Y + {{\mathrm{{div}}}}_Y(f)$ and observe that $K_Y'$ and $K_X'$ agree where $\pi$ is an isomorphism. $\square$

Definition A.7

(Relative canonical divisor) Choose $K_Y$ and $K_X$ as in Lemma A.6. We define the relative canonical divisor $K_{Y/X} := K_Y - \pi ^* K_X$, and observe it is exceptional and also independent of the choice of $K_Y$ and $K_X$. Note that if one chooses $\omega _X \cong \mathcal {O}_X$, then one may take $K_X = 0$ and so $K_Y = K_{Y/X}$ may be chosen to be exceptional.

Lemma A.8

Suppose that $(R, \mathfrak {m}, k)$ is a regular local Noetherian ring of dimension d and that $Y \xrightarrow {\ \ }X = {{\mathrm{{Spec}}}}R$ is the blowup of $\mathfrak {m}$. Then Y is regular, has prime exceptional divisor E with $\mathfrak {m}\mathcal {O}_Y =\mathcal {O}_Y(-E)$ and $K_{Y/X} = (d-1)E$.

Proof

This is well known, but because we do not know of a reference where it is phrased in this language outside of the context of varieties over a field, we include a quick geometric proof. Equivalent commutative algebra statements can be found for example in [26, 27, 44].

A direct computation shows that the exceptional divisor $E \cong \mathbb {P}^{d-1}_k$ lives in the regular scheme Y. The same computation also shows that $\mathcal {O}_X(-E)|_E = \mathcal {O}_E(1)$. Because we know⁹ that $(K_Y + E)|_E = K_E$ and that $\mathcal {O}_E(K_E) = \mathcal {O}_E(-d)$, if we write $K_Y = nE$, then $(K_Y + E)|_E = (nE + E)|_E = K_E$ and so $-(n+1) = -d$ and thus $n = d-1$ as claimed. $\square$