On some properties of the asymptotic Samuel function

Definition 1.1.Let $$A$$ be a Noetherian ring and let $$I\subset A$$ be a proper ideal. The asymptotic Samuel function at $$I$$, $$\bar{\nu }_I:A\rightarrow \mathbb {R}\cup \lbrace \infty \rbrace$$, is defined as follows:

Properties (A), (B), and (C) 1.3.In the following lines, we will revisit properties (A), (B), and (C) in the case of the asymptotic Samuel function.

• (A) In general, the asymptotic Samuel function is not a valuation and it is not hard to find examples.

Example 1.4.Let $$k$$ be a field. Consider the ring $$B=k[x,y,z]/\langle xy-z^3\rangle$$, with maximal ideal $${\mathfrak {m}}=\langle \overline{x},\overline{y},\overline{z}\rangle$$. We have that:

Proposition 2.6.Let $$(B,{\mathfrak {m}})$$ be an equidimensional excellent equicharacteristic local ring of dimension $$d\ge 1$$ containing a field $$k$$. Suppose $${\mathfrak {m}}$$ has a reduction generated by $$d$$ elements, $$y_1,\ldots, y_d\in {\mathfrak {m}}$$. Set $$A:=k[y_1,\ldots, y_d]_{\langle y_1,\ldots, y_d\rangle }\subset B$$. Then, for $$a\in A$$ and $$b\in B$$,

Example 1.5.Let $$k$$ be a field, and let $$B=\left(k[x,y,z]/\langle y^2 + zx^3\rangle \right)_{{\mathfrak {m}}}$$, where $${\mathfrak {m}}=\langle \overline{x},\overline{y},\overline{z}\rangle$$. Set $${\mathfrak {p}}=\langle \overline{y},\overline{z} \rangle$$. Notice that

Theorem 3.1.Let $$B$$ be an equidimensional excellent ring containing a field. Let $${\mathfrak {p}}_1\subset {\mathfrak {p}}_2\subset B$$ be two prime ideals such that $$e_{B_{\mathfrak {p}_1}}({\mathfrak {p}}_1B_{\mathfrak {p}_1})= e_{B_{\mathfrak {p}_2}}({\mathfrak {p}}_2B_{\mathfrak {p}_2})$$. Then $$\overline{\nu }_{{\mathfrak {p}}_1B_{\mathfrak {p}_1}}(b)\le \overline{\nu }_{{\mathfrak {p}}_2B_{\mathfrak {p}_2}}(b)$$ for $$b\in B$$.

The Samuel slope of a local ring 1.6.The notion of Samuel slope of a local ring was introduced in [2]. More precisely, for a Noetherian local ring $$(B, \mathfrak {m},k)$$ of dimension $$d$$ we consider the natural map:

Theorem 4.5.Let $$(B,{\mathfrak {m}},k)$$ be a non-regular reduced equicharacteristic equidimensional excellent local ring. Then, $$\mathcal {S}\!\operatorname{-sl}(B)\in \mathbb {Q}$$.

Theorem 4.7.Let $$(B,{\mathfrak {m}},k)$$ be a non-reduced equicharacteristic equidimensional excellent local ring. Then $$\mathcal {S}\!\operatorname{-sl}(B)=\mathcal {S}\!\operatorname{-sl}(B_{\text{{\rm red}}})$$.

Theorem 5.5.Let $$(B,{\mathfrak {m}},k)$$ be an equicharacteristic equidimensional excellent local ring, and let $$(B,{\mathfrak {m}})\rightarrow (B^{\prime },{\mathfrak {m}}^{\prime })$$ be a local-étale extension. Then, $$\mathcal {S}\!\operatorname{-sl}(B)=\mathcal {S}\!\operatorname{-sl}(B^{\prime })$$.

Definition 2.1.Let $$S\subset B$$ be a finite extension of excellent rings with $$S$$ regular and $$B$$ equidimensional and reduced. Let $$K$$ be the fraction field of $$S$$ and let $$L:=B\otimes _S K$$. Suppose that no non-zero element of $$S$$ is a zero divisor in $$B$$. We say that the projection $$\operatorname{Spec}(B)\rightarrow \operatorname{Spec}(S)$$ (or the extension $$S\subset B$$) is finite-transversal with respect to $${\mathfrak {p}}\in \operatorname{Spec}(B)$$ if:

Proposition 2.2. ([[34], Corollary 4.9])Let $$S\subset B$$ be a finite extension of excellent rings with $$S$$ regular, and $$B$$ equidimensional and reduced. Suppose that no non-zero element of $$S$$ is a zero divisor in $$B$$. Let $${\mathfrak {p}}\subset B$$ be a prime ideal, and let $$\mathfrak {q}=\mathfrak {p}\cap S$$. Then, the following are equivalent:

• (1) $$e_{B_{\mathfrak {p}}}(\mathfrak {p} B_\mathfrak {p})=[L:K]$$.
• (2) The following three conditions hold:
  • (i) $$\mathfrak {p}$$ is the only prime of $$B$$ dominating $$\mathfrak {q}$$,
  • (ii) $$k(\mathfrak {p})=B_{\mathfrak {p}}/\mathfrak {p}B_{\mathfrak {p}}=S_{\mathfrak {q}}/\mathfrak {q}S_{\mathfrak {q}}=k(\mathfrak {q})$$,
  • (iii) $$e_{B_{\mathfrak {p}}}({\mathfrak {q}}B_{\mathfrak {p}})=e_{B_{\mathfrak {p}}}({\mathfrak {p}}B_{\mathfrak {p}})$$.

Proposition 2.3. ([[34], Lemma 5.2])Let $$S\subset B$$ be a finite extension such that the non-zero elements of $$S$$ are non-zero divisors in $$B$$. Assume that $$S$$ is a regular ring and let $$K=K(S)$$ be the quotient field of $$S$$. Let $$\theta \in B$$ and let $$f(Z)\in K[Z]$$ be the monic polynomial of minimal degree such that $$f(\theta)=0$$. If $$S[\theta]$$ denotes the $$S$$-subalgebra of $$B$$ generated by $$\theta$$, then:

• (1) the coefficients of $$f$$ are in $$S$$, that is, $$f(Z)\in S[Z]$$, and
• (2) $$S[\theta]\cong S[Z]/\langle f(Z)\rangle$$.

Theorem 2.4. ([[19], Theorem 2.1])Let $$(B, {\mathfrak {m}})$$ be a Noetherian equicharacteristic equidimensional and excellent local ring of Krull dimension $$d$$. Assume that there is a faithfully flat extension $$(B, {\mathfrak {m}}) \rightarrow (\widetilde{B}, \widetilde{\mathfrak {m}})$$ with $${\mathfrak {m}}\widetilde{B}=\widetilde{\mathfrak {m}}$$ together with a finite-transversal morphism with respect to $$\widetilde{\mathfrak {m}}$$, $$S\subset \widetilde{B}$$. Let $$b\in B$$. If

Proof.Theorem 2.1 in [19] is stated in the case in which $$\widetilde{B}= \widehat{B}$$, and then a reduction to the domain case is considered. See [7, Theorem 11.6.8], where it is checked that Hickel's theorem holds in this more general setting.$$\Box$$

Constructing finite-transversal projections 2.5.Given an excellent reduced equidimensional ring $$B$$, and a point $${\mathfrak {p}}\in \operatorname{Spec}(B)$$, in general, there might not be a regular ring $$S$$ and a finite-transversal projection with respect to $${\mathfrak {p}}$$, $$\operatorname{Spec}(B)\rightarrow \operatorname{Spec}(S)$$. To start with, it is a necessary condition that $${\mathfrak {p}}$$ has a reduction generated by $$\dim (B_{\mathfrak {p}})$$-elements. But even if such condition is satisfied, the existence of the required finite projection is not guaranteed (see [7, Example 11.3.11]). However, finite-transversal projections can be constructed if we are allowed to extend our ring $$B$$. For instance, in [34] (see [9, Proposition 31.1]) it is proven that if $$B$$ is essentially of finite type over a perfect field $$k$$, then a finite-transversal projection can be constructed in some étale extension of $$B$$.

Proposition 2.6.Let $$(B,{\mathfrak {m}})$$ be an equidimensional excellent equicharacteristic local ring of dimension $$d\ge 1$$. Suppose $${\mathfrak {m}}$$ has a reduction generated by $$d$$ elements, $$y_1,\ldots, y_d\in {\mathfrak {m}}$$, and let $$k\subset B$$ be a field. Set $$A:=k[y_1,\ldots, y_d]_{\langle y_1,\ldots, y_d\rangle }\subset B$$. Then, for $$a\in A$$ and $$b\in B$$,

Proof.Using the arguments in Section  2.5, we can assume that $$B$$ is reduced and complete, and consider the finite-transversal extension

Proposition 2.7.Suppose that $$(S, {\mathfrak {n}})\subset (B,{\mathfrak {m}})$$ is finite-transversal with respect to $${\mathfrak {m}}$$. Let $${\mathfrak {p}}\subset B$$ be a prime ideal with $$e_{B_{\mathfrak {p}}}({\mathfrak {p}}B_{\mathfrak {p}})=e_B({\mathfrak {m}})$$. Let $${\mathfrak {q}}={\mathfrak {p}}\cap S$$. Then:

• (i) The extension $$S_{\mathfrak {q}}\subset B_{\mathfrak {p}}$$ is finite transversal with respect to $${\mathfrak {p}}$$;
• (ii) The local ring $$B/{\mathfrak {p}}$$ is regular if and only if $$S/{\mathfrak {q}}$$ is regular;
• (iii) If $$B/{\mathfrak {p}}$$ is regular then $$S/{\mathfrak {q}}=B/{\mathfrak {p}}$$.

Proof.Similar results were proven in [34, Corollary 5.9, Proposition 6.3] and [1, Corollary 2.8] in the context of algebraic varieties defined over perfect fields. Here, we check that the statement holds for more general rings under the hypotheses of the proposition.

• (i) Consider the following commutative diagram with vertical finite morphisms:
  
  Observe that the generic rank of the extension $$ S_{\mathfrak {q}}\rightarrow B\otimes _S S_{\mathfrak {q}}$$ is $$m=e_{B_{\mathfrak {p}}}({\mathfrak {p}}B_{\mathfrak {p}})$$. Hence, by definition, $$S_{\mathfrak {q}}\rightarrow B\otimes _S S_{\mathfrak {q}}$$ is finite-transversal with respect to $${\mathfrak {p}}$$. By Proposition 2.2 (2) (i), $$B\otimes _S S_{\mathfrak {q}}=B_{\mathfrak {p}}$$. In other words, $$S_{\mathfrak {q}}\subset B_{\mathfrak {p}}$$ is finite transversal with respect to $${\mathfrak {p}}$$.
• (ii) Since $$S_{\mathfrak {q}}\subset B_{\mathfrak {p}}$$ is finite-transversal with respect to $${\mathfrak {p}}$$, by Proposition 2.2 (2) (ii), $$k({\mathfrak {p}})=k({\mathfrak {q}})$$. Now, consider the commutative diagram with vertical finite extensions,
  
  Note that $$S/{\mathfrak {q}}\rightarrow B/{\mathfrak {p}}$$ is a finite extension of local rings. Since conditions (2) (i)–(iii) of Proposition 2.2 hold for $$(S, {\mathfrak {n}})\rightarrow (B,{\mathfrak {m}})$$, the same conditions hold for $$(S/{\mathfrak {q}}, {\mathfrak {n}}/{\mathfrak {q}})\rightarrow (B/{\mathfrak {p}}, {\mathfrak {m}}/{\mathfrak {p}})$$. Now, apply Zariski's multiplicity formula for finite projections (Theorem 2.2) to $$S/{\mathfrak {q}} \rightarrow B/{\mathfrak {p}}$$ to obtain,
  $$$$\begin{equation*} 1=e_{S/{\mathfrak {q}}}\cdot [k({\mathfrak {p}}): k({\mathfrak {q}})]= e_{B/{\mathfrak {p}}}\cdot [k({\mathfrak {m}}): k({\mathfrak {n}}))]=e_{B/{\mathfrak {p}}}, \end{equation*}$$$$
  from where the claim in (ii) follows.
• (iii) By (ii) if $$B/{\mathfrak {p}}$$ is regular, then $$S/{\mathfrak {q}}\subset B/{\mathfrak {p}}$$ is a finite extension of regular local rings with the same quotient field. Since $$S/{\mathfrak {q}}$$ is regular, it is normal, and hence $$S/{\mathfrak {q}}= B/{\mathfrak {p}}$$.

Proposition 2.8. (Presentations of finite-transversal extensions)Suppose that $$(S, {\mathfrak {n}})\subset (B,{\mathfrak {m}})$$ is finite-transversal with respect to $${\mathfrak {m}}$$. Let $${\mathfrak {p}}\subset B$$ be a prime ideal with $$e_{B_{\mathfrak {p}}}({\mathfrak {p}}B_{\mathfrak {p}})=e_B({\mathfrak {m}})$$, and assume in addition that $$B/{\mathfrak {p}}$$ is a regular local ring. Let $${\mathfrak {q}}={\mathfrak {p}}\cap S$$. There are $$\theta _1,\ldots, \theta _e\in {\mathfrak {p}}$$ such that:

• (i) $$B=S[\theta _1,\ldots, \theta _e]$$;
• (ii) $${\mathfrak {p}}={\mathfrak {q}}B+\langle \theta _1,\ldots,\theta _e\rangle$$.

• (iii) $${\mathfrak {q}}B$$ is a reduction of $${\mathfrak {p}}$$ (in $$B$$).

Proof.We follow ideas from [34, Lemma 6.4] for part (i), [2, Lemma 8.10] for part (ii) and [1, Lemma 3.6] for part (iii), where similar results were proven in the context of algebraic varieties defined over perfect fields. To facilitate the reading of the paper we check here that the proofs can be adapted to cover a wider class of rings under the hypotheses of the proposition.

• (i) Write $$B=S[\theta _1^{\prime },\ldots, \theta _e^{\prime }]$$. By Proposition 2.7 (iii), $$S/{\mathfrak {q}}= B/{\mathfrak {p}}$$, therefore, for each $$i\in \lbrace 1,\ldots, e\rbrace$$, there is some $$s_i\in S$$ such that $$\theta _i^{\prime }-s_i\in {\mathfrak {p}}$$. Set $$\theta _i:=\theta _i^{\prime }-s_i$$ for $$i=1\ldots, e$$. Then, $$B=S[\theta _1,\ldots, \theta _e]$$.
• (ii) By (i), we can write $$B=S[\theta _1,\ldots, \theta _e]$$ with $$\theta _i\in {\mathfrak {p}}$$ for $$i=1,\ldots, e$$. Since $$S\subset B$$ is finite-transversal at $${\mathfrak {m}}$$, $$S/{\mathfrak {n}}=B/{\mathfrak {m}}$$, therefore,
  $$$$\begin{equation} {\mathfrak {m}}={\mathfrak {n}}+\langle \theta _1,\ldots, \theta _e\rangle . \end{equation}$$$$()
  If $${\mathfrak {p}}={\mathfrak {m}}$$ we are done. Otherwise, since $${\mathfrak {q}} \subset S$$ defines a regular subscheme, there is a regular system of parameters in $$S$$, $$y_1,\ldots, y_d$$, such that $${\mathfrak {q}}=\langle y_1,\ldots, y_r\rangle$$ for some $$r\in \lbrace 1,\ldots, d-1\rbrace$$. Then
  $$$$\begin{equation} {\mathfrak {q}}B+\langle \theta _1,\ldots, \theta _e\rangle \subset {\mathfrak {p}}. \end{equation}$$$$()
  Now,
  $$$$\begin{equation*} d-r=\dim (S/{\mathfrak {q}})=\dim (B/{\mathfrak {p}})\le \dim (B/({\mathfrak {q}}B+\langle \theta _1,\ldots, \theta _e\rangle))\le d-r, \end{equation*}$$$$
  where the last inequality follows because by (2.12),
  $$$$\begin{equation*} {\mathfrak {m}}/({\mathfrak {q}}+\langle \theta _1+\cdots +\theta _e\rangle)= ({\mathfrak {n}}+\langle \theta _1,\ldots, \theta _e\rangle)/ ({\mathfrak {q}}+\langle \theta _1+\cdots +\theta _e\rangle), \end{equation*}$$$$
  and therefore, the maximal ideal $${\mathfrak {m}}/({\mathfrak {q}}+\langle \theta _1,\ldots, \theta _e\rangle)$$ can be generated by $$d-r$$ elements. Since $$B/{\mathfrak {p}}$$ is an integral domain necessarily the containment in (2.13) is an equality.
• (iii)

  By (i) we can assume that $$B=S[\theta _1,\ldots, \theta _e]$$ with $$\theta _i\in {\mathfrak {p}}$$ for $$i=1,\ldots, e$$. By Proposition 2.7 (i) and by Proposition 2.2, we have that $${\mathfrak {q}}B_{\mathfrak {p}}$$ is a reduction of $${\mathfrak {p}}B_{\mathfrak {p}}$$. To see that $${\mathfrak {p}}$$ is the integral closure of $${\mathfrak {q}}B$$ in $$B$$ it suffices to check this condition at all the maximal ideals containing $${\mathfrak {p}}$$. Since $$B$$ is local, this amounts to checking this condition at $$B$$.

  Since $$B/{\mathfrak {p}}$$ is a regular local ring, by Proposition 2.7 (iii), $$S/{\mathfrak {q}}$$ is also a regular local ring. Observe that the multiplicity of $$S$$ is 1, and so is the multiplicity of $$S_{\mathfrak {q}}$$. Hence, by Theorem 2.9, $$\text{ht}({\mathfrak {q}})=l({\mathfrak {q}})$$ in $$S$$.

  Now, since the extension $$S\subset B$$ is finite, $$\text{ht}({\mathfrak {q}}B)=\text{ht}({\mathfrak {q}})$$, and the blow up of $$B$$ at $${\mathfrak {q}}B$$ is finite over the blow up of $$S$$ at $${\mathfrak {q}}$$. Hence, the fibers over the closed points have the same dimensions, and therefore, $$l({\mathfrak {q}}B)=l({\mathfrak {q}})$$. Therefore, $$l({\mathfrak {q}}B)=\text{ht}({\mathfrak {q}}B)$$ in $$B$$. Recall that $$e_{B_{\mathfrak {p}}}({\mathfrak {q}}B_{\mathfrak {p}})=e_{B_{\mathfrak {p}}}({\mathfrak {p}}B_{\mathfrak {p}})$$, by Proposition 2.2 (iii). Finally, since $${\mathfrak {p}}$$ is the only minimal prime of $${\mathfrak {q}}B$$, the statement follows from Theorem 2.10.$$\Box$$

Theorem 2.9. (Hironaka–Schickhoff, [[24], Corollary 3, p. 121])Let $$(A,M)$$ be a formally equidimensional local ring, and let $${\mathfrak {p}}\subset A$$ be a prime ideal so that $$A/{\mathfrak {p}}$$ is regular. Then, $$\text{ht}({\mathfrak {p}})=l({\mathfrak {p}})$$ in $$A$$ if and only if the local rings $$A$$ and $$A_{\mathfrak {p}}$$ have the same multiplicity.

Theorem 2.10. (Böger, [[24], Theorems 2 and 3, pp. 115–116], [[32], Corollary 11.3.2])Let $$(A,M)$$ be a formally equidimensional local ring. Fix an ideal $$I\subset A$$ so that $$\text{ht}(I)=l(I)$$. Consider an ideal $$J\subset A$$ so that $$I\subset J\subset \sqrt {I}$$. Then, $$I$$ is a reduction of $$J$$ if and only if $$e_{A_{\mathfrak {q}}}(IA_{\mathfrak {q}})=e_{A_{\mathfrak {q}}}(JA_{\mathfrak {q}})$$ for each minimal prime ideal $${\mathfrak {q}}$$ of $$I$$.

Proposition 2.11. (Intermediate extensions of finite-transversal extensions)Suppose that $$(S, {\mathfrak {n}})\subset (B,{\mathfrak {m}})$$ is finite-transversal with respect to $${\mathfrak {m}}$$. Let $${\mathfrak {p}}\subset B$$ be a prime ideal with $$e_{B_{\mathfrak {p}}}({\mathfrak {p}}B_{\mathfrak {p}})=e_B({\mathfrak {m}})$$. Let $$B^{\prime }\subset B$$ be an intermediate extension, that is, $$S\subset B^{\prime }\subset B$$, and consider the diagram:

• (i) The ring $$B^{\prime }$$ is local with maximal ideal $${\mathfrak {m}}^{\prime }$$;
• (ii) The extension $$S\subset B^{\prime }$$ is finite-transversal with respect to $${\mathfrak {m}}^{\prime }$$ of generic rank $$m^{\prime }=e_{B^{\prime }}({\mathfrak {m}}^{\prime })$$;
• (iii) The extension $$S_{\mathfrak {q}}\subset B^{\prime }_{{\mathfrak {p}}^{\prime }}$$ is finite-transversal with respect to $${\mathfrak {p}}^{\prime }$$ and $$e_{B^{\prime }_{{\mathfrak {p}}^{\prime }}}({\mathfrak {p}}^{\prime }B^{\prime }_{{\mathfrak {p}}^{\prime }})=m^{\prime }=e_{B^{\prime }}({\mathfrak {m}}^{\prime })$$;
• (iv) For $$b\in B^{\prime }$$,
  $$$$\begin{equation} \overline{\nu }_{\mathfrak {m}}(b)=\overline{\nu }_{{\mathfrak {m}}^{\prime }}(b) \end{equation}$$$$()
  and
  $$$$\begin{equation} \overline{\nu }_{{\mathfrak {p}}B_{\mathfrak {p}}}(b)=\overline{\nu }_{{\mathfrak {p}}^{\prime }B_{{\mathfrak {p}}^{\prime }}}(b); \end{equation}$$$$()
• (v) If $$B/{\mathfrak {p}}$$ is regular, then $$B^{\prime }/{\mathfrak {p}}^{\prime }$$ is regular, and in such case, $${\mathfrak {q}}B^{\prime }$$ is a reduction of $${\mathfrak {p}}^{\prime }$$ in $$B^{\prime }$$.