The Stacks project

Proof. These statements hold because taking integral closure commutes with localization, see Lemma 10.36.11. $\square$

Proof. Assume $R_{f_ i}$ is N-2 (or N-1). Let $L$ be a finite extension of the fraction field of $R$ (equal to the fraction field in the N-1 case). Let $S$ be the integral closure of $R$ in $L$. By Lemma 10.36.11 we see that $S_{f_ i}$ is the integral closure of $R_{f_ i}$ in $L$. Hence $S_{f_ i}$ is finite over $R_{f_ i}$ by assumption. Thus $S$ is finite over $R$ by Lemma 10.23.2. $\square$

Proof. Let $L/K$ be the induced extension of fraction fields. Note that this is a finite field extension (for example by Lemma 10.122.2 (2) applied to the fibre $S \otimes _ R K$, and the definition of a quasi-finite ring map). Let $S'$ be the integral closure of $R$ in $S$. Then $S'$ is contained in the integral closure of $R$ in $L$ which is finite over $R$ by assumption. As $R$ is Noetherian this implies $S'$ is finite over $R$. By Lemma 10.123.14 there exist elements $g_1, \ldots , g_ n \in S'$ such that $S'_{g_ i} \cong S_{g_ i}$ and such that $g_1, \ldots , g_ n$ generate the unit ideal in $S$. Hence it suffices to show that $S'$ is N-2 by Lemmas 10.161.3 and 10.161.4. Thus we have reduced to the case where $S$ is finite over $R$.

Assume $R \subset S$ with hypotheses as in the lemma and moreover that $S$ is finite over $R$. Let $M$ be a finite field extension of the fraction field of $S$. Then $M$ is also a finite field extension of $K$ and we conclude that the integral closure $T$ of $R$ in $M$ is finite over $R$. By Lemma 10.36.16 we see that $T$ is also the integral closure of $S$ in $M$ and we win by Lemma 10.36.15. $\square$

Proof. Let $R'$ be the integral closure of $R$ in its field of fractions $K$. Let $S'$ be the integral closure of $R[z, z^{-1}]$ in its field of fractions. Clearly $R' \subset S'$. Since $K[z, z^{-1}]$ is a normal domain we see that $S' \subset K[z, z^{-1}]$. Suppose that $f_1, \ldots , f_ n \in S'$ generate $S'$ as $R[z, z^{-1}]$-module. Say $f_ i = \sum a_{ij}z^ j$ (finite sum), with $a_{ij} \in K$. For any $x \in R'$ we can write

with $h_ i \in R[z, z^{-1}]$. Thus we see that $R'$ is contained in the finite $R$-submodule $\sum Ra_{ij} \subset K$. Since $R$ is Noetherian we conclude that $R'$ is a finite $R$-module. $\square$

Proof. Omitted. (Hint: Integral closures of $R$ in extension fields are contained in integral closures of $S$ in extension fields.) $\square$

Proof. Consider the trace pairing (Fields, Definition 9.20.6)

Since $L/K$ is separable this is nondegenerate (Fields, Lemma 9.20.7). Moreover, if $x \in L$ is integral over $R$, then $\text{Trace}_{L/K}(x)$ is in $R$. This is true because the minimal polynomial of $x$ over $K$ has coefficients in $R$ (Lemma 10.38.6) and because $\text{Trace}_{L/K}(x)$ is an integer multiple of one of these coefficients (Fields, Lemma 9.20.3). Pick $x_1, \ldots , x_ n \in L$ which are integral over $R$ and which form a $K$-basis of $L$. Then the integral closure $S \subset L$ is contained in the $R$-module

By linear algebra we see that $M \cong R^{\oplus n}$ as an $R$-module. Hence $S \subset R^{\oplus n}$ is a finitely generated $R$-module as $R$ is Noetherian. $\square$

Proof. After replacing $x$ by $fx$ and $a$ by $f^ pa$ for some $f \in R$ we may assume $a \in R$. Hence also $D(a) \in R$. We will show by induction on $i \leq p - 1$ that if

is integral over $R$, then $D(a)^ i a_ j \in R$. Thus the integral closure is contained in the finite $R$-module with basis $D(a)^{-p + 1}x^ j$, $j = 0, \ldots , p - 1$. Since $R$ is Noetherian this proves the lemma.

If $i = 0$, then $y = a_0$ is integral over $R$ if and only if $a_0 \in R$ and the statement is true. Suppose the statement holds for some $i < p - 1$ and suppose that

is integral over $R$. Then

is an element of $R$ (as it is in $K$ and integral over $R$). Applying $D$ we obtain

is in $R$. Hence it follows that

is integral over $R$. By induction we find $D(a)^{i + 1}a_ j \in R$ for $j = 1, \ldots , i + 1$. (Here we use that $1, \ldots , i + 1$ are invertible.) Hence $D(a)^{i + 1}a_0$ is also in $R$ because it is the difference of $y$ and $\sum _{j > 0} D(a)^{i + 1}a_ jx^ j$ which are integral over $R$ (since $x$ is integral over $R$ as $a \in R$). $\square$

Proof. This is clear from Lemma 10.161.8 since every field extension in characteristic zero is separable. $\square$

Proof. Assume the integral closure of $R$ in every finite purely inseparable field extension of $K$ is finite. Let $L/K$ be any finite extension. We have to show the integral closure of $R$ in $L$ is finite over $R$. Choose a finite normal field extension $M/K$ containing $L$. As $R$ is Noetherian it suffices to show that the integral closure of $R$ in $M$ is finite over $R$. By Fields, Lemma 9.27.3 there exists a subextension $M/M_{insep}/K$ such that $M_{insep}/K$ is purely inseparable, and $M/M_{insep}$ is separable. By assumption the integral closure $R'$ of $R$ in $M_{insep}$ is finite over $R$. By Lemma 10.161.8 the integral closure $R''$ of $R'$ in $M$ is finite over $R'$. Then $R''$ is finite over $R$ by Lemma 10.7.3. Since $R''$ is also the integral closure of $R$ in $M$ (see Lemma 10.36.16) we win. $\square$

Proof. Assume $R$ is N-1. Let $R'$ be the integral closure of $R$ which is finite over $R$. Hence also $R'[x]$ is finite over $R[x]$. The ring $R'[x]$ is normal (see Lemma 10.37.8), hence N-1. This proves the first assertion.

For the second assertion, by Lemma 10.161.7 it suffices to show that $R'[x]$ is N-2. In other words we may and do assume that $R$ is a normal N-2 domain. In characteristic zero we are done by Lemma 10.161.11. In characteristic $p > 0$ we have to show that the integral closure of $R[x]$ is finite in any finite purely inseparable extension of $L/K(x)$ where $K$ is the fraction field of $R$. There exists a finite purely inseparable field extension $L'/K$ and $q = p^ e$ such that $L \subset L'(x^{1/q})$; some details omitted. As $R[x]$ is Noetherian it suffices to show that the integral closure of $R[x]$ in $L'(x^{1/q})$ is finite over $R[x]$. And this integral closure is equal to $R'[x^{1/q}]$ with $R \subset R' \subset L'$ the integral closure of $R$ in $L'$. Since $R$ is N-2 we see that $R'$ is finite over $R$ and hence $R'[x^{1/q}]$ is finite over $R[x]$. $\square$

Proof. It is clear that the standard open $D(f)$ is contained in $U$. By Serre's criterion Lemma 10.157.4 we see that $\mathfrak p \not\in U$ implies that for some $\mathfrak q \subset \mathfrak p$ we have either

1. Case I: $\text{depth}(R_{\mathfrak q}) < 2$ and $\dim (R_{\mathfrak q}) \geq 2$, and

2. Case II: $R_{\mathfrak q}$ is not regular and $\dim (R_{\mathfrak q}) = 1$.

This in particular also means that $R_{\mathfrak q}$ is not normal, and hence $f \in \mathfrak q$. In case I we see that $\text{depth}(R_{\mathfrak q}) = \text{depth}(R_{\mathfrak q}/fR_{\mathfrak q}) + 1$. Hence such a prime $\mathfrak q$ is the same thing as an embedded associated prime of $R/fR$. In case II $\mathfrak q$ is an associated prime of $R/fR$ of height 1. Thus there is a finite set $E$ of such primes $\mathfrak q$ (see Lemma 10.63.5) and

as desired. $\square$

Proof. First assume $R$ is N-1. Let $R'$ be the integral closure of $R$ in its field of fractions $K$. By assumption we can find $x_1, \ldots , x_ n$ in $R'$ which generate $R'$ as an $R$-module. Since $R' \subset K$ we can find $f_ i \in R$ nonzero such that $f_ i x_ i \in R$. Then $R_ f \cong R'_ f$ where $f = f_1 \ldots f_ n$. Hence $R_ f$ is normal and we have (1). Part (2) follows from Lemma 10.161.3.

Assume (1) and (2). Let $K$ be the fraction field of $R$. Suppose that $R \subset R' \subset K$ is a finite extension of $R$ contained in $K$. Note that $R_ f = R'_ f$ since $R_ f$ is already normal. Hence by Lemma 10.161.14 the set of primes $\mathfrak p' \in \mathop{\mathrm{Spec}}(R')$ with $R'_{\mathfrak p'}$ non-normal is closed in $\mathop{\mathrm{Spec}}(R')$. Since $\mathop{\mathrm{Spec}}(R') \to \mathop{\mathrm{Spec}}(R)$ is closed the image of this set is closed in $\mathop{\mathrm{Spec}}(R)$. For such a ring $R'$ denote $Z_{R'} \subset \mathop{\mathrm{Spec}}(R)$ this image.

Pick a maximal ideal $\mathfrak m \subset R$. Let $R_{\mathfrak m} \subset R_{\mathfrak m}'$ be the integral closure of the local ring in $K$. By assumption this is a finite ring extension. By Lemma 10.36.11 we can find finitely many elements $x_1, \ldots , x_ n \in K$ integral over $R$ such that $R_{\mathfrak m}'$ is generated by $x_1, \ldots , x_ n$ over $R_{\mathfrak m}$. Let $R' = R[x_1, \ldots , x_ n] \subset K$. With this choice it is clear that $\mathfrak m \not\in Z_{R'}$.

As $\mathop{\mathrm{Spec}}(R)$ is quasi-compact, the above shows that we can find a finite collection $R \subset R'_ i \subset K$ such that $\bigcap Z_{R'_ i} = \emptyset$. Let $R'$ be the subring of $K$ generated by all of these. It is finite over $R$. Also $Z_{R'} = \emptyset$. Namely, every prime $\mathfrak p'$ lies over a prime $\mathfrak p'_ i$ such that $(R'_ i)_{\mathfrak p'_ i}$ is normal. This implies that $R'_{\mathfrak p'} = (R'_ i)_{\mathfrak p'_ i}$ is normal too. Hence $R'$ is normal, in other words $R'$ is the integral closure of $R$ in $K$. $\square$

Proof. We may assume $x \not= 0$ since otherwise the lemma is trivial. Let $K$ be the fraction field of $R$. If the characteristic of $K$ is zero the lemma follows from (1), see Lemma 10.161.11. Hence we may assume that the characteristic of $K$ is $p > 0$, and we may apply Lemma 10.161.12. Thus given $L/K$ a finite purely inseparable field extension we have to show that the integral closure $S$ of $R$ in $L$ is finite over $R$.

Let $q$ be a power of $p$ such that $L^ q \subset K$. By enlarging $L$ if necessary we may assume there exists an element $y \in L$ such that $y^ q = x$. Since $R \to S$ induces a homeomorphism of spectra (see Lemma 10.46.7) there is a unique prime ideal $\mathfrak q \subset S$ lying over the prime ideal $\mathfrak p = xR$. It is clear that

since $y^ q = x$. Observe that $R_{\mathfrak p}$ is a discrete valuation ring by Lemma 10.119.7. Then $S_{\mathfrak q}$ is Noetherian by Krull-Akizuki (Lemma 10.119.12). Whereupon we conclude $S_{\mathfrak q}$ is a discrete valuation ring by Lemma 10.119.7 once again. By Lemma 10.119.10 we see that $\kappa (\mathfrak q)/\kappa (\mathfrak p)$ is a finite field extension. Hence the integral closure $S' \subset \kappa (\mathfrak q)$ of $R/xR$ is finite over $R/xR$ by assumption (2). Since $S/yS \subset S'$ this implies that $S/yS$ is finite over $R$. Note that $S/y^ nS$ has a finite filtration whose subquotients are the modules $y^ iS/y^{i + 1}S \cong S/yS$. Hence we see that each $S/y^ nS$ is finite over $R$. In particular $S/xS$ is finite over $R$. Also, it is clear that $\bigcap x^ nS = (0)$ since an element in the intersection has $q$th power contained in $\bigcap x^ nR = (0)$ (Lemma 10.51.4). Thus we may apply Lemma 10.96.12 to conclude that $S$ is finite over $R$, and we win. $\square$

Proof. Observe that $R[[x]]$ is Noetherian by Lemma 10.31.2. Let $R' \supset R$ be the integral closure of $R$ in its fraction field. Because $R$ is N-2 this is finite over $R$. Hence $R'[[x]]$ is finite over $R[[x]]$. By Lemma 10.37.9 we see that $R'[[x]]$ is a normal domain. Apply Lemma 10.161.16 to the element $x \in R'[[x]]$ to see that $R'[[x]]$ is N-2. Then Lemma 10.161.7 shows that $R[[x]]$ is N-2. $\square$