The Stacks project

Proof. As $S$ is essentially of finite type over $R$ it is Noetherian and has finitely many minimal primes $\mathfrak q_1, \ldots , \mathfrak q_ m$, see Lemma 10.31.6. Since $S$ is reduced we have $S \subset \prod S_{\mathfrak q_ i}$ and each $S_{\mathfrak q_ i} = K_ i$ is a field, see Lemmas 10.25.4 and 10.25.1. It suffices to show that the integral closure $A_ i'$ of $R$ in each $K_ i$ is finite over $R$. This is true because $R$ is Noetherian and $A \subset \prod A_ i'$. Let $\mathfrak p_ i \subset R$ be the prime of $R$ corresponding to $\mathfrak q_ i$. As $S$ is essentially of finite type over $R$ we see that $K_ i = S_{\mathfrak q_ i} = \kappa (\mathfrak q_ i)$ is a finitely generated field extension of $\kappa (\mathfrak p_ i)$. Hence the algebraic closure $L_ i$ of $\kappa (\mathfrak p_ i)$ in $K_ i$ is finite over $\kappa (\mathfrak p_ i)$, see Fields, Lemma 9.26.11. It is clear that $A_ i'$ is the integral closure of $R/\mathfrak p_ i$ in $L_ i$, and hence we win by definition of a Nagata ring. $\square$

Proof. Namely, assume the condition of the lemma. Let $R \to S$ be a finite type ring map with $S$ a domain. Let $L$ be a finite extension of the fraction field of $S$. Then there exists a finite ring extension $S \subset S' \subset L$ such that $L$ is the fraction field of $S'$. By assumption $S'$ is N-1, and hence the integral closure $S''$ of $S'$ in $L$ is finite over $S'$. Thus $S''$ is finite over $S$ (Lemma 10.7.3) and $S''$ is the integral closure of $S$ in $L$ (Lemma 10.36.16). We conclude that $R$ is universally Japanese. $\square$

Proof. The case of an algebra of finite type over $R$ is immediate from the definition. The general case follows on applying Lemma 10.161.3. $\square$

Proof. First note that $S$ is Noetherian as $R$ is Noetherian and a quasi-finite ring map is of finite type. Let $\mathfrak q \subset S$ be a prime ideal, and set $\mathfrak p = R \cap \mathfrak q$. Then $R/\mathfrak p \subset S/\mathfrak q$ is quasi-finite and hence we conclude that $S/\mathfrak q$ is N-2 by Lemma 10.161.5 as desired. $\square$

Proof. Clear from Lemma 10.161.3. $\square$

Proof. Let $\varphi : R \to S$ be a finite type ring map so that $S$ is a domain. Then $\varphi (f_1), \ldots , \varphi (f_ n)$ generate the unit ideal in $S$. Hence if each $S_{f_ i} = S_{\varphi (f_ i)}$ is N-1 then so is $S$, see Lemma 10.161.4. This proves (1).

If each $R_{f_ i}$ is Nagata, then each $R_{f_ i}$ is Noetherian and hence $R$ is Noetherian, see Lemma 10.23.2. And if $\mathfrak p \subset R$ is a prime, then we see each $R_{f_ i}/\mathfrak pR_{f_ i} = (R/\mathfrak p)_{f_ i}$ is N-2 and hence we conclude $R/\mathfrak p$ is N-2 by Lemma 10.161.4. This proves (2). $\square$

Proof. Let $R$ be a complete local Noetherian ring. Let $\mathfrak p \subset R$ be a prime. Then $R/\mathfrak p$ is also a complete local Noetherian ring, see Lemma 10.160.2. Hence it suffices to show that a Noetherian complete local domain $R$ is N-2. By Lemmas 10.161.5 and 10.160.11 we reduce to the case $R = k[[X_1, \ldots , X_ d]]$ where $k$ is a field or $R = \Lambda [[X_1, \ldots , X_ d]]$ where $\Lambda $ is a Cohen ring.

In the case $k[[X_1, \ldots , X_ d]]$ we reduce to the statement that a field is N-2 by Lemma 10.161.17. This is clear. In the case $\Lambda [[X_1, \ldots , X_ d]]$ we reduce to the statement that a Cohen ring $\Lambda $ is N-2. Applying Lemma 10.161.16 once more with $x = p \in \Lambda $ we reduce yet again to the case of a field. Thus we win. $\square$

Proof. In this proof we will use the remarks immediately following Definition 10.162.9. As $R \to R^\wedge $ is a faithfully flat local ring homomorphism it is injective and (1) follows.

Let $\mathfrak q$ be a minimal prime of $R$, and assume $R$ is analytically unramified. Then $\mathfrak q$ is an associated prime of $R$ (see Proposition 10.63.6). Hence there exists an $f \in R$ such that $\{ x \in R \mid fx = 0\} = \mathfrak q$. Note that $(R/\mathfrak q)^\wedge = R^\wedge /\mathfrak q^\wedge $, and that $\{ x \in R^\wedge \mid fx = 0\} = \mathfrak q^\wedge $, because completion is exact (Lemma 10.97.2). If $x \in R^\wedge $ is such that $x^2 \in \mathfrak q^\wedge $, then $fx^2 = 0$ hence $(fx)^2 = 0$ hence $fx = 0$ hence $x \in \mathfrak q^\wedge $. Thus $\mathfrak q$ is analytically unramified and (2) holds.

Assume $R$ is reduced with minimal primes $\mathfrak q_1, \ldots , \mathfrak q_ t$, and each $\mathfrak q_ i$ is analytically unramified. Then $R \to R/\mathfrak q_1 \times \ldots \times R/\mathfrak q_ t$ is injective. Since completion is exact (see Lemma 10.97.2) we see that $R^\wedge \subset (R/\mathfrak q_1)^\wedge \times \ldots \times (R/\mathfrak q_ t)^\wedge $. Hence (3) is clear.

Assume $R$ is analytically unramified. Let $\mathfrak p_1, \ldots , \mathfrak p_ s$ be the minimal primes of $R^\wedge $. Then we see that

with each $R^\wedge _{\mathfrak p_ i}$ a field as $R^\wedge $ is reduced (see Lemma 10.25.4). Hence the integral closure $S$ of $R^\wedge $ in $Q(R^\wedge )$ is equal to $S = S_1 \times \ldots \times S_ s$ with $S_ i$ the integral closure of $R^\wedge /\mathfrak p_ i$ in its fraction field. In particular $S$ is finite over $R^\wedge $. Denote $R'$ the integral closure of $R$ in $Q(R)$. As $R \to R^\wedge $ is flat we see that $R' \otimes _ R R^\wedge \subset Q(R) \otimes _ R R^\wedge \subset Q(R^\wedge )$. Moreover $R' \otimes _ R R^\wedge $ is integral over $R^\wedge $ (Lemma 10.36.13). Hence $R' \otimes _ R R^\wedge \subset S$ is a $R^\wedge $-submodule. As $R^\wedge $ is Noetherian it is a finite $R^\wedge $-module. Thus we may find $f_1, \ldots , f_ n \in R'$ such that $R' \otimes _ R R^\wedge $ is generated by the elements $f_ i \otimes 1$ as a $R^\wedge $-module. By faithful flatness we see that $R'$ is generated by $f_1, \ldots , f_ n$ as an $R$-module. This proves (4).

Part (5) is a special case of part (4). $\square$

Proof. Assumption (2) says that $R^\wedge /\mathfrak pR^\wedge $ is a reduced ring. Hence an associated prime $\mathfrak q \subset R^\wedge $ of $R^\wedge /\mathfrak pR^\wedge $ is the same thing as a minimal prime over $\mathfrak pR^\wedge $. In particular we see that the maximal ideal of $(R^\wedge )_{\mathfrak q}$ is $\mathfrak p(R^\wedge )_{\mathfrak q}$. Choose $x \in R$ such that $xR_{\mathfrak p} = \mathfrak pR_{\mathfrak p}$. By the above we see that $x \in (R^\wedge )_{\mathfrak q}$ generates the maximal ideal. As $R \to R^\wedge $ is faithfully flat we see that $x$ is a nonzerodivisor in $(R^\wedge )_{\mathfrak q}$. Hence we win. $\square$

Proof. Let $\mathfrak p_1, \ldots , \mathfrak p_ t$ be the associated primes of the $R$-module $R/xR$. Since $R/xR$ has no embedded primes we see that each $\mathfrak p_ i$ has height $1$, and is a minimal prime over $(x)$. For each $i$, let $\mathfrak q_{i1}, \ldots , \mathfrak q_{is_ i}$ be the associated primes of the $R^\wedge $-module $R^\wedge /\mathfrak p_ iR^\wedge $. By Lemma 10.162.11 we see that $(R^\wedge )_{\mathfrak q_{ij}}$ is regular. By Lemma 10.65.3 we see that

Let $y \in R^\wedge $ with $y^2 = 0$. As $(R^\wedge )_{\mathfrak q_{ij}}$ is regular, and hence a domain (Lemma 10.106.2) we see that $y$ maps to zero in $(R^\wedge )_{\mathfrak q_{ij}}$. Hence $y$ maps to zero in $R^\wedge /xR^\wedge $ by Lemma 10.63.19. Hence $y = xy'$. Since $x$ is a nonzerodivisor (as $R \to R^\wedge $ is flat) we see that $(y')^2 = 0$. Hence we conclude that $y \in \bigcap x^ nR^\wedge = (0)$ (Lemma 10.51.4). $\square$

Proof. By induction on $\dim (R)$. The case $\dim (R) = 0$ is trivial. Hence we assume $\dim (R) = d$ and that the lemma holds for all Noetherian Nagata domains of dimension $< d$.

Let $R \subset S$ be the integral closure of $R$ in the field of fractions of $R$. By assumption $S$ is a finite $R$-module. By Lemma 10.162.5 we see that $S$ is Nagata. By Lemma 10.112.4 we see $\dim (R) = \dim (S)$. Let $\mathfrak m_1, \ldots , \mathfrak m_ t$ be the maximal ideals of $S$. Each of these lies over the maximal ideal $\mathfrak m$ of $R$. Moreover

for sufficiently large $n$ as $S/\mathfrak mS$ is Artinian. By Lemma 10.97.2 $R^\wedge \to S^\wedge $ is an injective map, and by the Chinese Remainder Lemma 10.15.4 combined with Lemma 10.96.9 we have $S^\wedge = \prod S^\wedge _ i$ where $S^\wedge _ i$ is the completion of $S$ with respect to the maximal ideal $\mathfrak m_ i$. Hence it suffices to show that $S_{\mathfrak m_ i}$ is analytically unramified. In other words, we have reduced to the case where $R$ is a Noetherian normal Nagata domain.

Assume $R$ is a Noetherian, normal, local Nagata domain. Pick a nonzero $x \in \mathfrak m$ in the maximal ideal. We are going to apply Lemma 10.162.12. We have to check properties (1), (2), (3)(a) and (3)(b). Property (1) is clear. We have that $R/xR$ has no embedded primes by Lemma 10.157.6. Thus property (2) holds. The same lemma also tells us each associated prime $\mathfrak p$ of $R/xR$ has height $1$. Hence $R_{\mathfrak p}$ is a $1$-dimensional normal domain hence regular (Lemma 10.119.7). Thus (3)(a) holds. Finally (3)(b) holds by induction hypothesis, since $R/\mathfrak p$ is Nagata (by Lemma 10.162.5 or directly from the definition). Thus we conclude $R$ is analytically unramified. $\square$

Proof. Assume $R$ is Nagata and let $R \to S$ and $\mathfrak m' \subset S$ be as in (2). Then $S$ is Nagata by Lemma 10.162.5. Hence the local ring $S_{\mathfrak m'}$ is Nagata (Lemma 10.162.6). Thus it is analytically unramified by Lemma 10.162.13. It is clear that (2) implies (3).

Assume (3) holds. Let $\mathfrak p \subset R$ be a prime ideal and let $L/\kappa (\mathfrak p)$ be a finite extension of fields. To prove (1) we have to show that the integral closure of $R/\mathfrak p$ is finite over $R/\mathfrak p$. Choose $x_1, \ldots , x_ n \in L$ which generate $L$ over $\kappa (\mathfrak p)$. For each $i$ let $P_ i(T) = T^{d_ i} + a_{i, 1} T^{d_ i - 1} + \ldots + a_{i, d_ i}$ be the minimal polynomial for $x_ i$ over $\kappa (\mathfrak p)$. After replacing $x_ i$ by $f_ i x_ i$ for a suitable $f_ i \in R$, $f_ i \not\in \mathfrak p$ we may assume $a_{i, j} \in R/\mathfrak p$. In fact, after further multiplying by elements of $\mathfrak m$, we may assume $a_{i, j} \in \mathfrak m/\mathfrak p \subset R/\mathfrak p$ for all $i, j$. Having done this let $S = R/\mathfrak p[x_1, \ldots , x_ n] \subset L$. Then $S$ is finite over $R$, a domain, and $S/\mathfrak m S$ is a quotient of $R/\mathfrak m[T_1, \ldots , T_ n]/(T_1^{d_1}, \ldots , T_ n^{d_ n})$. Hence $S$ is local. By (3) $S$ is analytically unramified and by Lemma 10.162.10 we find that its integral closure $S'$ in $L$ is finite over $S$. Since $S'$ is also the integral closure of $R/\mathfrak p$ in $L$ we win. $\square$

Proof. It is clear that a Noetherian universally Japanese ring is universally Nagata (i.e., condition (2) holds). Let $R$ be a Nagata ring. We will show that any finitely generated $R$-algebra $S$ is Nagata. This will prove the proposition.

Step 1. There exists a sequence of ring maps $R = R_0 \to R_1 \to R_2 \to \ldots \to R_ n = S$ such that each $R_ i \to R_{i + 1}$ is generated by a single element. Hence by induction it suffices to prove $S$ is Nagata if $S \cong R[x]/I$.

Step 2. Let $\mathfrak q \subset S$ be a prime of $S$, and let $\mathfrak p \subset R$ be the corresponding prime of $R$. We have to show that $S/\mathfrak q$ is N-2. Hence we have reduced to the proving the following: (*) Given a Nagata domain $R$ and a monogenic extension $R \subset S$ of domains then $S$ is N-2.

Step 3. Let $R$ be a Nagata domain and $R \subset S$ a monogenic extension of domains. Let $R \subset R'$ be the integral closure of $R$ in its fraction field. Let $S'$ be the subring of the fraction field of $S$ generated by $R'$ and $S$. As $R'$ is finite over $R$ (by the Nagata property) also $S'$ is finite over $S$. Since $S$ is Noetherian it suffices to prove that $S'$ is N-2 (Lemma 10.161.7). Hence we have reduced to proving the following: (**) Given a normal Nagata domain $R$ and a monogenic extension $R \subset S$ of domains then $S$ is N-2.

Step 4: Let $R$ be a normal Nagata domain and let $R \subset S$ be a monogenic extension of domains. Suppose the induced extension of fraction fields of $R$ and $S$ is purely transcendental. In this case $S = R[x]$. By Lemma 10.161.13 we see that $S$ is N-2. Hence we have reduced to proving the following: (**) Given a normal Nagata domain $R$ and a monogenic extension $R \subset S$ of domains inducing a finite extension of fraction fields then $S$ is N-2.

Step 5. Let $R$ be a normal Nagata domain and let $R \subset S$ be a monogenic extension of domains inducing a finite extension of fraction fields $L/K$. Choose an element $x \in S$ which generates $S$ as an $R$-algebra. Let $M/L$ be a finite extension of fields. Let $R'$ be the integral closure of $R$ in $M$. Then the integral closure $S'$ of $S$ in $M$ is equal to the integral closure of $R'[x]$ in $M$. Also the fraction field of $R'$ is $M$ and $R \subset R'$ is finite (by the Nagata property of $R$). This implies that $R'$ is a Nagata ring (Lemma 10.162.5). To show that $S'$ is finite over $S$ is the same as showing that $S'$ is finite over $R'[x]$. Replace $R$ by $R'$ and $S$ by $R'[x]$ to reduce to the following statement: (***) Given a normal Nagata domain $R$ with fraction field $K$, and $x \in K$, the ring $S \subset K$ generated by $R$ and $x$ is N-1.

Step 6. Let $R$ be a normal Nagata domain with fraction field $K$. Let $x = b/a \in K$. We have to show that the ring $S \subset K$ generated by $R$ and $x$ is N-1. Note that $S_ a \cong R_ a$ is normal. Hence by Lemma 10.161.15 it suffices to show that $S_{\mathfrak m}$ is N-1 for every maximal ideal $\mathfrak m$ of $S$.

With assumptions as in the preceding paragraph, pick such a maximal ideal and set $\mathfrak n = R \cap \mathfrak m$. The residue field extension $\kappa (\mathfrak m)/\kappa (\mathfrak n)$ is finite (Theorem 10.34.1) and generated by the image of $x$. Hence there exists a monic polynomial $f(X) = X^ d + \sum _{i = 1, \ldots , d} a_ iX^{d -i}$ with $f(x) \in \mathfrak m$. Let $K''/K$ be a finite extension of fields such that $f(X)$ splits completely in $K''[X]$. Let $R'$ be the integral closure of $R$ in $K''$. Let $S' \subset K''$ be the subring generated by $R'$ and $x$. As $R$ is Nagata we see $R'$ is finite over $R$ and Nagata (Lemma 10.162.5). Moreover, $S'$ is finite over $S$. If for every maximal ideal $\mathfrak m'$ of $S'$ the local ring $S'_{\mathfrak m'}$ is N-1, then $S'_{\mathfrak m}$ is N-1 by Lemma 10.161.15, which in turn implies that $S_{\mathfrak m}$ is N-1 by Lemma 10.161.7. After replacing $R$ by $R'$ and $S$ by $S'$, and $\mathfrak m$ by any of the maximal ideals $\mathfrak m'$ lying over $\mathfrak m$ we reach the situation where the polynomial $f$ above split completely: $f(X) = \prod _{i = 1, \ldots , d} (X - a_ i)$ with $a_ i \in R$. Since $f(x) \in \mathfrak m$ we see that $x - a_ i \in \mathfrak m$ for some $i$. Finally, after replacing $x$ by $x - a_ i$ we may assume that $x \in \mathfrak m$.

To recapitulate: $R$ is a normal Nagata domain with fraction field $K$, $x \in K$ and $S$ is the subring of $K$ generated by $x$ and $R$, finally $\mathfrak m \subset S$ is a maximal ideal with $x \in \mathfrak m$. We have to show $S_{\mathfrak m}$ is N-1.

We will show that Lemma 10.162.12 applies to the local ring $S_{\mathfrak m}$ and the element $x$. This will imply that $S_{\mathfrak m}$ is analytically unramified, whereupon we see that it is N-1 by Lemma 10.162.10.

We have to check properties (1), (2), (3)(a) and (3)(b). Property (1) is trivial. Let $I = \mathop{\mathrm{Ker}}(R[X] \to S)$ where $X \mapsto x$. We claim that $I$ is generated by all linear forms $aX - b$ such that $ax = b$ in $K$. Clearly all these linear forms are in $I$. If $g = a_ d X^ d + \ldots a_1 X + a_0 \in I$, then we see that $a_ dx$ is integral over $R$ (Lemma 10.123.1) and hence $b := a_ dx \in R$ as $R$ is normal. Then $g - (a_ dX - b)X^{d - 1} \in I$ and we win by induction on the degree. As a consequence we see that

where

By Lemma 10.157.6 we see that $S/xS = R/J$ has no embedded primes as an $R$-module, hence as an $R/J$-module, hence as an $S/xS$-module, hence as an $S$-module. This proves property (2). Take such an associated prime $\mathfrak q \subset S$ with the property $\mathfrak q \subset \mathfrak m$ (so that it is an associated prime of $S_{\mathfrak m}/xS_{\mathfrak m}$ – it does not matter for the arguments). Then $\mathfrak q$ is minimal over $xS$ and hence has height $1$. By the sequence of equalities above we see that $\mathfrak p = R \cap \mathfrak q$ is an associated prime of $R/J$, and so has height $1$ (see Lemma 10.157.6). Thus $R_{\mathfrak p}$ is a discrete valuation ring and therefore $R_{\mathfrak p} \subset S_{\mathfrak q}$ is an equality. This shows that $S_{\mathfrak q}$ is regular. This proves property (3)(a). Finally, $(S/\mathfrak q)_{\mathfrak m}$ is a localization of $S/\mathfrak q$, which is a quotient of $S/xS = R/J$. Hence $(S/\mathfrak q)_{\mathfrak m}$ is a localization of a quotient of the Nagata ring $R$, hence Nagata (Lemmas 10.162.5 and 10.162.6) and hence analytically unramified (Lemma 10.162.13). This shows (3)(b) holds and we are done. $\square$

Proof. The Noetherian complete local ring case is Lemma 10.162.8. In the other cases you just check if $R/\mathfrak p$ is N-2 for every prime ideal $\mathfrak p$ of the ring. This is clear whenever $R/\mathfrak p$ is a field, i.e., $\mathfrak p$ is maximal. Hence for the Dedekind ring case we only need to check it when $\mathfrak p = (0)$. But since we assume the fraction field has characteristic zero Lemma 10.161.11 kicks in. $\square$

Proof. Consider the ring extension $A \subset B = A[x]/(x^ p - a)$. If $a$ does not have a $p$th root in $A$, then $B$ is a domain whose completion isn't reduced. This contradicts our earlier results, as $B$ is a Nagata ring (Proposition 10.162.15) and hence analytically unramified by Lemma 10.162.13. $\square$