## Tag `032E`

## 10.156. Nagata rings

Here is the definition.

Definition 10.156.1. Let $R$ be a ring.

- We say $R$ is
universally Japaneseif for any finite type ring map $R \to S$ with $S$ a domain we have that $S$ is N-2 (i.e., Japanese).- We say that $R$ is a
Nagata ringif $R$ is Noetherian and for every prime ideal $\mathfrak p$ the ring $R/\mathfrak p$ is N-2.

It is clear that a Noetherian universally Japanese ring is a Nagata ring. It is our goal to show that a Nagata ring is universally Japanese. This is not obvious at all, and requires some work. But first, here is a useful lemma.

Lemma 10.156.2. Let $R$ be a Nagata ring. Let $R \to S$ be essentially of finite type with $S$ reduced. Then the integral closure of $R$ in $S$ is finite over $R$.

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 10.30.6. 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 10.24.4 and 10.24.1. 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 $\subset K_i$ is finite over $\kappa(\mathfrak p_i)$, see Fields, Lemma 9.26.10. 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. $\square$Lemma 10.156.3. Let $R$ be a ring. To check that $R$ is universally Japanese it suffices to show: If $R \to S$ is of finite type, and $S$ a domain then $S$ is N-1.

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 10.35.16). We conclude that $R$ is universally Japanese. $\square$Lemma 10.156.4. If $R$ is universally Japanese then any algebra essentially of finite type over $R$ is universally Japanese.

Proof.The case of an algebra of finite type over $R$ is immediate from the definition. The general case follows on applying Lemma 10.155.3. $\square$Lemma 10.156.5. Let $R$ be a Nagata ring. If $R \to S$ is a quasi-finite ring map (for example finite) then $S$ is a Nagata ring also.

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 10.155.5 as desired. $\square$Lemma 10.156.6. A localization of a Nagata ring is a Nagata ring.

Proof.Clear from Lemma 10.155.3. $\square$Lemma 10.156.7. Let $R$ be a ring. Let $f_1, \ldots, f_n \in R$ generate the unit ideal.

- If each $R_{f_i}$ is universally Japanese then so is $R$.
- If each $R_{f_i}$ is Nagata then so is $R$.

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 10.155.4. 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 10.155.4. This proves (2). $\square$

Lemma 10.156.8. A Noetherian complete local ring is a Nagata ring.

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 10.154.2. Hence it suffices to show that a Noetherian complete local domain $R$ is N-2. By Lemmas 10.155.5 and 10.154.11 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 10.155.17. 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 10.155.16 once more with $x = p \in \Lambda$ we reduce yet again to the case of a field. Thus we win. $\square$

Definition 10.156.9. Let $(R, \mathfrak m)$ be a Noetherian local ring. We say $R$ is

analytically unramifiedif its completion $R^\wedge = \mathop{\rm lim}\nolimits_n R/\mathfrak m^n$ is reduced. A prime ideal $\mathfrak p \subset R$ is said to beanalytically unramifiedif $R/\mathfrak p$ is analytically unramified.At this point we know the following are true for any Noetherian local ring $R$: The map $R \to R^\wedge$ is a faithfully flat local ring homomorphism (Lemma 10.96.3). The completion $R^\wedge$ is Noetherian (Lemma 10.96.5) and complete (Lemma 10.96.4). Hence the completion $R^\wedge$ is a Nagata ring (Lemma 10.156.8). Moreover, we have seen in Section 10.154 that $R^\wedge$ is a quotient of a regular local ring (Theorem 10.154.8), and hence universally catenary (Remark 10.154.9).

Lemma 10.156.10. Let $(R, \mathfrak m)$ be a Noetherian local ring.

- If $R$ is analytically unramified, then $R$ is reduced.
- If $R$ is analytically unramified, then each minimal prime of $R$ is analytically unramified.
- If $R$ is reduced with minimal primes $\mathfrak q_1, \ldots, \mathfrak q_t$, and each $\mathfrak q_i$ is analytically unramified, then $R$ is analytically unramified.
- If $R$ is analytically unramified, then the integral closure of $R$ in its total ring of fractions $Q(R)$ is finite over $R$.
- If $R$ is a domain and analytically unramified, then $R$ is N-1.

Proof.In this proof we will use the remarks immediately following Definition 10.156.9. 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 10.62.6). 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 10.96.2). 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 10.96.2) 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 $$ Q(R^\wedge) = R^\wedge_{\mathfrak p_1} \times \ldots \times R^\wedge_{\mathfrak p_s} $$ with each $R^\wedge_{\mathfrak p_i}$ a field as $R^\wedge$ is reduced (see Lemma 10.24.4). 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 10.35.13). 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). $\square$

Lemma 10.156.11. Let $R$ be a Noetherian local ring. Let $\mathfrak p \subset R$ be a prime. Assume

- $R_{\mathfrak p}$ is a discrete valuation ring, and
- $\mathfrak p$ is analytically unramified.
Then for any associated prime $\mathfrak q$ of $R^\wedge/\mathfrak pR^\wedge$ the local ring $(R^\wedge)_{\mathfrak q}$ is a discrete valuation ring.

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. $\square$Lemma 10.156.12. Let $(R, \mathfrak m)$ be a Noetherian local domain. Let $x \in \mathfrak m$. Assume

- $x \not = 0$,
- $R/xR$ has no embedded primes, and
- for each associated prime $\mathfrak p \subset R$ of $R/xR$ we have

- the local ring $R_{\mathfrak p}$ is regular, and
- $\mathfrak p$ is analytically unramified.
Then $R$ is analytically unramified.

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 10.156.11 we see that $(R^\wedge)_{\mathfrak q_{ij}}$ is regular. By Lemma 10.64.3 we see that $$ \text{Ass}_{R^\wedge}(R^\wedge/xR^\wedge) = \bigcup\nolimits_{\mathfrak p \in \text{Ass}_R(R/xR)} \text{Ass}_{R^\wedge}(R^\wedge/\mathfrak pR^\wedge) = \{\mathfrak q_{ij}\}. $$ Let $y \in R^\wedge$ with $y^2 = 0$. As $(R^\wedge)_{\mathfrak q_{ij}}$ is regular, and hence a domain (Lemma 10.105.2) 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 10.62.19. 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 10.50.4). $\square$Lemma 10.156.13. Let $(R, \mathfrak m)$ be a local ring. If $R$ is Noetherian, a domain, and Nagata, then $R$ is analytically unramified.

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 10.156.5 we see that $S$ is Nagata. By Lemma 10.111.4 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 $$ (\mathfrak m_1 \cap \ldots \cap \mathfrak m_t)^n \subset \mathfrak mS $$ for sufficiently large $n$ as $S/\mathfrak mS$ is Artinian. By Lemma 10.96.2 $R^\wedge \to S^\wedge$ is an injective map, and by the Chinese Remainder Lemma 10.14.3 combined with Lemma 10.95.9 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 10.156.12. 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 10.151.6. 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 10.118.7). Thus (3)(a) holds. Finally (3)(b) holds by induction hypothesis, since $R/\mathfrak p$ is Nagata (by Lemma 10.156.5 or directly from the definition). Thus we conclude $R$ is analytically unramified. $\square$

Lemma 10.156.14. Let $(R, \mathfrak m)$ be a Noetherian local ring. The following are equivalent

- $R$ is Nagata,
- for $R \to S$ finite with $S$ a domain and $\mathfrak m' \subset S$ maximal the local ring $S_{\mathfrak m'}$ is analytically unramified,
- for $(R, \mathfrak m) \to (S, \mathfrak m')$ finite local homomorphism with $S$ a domain, then $S$ is analytically unramified.

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 10.156.5. Hence the local ring $S_{\mathfrak m'}$ is Nagata (Lemma 10.156.6). Thus it is analytically unramified by Lemma 10.156.13. 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 10.156.10 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. $\square$

The following proposition says in particular that an algebra of finite type over a Nagata ring is a Nagata ring.

Proposition 10.156.15 (Nagata). Let $R$ be a ring. The following are equivalent:

- $R$ is a Nagata ring,
- any finite type $R$-algebra is Nagata, and
- $R$ is universally Japanese and Noetherian.

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 10.155.7). 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 10.155.13 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 $L \subset M$ 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 10.156.5). 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 10.155.15 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 $\kappa(\mathfrak n) \subset \kappa(\mathfrak m)$ is finite (Theorem 10.33.1) 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 $K \subset K''$ 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 10.156.5). 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 10.155.15, which in turn implies that $S_{\mathfrak m}$ is N-1 by Lemma 10.155.7. 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 10.156.12 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 10.156.10.

We have to check properties (1), (2), (3)(a) and (3)(b). Property (1) is trivial. Let $I = \mathop{\rm Ker}(R[X] \to S)$ 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 10.122.1) 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 $$ S/xS = R[X]/(X, I) = R/J $$ where $$ J = \{b \in R \mid ax = b \text{ for some }a \in R\} = xR \cap R $$ By Lemma 10.151.6 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 10.151.6). 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 10.156.5 and 10.156.6) and hence analytically unramified (Lemma 10.156.13). This shows (3)(b) holds and we are done. $\square$

Proposition 10.156.16. The following types of rings are Nagata and in particular universally Japanese:

- fields,
- Noetherian complete local rings,
- $\mathbf{Z}$,
- Dedekind domains with fraction field of characteristic zero,
- finite type ring extensions of any of the above.

Proof.The Noetherian complete local ring case is Lemma 10.156.8. 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 10.155.11 kicks in. $\square$Example 10.156.17. A discrete valuation ring is Nagata if and only if it is N-2 (this follows immediately from the definition). The discrete valuation ring $A$ of Example 10.118.5 is not Nagata, i.e., it is not N-2. Namely, the finite extension $A \subset R = A[f]$ is not N-1. To see this say $f = \sum a_i x^i$. For every $n \geq 1$ set $g_n = \sum_{i < n} a_i x^i \in A$. Then $h_n = (f - g_n)/x^n$ is an element of the fraction field of $R$ and $h_n^p \in k^p[[x]] \subset A$. Hence the integral closure $R'$ of $R$ contains $h_1, h_2, h_3, \ldots$. Now, if $R'$ were finite over $R$ and hence $A$, then $f = x^n h_n + g_n$ would be contained in the submodule $A + x^nR'$ for all $n$. By Artin-Rees this would imply $f \in A$ (Lemma 10.50.4), a contradiction.

Lemma 10.156.18. Let $(A, \mathfrak m)$ be a Noetherian local domain which is Nagata and has fraction field of characteristic $p$. If $a \in A$ has a $p$th root in $A^\wedge$, then $a$ has a $p$th root in $A$.

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 10.156.15) and hence analytically unramified by Lemma 10.156.13. $\square$

The code snippet corresponding to this tag is a part of the file `algebra.tex` and is located in lines 43032–43710 (see updates for more information).

```
\section{Nagata rings}
\label{section-nagata}
\noindent
Here is the definition.
\begin{definition}
\label{definition-nagata}
Let $R$ be a ring.
\begin{enumerate}
\item We say $R$ is {\it universally Japanese} if for any finite
type ring map $R \to S$ with $S$ a domain we have that $S$ is N-2
(i.e., Japanese).
\item We say that $R$ is a {\it Nagata ring} if $R$ is Noetherian and
for every prime ideal $\mathfrak p$ the ring $R/\mathfrak p$ is N-2.
\end{enumerate}
\end{definition}
\noindent
It is clear that a Noetherian universally Japanese ring is a Nagata ring.
It is our goal to show that a Nagata ring is universally Japanese. This is
not obvious at all, and requires some work. But first, here is a useful
lemma.
\begin{lemma}
\label{lemma-nagata-in-reduced-finite-type-finite-integral-closure}
Let $R$ be a Nagata ring.
Let $R \to S$ be essentially of finite type with $S$ reduced.
Then the integral closure of $R$ in $S$ is finite over $R$.
\end{lemma}
\begin{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 \ref{lemma-Noetherian-irreducible-components}.
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 \ref{lemma-total-ring-fractions-no-embedded-points}
and \ref{lemma-minimal-prime-reduced-ring}.
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 $\subset K_i$
is finite over $\kappa(\mathfrak p_i)$, see
Fields, Lemma \ref{fields-lemma-algebraic-closure-in-finitely-generated}.
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.
\end{proof}
\begin{lemma}
\label{lemma-check-universally-japanese}
Let $R$ be a ring.
To check that $R$ is universally Japanese it suffices to show:
If $R \to S$ is of finite type, and $S$ a domain then $S$ is N-1.
\end{lemma}
\begin{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 \ref{lemma-finite-transitive})
and $S''$ is the integral closure of $S$ in $L$
(Lemma \ref{lemma-integral-closure-transitive}).
We conclude that $R$ is universally Japanese.
\end{proof}
\begin{lemma}
\label{lemma-universally-japanese}
If $R$ is universally Japanese then any algebra essentially of finite type
over $R$ is universally Japanese.
\end{lemma}
\begin{proof}
The case of an algebra of finite type over $R$ is immediate from
the definition. The general case follows on applying
Lemma \ref{lemma-localize-N}.
\end{proof}
\begin{lemma}
\label{lemma-quasi-finite-over-nagata}
Let $R$ be a Nagata ring.
If $R \to S$ is a quasi-finite ring map (for example finite)
then $S$ is a Nagata ring also.
\end{lemma}
\begin{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 \ref{lemma-quasi-finite-over-Noetherian-japanese}
as desired.
\end{proof}
\begin{lemma}
\label{lemma-nagata-localize}
A localization of a Nagata ring is a Nagata ring.
\end{lemma}
\begin{proof}
Clear from Lemma \ref{lemma-localize-N}.
\end{proof}
\begin{lemma}
\label{lemma-nagata-local}
Let $R$ be a ring. Let $f_1, \ldots, f_n \in R$ generate the
unit ideal.
\begin{enumerate}
\item If each $R_{f_i}$ is universally Japanese then so is $R$.
\item If each $R_{f_i}$ is Nagata then so is $R$.
\end{enumerate}
\end{lemma}
\begin{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 \ref{lemma-Japanese-local}. This proves (1).
\medskip\noindent
If each $R_{f_i}$ is Nagata, then each $R_{f_i}$ is Noetherian and
hence $R$ is Noetherian, see Lemma \ref{lemma-cover}. 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 \ref{lemma-Japanese-local}. This proves (2).
\end{proof}
\begin{lemma}
\label{lemma-Noetherian-complete-local-Nagata}
A Noetherian complete local ring is a Nagata ring.
\end{lemma}
\begin{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 \ref{lemma-quotient-complete-local}.
Hence it suffices to show that a Noetherian complete local
domain $R$ is N-2. By
Lemmas \ref{lemma-quasi-finite-over-Noetherian-japanese}
and \ref{lemma-complete-local-Noetherian-domain-finite-over-regular}
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.
\medskip\noindent
In the case $k[[X_1, \ldots, X_d]]$ we reduce to the statement that a
field is N-2 by Lemma \ref{lemma-power-series-over-N-2}. 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 \ref{lemma-tate-japanese}
once more with $x = p \in \Lambda$ we reduce yet again to the case
of a field. Thus we win.
\end{proof}
\begin{definition}
\label{definition-analytically-unramified}
Let $(R, \mathfrak m)$ be a Noetherian local ring.
We say $R$ is {\it analytically unramified} if its completion
$R^\wedge = \lim_n R/\mathfrak m^n$ is reduced.
A prime ideal $\mathfrak p \subset R$ is said to be
{\it analytically unramified} if $R/\mathfrak p$ is analytically
unramified.
\end{definition}
\noindent
At this point we know
the following are true for any Noetherian local ring $R$:
The map $R \to R^\wedge$ is a faithfully flat local ring homomorphism
(Lemma \ref{lemma-completion-faithfully-flat}).
The completion $R^\wedge$ is Noetherian
(Lemma \ref{lemma-completion-Noetherian})
and complete (Lemma \ref{lemma-completion-complete}).
Hence the completion $R^\wedge$ is a Nagata ring
(Lemma \ref{lemma-Noetherian-complete-local-Nagata}).
Moreover, we have seen in Section \ref{section-cohen-structure-theorem}
that $R^\wedge$ is
a quotient of a regular local ring
(Theorem \ref{theorem-cohen-structure-theorem}), and hence
universally catenary
(Remark \ref{remark-Noetherian-complete-local-ring-universally-catenary}).
\begin{lemma}
\label{lemma-analytically-unramified-easy}
Let $(R, \mathfrak m)$ be a Noetherian local ring.
\begin{enumerate}
\item If $R$ is analytically unramified, then $R$ is reduced.
\item If $R$ is analytically unramified, then each minimal prime of
$R$ is analytically unramified.
\item If $R$ is reduced with minimal primes
$\mathfrak q_1, \ldots, \mathfrak q_t$, and each $\mathfrak q_i$
is analytically unramified, then $R$ is analytically unramified.
\item If $R$ is analytically unramified, then the integral closure
of $R$ in its total ring of fractions $Q(R)$ is finite over $R$.
\item If $R$ is a domain and analytically unramified, then $R$ is N-1.
\end{enumerate}
\end{lemma}
\begin{proof}
In this proof we will use the remarks immediately following
Definition \ref{definition-analytically-unramified}.
As $R \to R^\wedge$ is a faithfully flat local ring homomorphism
it is injective and (1) follows.
\medskip\noindent
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 \ref{proposition-minimal-primes-associated-primes}).
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 \ref{lemma-completion-flat}).
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.
\medskip\noindent
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 \ref{lemma-completion-flat})
we see that
$R^\wedge \subset (R/\mathfrak q_1)^\wedge \times \ldots \times
(R/\mathfrak q_t)^\wedge$. Hence (3) is clear.
\medskip\noindent
Assume $R$ is analytically unramified.
Let $\mathfrak p_1, \ldots, \mathfrak p_s$ be the minimal primes
of $R^\wedge$. Then we see that
$$
Q(R^\wedge) =
R^\wedge_{\mathfrak p_1} \times \ldots \times R^\wedge_{\mathfrak p_s}
$$
with each $R^\wedge_{\mathfrak p_i}$ a field
as $R^\wedge$ is reduced (see
Lemma \ref{lemma-total-ring-fractions-no-embedded-points}).
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 \ref{lemma-base-change-integral}).
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).
\medskip\noindent
Part (5) is a special case of part (4).
\end{proof}
\begin{lemma}
\label{lemma-codimension-1-analytically-unramified}
Let $R$ be a Noetherian local ring.
Let $\mathfrak p \subset R$ be a prime.
Assume
\begin{enumerate}
\item $R_{\mathfrak p}$ is a discrete valuation ring, and
\item $\mathfrak p$ is analytically unramified.
\end{enumerate}
Then for any associated prime $\mathfrak q$ of $R^\wedge/\mathfrak pR^\wedge$
the local ring $(R^\wedge)_{\mathfrak q}$ is a discrete valuation ring.
\end{lemma}
\begin{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.
\end{proof}
\begin{lemma}
\label{lemma-criterion-analytically-unramified}
Let $(R, \mathfrak m)$ be a Noetherian local domain.
Let $x \in \mathfrak m$. Assume
\begin{enumerate}
\item $x \not = 0$,
\item $R/xR$ has no embedded primes, and
\item for each associated prime $\mathfrak p \subset R$
of $R/xR$ we have
\begin{enumerate}
\item the local ring $R_{\mathfrak p}$ is regular, and
\item $\mathfrak p$ is analytically unramified.
\end{enumerate}
\end{enumerate}
Then $R$ is analytically unramified.
\end{lemma}
\begin{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 \ref{lemma-codimension-1-analytically-unramified}
we see that $(R^\wedge)_{\mathfrak q_{ij}}$ is regular.
By Lemma \ref{lemma-bourbaki} we see that
$$
\text{Ass}_{R^\wedge}(R^\wedge/xR^\wedge)
=
\bigcup\nolimits_{\mathfrak p \in \text{Ass}_R(R/xR)}
\text{Ass}_{R^\wedge}(R^\wedge/\mathfrak pR^\wedge)
=
\{\mathfrak q_{ij}\}.
$$
Let $y \in R^\wedge$ with $y^2 = 0$.
As $(R^\wedge)_{\mathfrak q_{ij}}$ is regular, and hence a domain
(Lemma \ref{lemma-regular-domain})
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 \ref{lemma-zero-at-ass-zero}.
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 \ref{lemma-intersect-powers-ideal-module-zero}).
\end{proof}
\begin{lemma}
\label{lemma-local-nagata-domain-analytically-unramified}
Let $(R, \mathfrak m)$ be a local ring.
If $R$ is Noetherian, a domain, and Nagata, then $R$ is
analytically unramified.
\end{lemma}
\begin{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 $< d$.
\medskip\noindent
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 \ref{lemma-quasi-finite-over-nagata} we see that
$S$ is Nagata. By Lemma \ref{lemma-integral-sub-dim-equal} 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
$$
(\mathfrak m_1 \cap \ldots \cap \mathfrak m_t)^n \subset \mathfrak mS
$$
for sufficiently large $n$ as $S/\mathfrak mS$ is Artinian.
By Lemma \ref{lemma-completion-flat} $R^\wedge \to S^\wedge$
is an injective map, and by the Chinese Remainder
Lemma \ref{lemma-chinese-remainder} combined with
Lemma \ref{lemma-change-ideal-completion} 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.
\medskip\noindent
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 \ref{lemma-criterion-analytically-unramified}.
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 \ref{lemma-normal-domain-intersection-localizations-height-1}.
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 \ref{lemma-characterize-dvr}). Thus (3)(a) holds.
Finally (3)(b) holds by induction hypothesis, since
$R/\mathfrak p$ is Nagata (by Lemma \ref{lemma-quasi-finite-over-nagata}
or directly from the definition).
Thus we conclude $R$ is analytically unramified.
\end{proof}
\begin{lemma}
\label{lemma-local-nagata-and-analytically-unramified}
Let $(R, \mathfrak m)$ be a Noetherian local ring. The following
are equivalent
\begin{enumerate}
\item $R$ is Nagata,
\item for $R \to S$ finite with $S$ a domain and $\mathfrak m' \subset S$
maximal the local ring $S_{\mathfrak m'}$ is analytically unramified,
\item for $(R, \mathfrak m) \to (S, \mathfrak m')$ finite
local homomorphism with $S$ a domain, then $S$ is analytically
unramified.
\end{enumerate}
\end{lemma}
\begin{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 \ref{lemma-quasi-finite-over-nagata}.
Hence the local ring $S_{\mathfrak m'}$ is Nagata
(Lemma \ref{lemma-nagata-localize}). Thus it is analytically
unramified by Lemma \ref{lemma-local-nagata-domain-analytically-unramified}.
It is clear that (2) implies (3).
\medskip\noindent
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 \ref{lemma-analytically-unramified-easy}
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.
\end{proof}
\noindent
The following proposition says in particular that an algebra of finite
type over a Nagata ring is a Nagata ring.
\begin{proposition}[Nagata]
\label{proposition-nagata-universally-japanese}
Let $R$ be a ring. The following are equivalent:
\begin{enumerate}
\item $R$ is a Nagata ring,
\item any finite type $R$-algebra is Nagata, and
\item $R$ is universally Japanese and Noetherian.
\end{enumerate}
\end{proposition}
\begin{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.
\medskip\noindent
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
$S \cong R[x]/I$.
\medskip\noindent
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.
\medskip\noindent
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 \ref{lemma-finite-extension-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 then $S$ is N-2.
\medskip\noindent
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 \ref{lemma-polynomial-ring-N-2} 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.
\medskip\noindent
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 $L \subset M$
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 \ref{lemma-quasi-finite-over-nagata}).
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.
\medskip\noindent
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 \ref{lemma-characterize-N-1} it suffices to show that
$S_{\mathfrak m}$ is N-1 for every maximal ideal $\mathfrak m$ of $S$.
\medskip\noindent
With assumptions as in the preceding paragraph, pick such a maximal
ideal and set $\mathfrak n = R \cap \mathfrak m$. The residue field
extension $\kappa(\mathfrak n) \subset \kappa(\mathfrak m)$ is finite
(Theorem \ref{theorem-nullstellensatz}) 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 $K \subset K''$ 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 \ref{lemma-quasi-finite-over-nagata}).
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 \ref{lemma-characterize-N-1}, which in turn
implies that $S_{\mathfrak m}$ is N-1 by
Lemma \ref{lemma-finite-extension-N-2}.
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 $x \in \mathfrak m$.
\medskip\noindent
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.
\medskip\noindent
We will show that Lemma \ref{lemma-criterion-analytically-unramified}
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 \ref{lemma-analytically-unramified-easy}.
\medskip\noindent
We have to check properties (1), (2), (3)(a) and (3)(b).
Property (1) is trivial.
Let $I = \Ker(R[X] \to S)$ 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 \ref{lemma-make-integral-trivial})
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
$$
S/xS = R[X]/(X, I) = R/J
$$
where
$$
J = \{b \in R \mid ax = b \text{ for some }a \in R\} = xR \cap R
$$
By Lemma \ref{lemma-normal-domain-intersection-localizations-height-1}
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 \ref{lemma-normal-domain-intersection-localizations-height-1}).
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 \ref{lemma-quasi-finite-over-nagata}
and \ref{lemma-nagata-localize})
and hence analytically unramified
(Lemma \ref{lemma-local-nagata-domain-analytically-unramified}).
This shows (3)(b) holds and we are done.
\end{proof}
\begin{proposition}
\label{proposition-ubiquity-nagata}
The following types of rings are Nagata and in particular universally Japanese:
\begin{enumerate}
\item fields,
\item Noetherian complete local rings,
\item $\mathbf{Z}$,
\item Dedekind domains with fraction field of characteristic zero,
\item finite type ring extensions of any of the above.
\end{enumerate}
\end{proposition}
\begin{proof}
The Noetherian complete local ring case is
Lemma \ref{lemma-Noetherian-complete-local-Nagata}.
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 \ref{lemma-domain-char-zero-N-1-2} kicks in.
\end{proof}
\begin{example}
\label{example-nonjapanese-dvr}
A discrete valuation ring is Nagata if and only if it is N-2
(this follows immediately from the definition). The discrete valuation
ring $A$ of Example \ref{example-bad-dvr-char-p} is not Nagata, i.e.,
it is not N-2. Namely, the finite extension
$A \subset R = A[f]$ is not N-1. To see this say $f = \sum a_i x^i$.
For every $n \geq 1$ set $g_n = \sum_{i < n} a_i x^i \in A$.
Then $h_n = (f - g_n)/x^n$ is an element of the fraction field of $R$
and $h_n^p \in k^p[[x]] \subset A$. Hence the integral closure $R'$
of $R$ contains $h_1, h_2, h_3, \ldots$. Now, if $R'$
were finite over $R$ and hence $A$, then $f = x^n h_n + g_n$
would be contained in the submodule $A + x^nR'$ for all $n$. By
Artin-Rees this would imply $f \in A$
(Lemma \ref{lemma-intersect-powers-ideal-module-zero}), a contradiction.
\end{example}
\begin{lemma}
\label{lemma-nagata-pth-roots}
Let $(A, \mathfrak m)$ be a Noetherian local domain which is Nagata
and has fraction field of characteristic $p$. If $a \in A$ has a
$p$th root in $A^\wedge$, then $a$ has a $p$th root in $A$.
\end{lemma}
\begin{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 \ref{proposition-nagata-universally-japanese})
and hence analytically unramified by
Lemma \ref{lemma-local-nagata-domain-analytically-unramified}.
\end{proof}
```

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