Proposition 10.162.15 (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.
Proposition 10.162.15 (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 10.161.7). 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
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