Loading web-font TeX/Main/Regular

The Stacks project

Lemma 10.162.10. Let (R, \mathfrak m) be a Noetherian local ring.

  1. If R is analytically unramified, then R is reduced.

  2. If R is analytically unramified, then each minimal prime of R is analytically unramified.

  3. 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.

  4. If R is analytically unramified, then the integral closure of R in its total ring of fractions Q(R) is finite over R.

  5. 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 10.162.9. 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

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 10.25.4). 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


Comments (0)

There are also:

  • 2 comment(s) on Section 10.162: Nagata rings

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.