Proposition 15.49.10. Let $R$ be a G-ring. If $R \to S$ is essentially of finite type then $S$ is a G-ring.

Proof. Since being a G-ring is a property of the local rings it is clear that a localization of a G-ring is a G-ring. Conversely, if every localization at a prime is a G-ring, then the ring is a G-ring. Thus it suffices to show that $S_\mathfrak q$ is a G-ring for every finite type $R$-algebra $S$ and every prime $\mathfrak q$ of $S$. Writing $S$ as a quotient of $R[x_1, \ldots , x_ n]$ we see from Lemma 15.49.3 that it suffices to prove that $R[x_1, \ldots , x_ n]$ is a G-ring. By induction on $n$ it suffices to prove that $R[x]$ is a G-ring. Let $\mathfrak q \subset R[x]$ be a maximal ideal. By Lemma 15.49.7 it suffices to show that

$R[x]_\mathfrak q \longrightarrow R[x]_\mathfrak q^\wedge$

is regular. If $\mathfrak q$ lies over $\mathfrak p \subset R$, then we may replace $R$ by $R_\mathfrak p$. Hence we may assume that $R$ is a Noetherian local G-ring with maximal ideal $\mathfrak m$ and that $\mathfrak q \subset R[x]$ lies over $\mathfrak m$. Note that there is a unique prime $\mathfrak q' \subset R^\wedge [x]$ lying over $\mathfrak q$. Consider the diagram

$\xymatrix{ R[x]_\mathfrak q^\wedge \ar[r] & (R^\wedge [x]_{\mathfrak q'})^\wedge \\ R[x]_\mathfrak q \ar[r] \ar[u] & R^\wedge [x]_{\mathfrak q'} \ar[u] }$

Since $R$ is a G-ring the lower horizontal arrow is regular (as a localization of a base change of the regular ring map $R \to R^\wedge$). Suppose we can prove the right vertical arrow is regular. Then it follows that the composition $R[x]_\mathfrak q \to (R^\wedge [x]_{\mathfrak q'})^\wedge$ is regular, and hence the left vertical arrow is regular by Lemma 15.40.7. Hence we see that we may assume $R$ is a Noetherian complete local ring and $\mathfrak q$ a prime lying over the maximal ideal of $R$.

Let $R$ be a Noetherian complete local ring and let $\mathfrak q \subset R[x]$ be a maximal ideal lying over the maximal ideal of $R$. Let $\mathfrak r \subset \mathfrak q$ be a prime ideal. We want to show that $R[x]_\mathfrak q^\wedge \otimes _{R[x]} \kappa (\mathfrak r)$ is a geometrically regular algebra over $\kappa (\mathfrak r)$. Set $\mathfrak p = R \cap \mathfrak r$. Then we can replace $R$ by $R/\mathfrak p$ and $\mathfrak q$ and $\mathfrak r$ by their images in $R/\mathfrak p[x]$, see Lemma 15.49.2. Hence we may assume that $R$ is a domain and that $\mathfrak r \cap R = (0)$.

By Algebra, Lemma 10.154.11 we can find $R_0 \subset R$ which is regular and such that $R$ is finite over $R_0$. Applying Lemma 15.49.3 we see that it suffices to prove $R[x]_\mathfrak q^\wedge \otimes _{R[x]} \kappa (\mathfrak r)$ is geometrically regular over $\kappa (r)$ when, in addition to the above, $R$ is a regular complete local ring.

Now $R$ is a regular complete local ring, we have $\mathfrak q \subset \mathfrak r \subset R[x]$, we have $(0) = R \cap \mathfrak r$ and $\mathfrak q$ is a maximal ideal lying over the maximal ideal of $R$. Since $R$ is regular the ring $R[x]$ is regular (Algebra, Lemma 10.157.10). Hence the localization $R[x]_\mathfrak q$ is regular. Hence the completions $R[x]_\mathfrak q^\wedge$ are regular, see Lemma 15.42.4. Hence the fibre $R[x]_{\mathfrak q}^\wedge \otimes _{R[x]} \kappa (\mathfrak r)$ is, as a localization of $R[x]_\mathfrak q^\wedge$, also regular. Thus we are done if the characteristic of the fraction field of $R$ is $0$.

If the characteristic of $R$ is positive, then $R = k[[x_1, \ldots , x_ n]]$. In this case we split the argument in two subcases:

1. The case $\mathfrak r = (0)$. The result is a direct consequence of Lemma 15.49.5.

2. The case $\mathfrak r \not= (0)$. This is Lemma 15.49.9.

$\square$

## Comments (2)

Comment #1405 by Matthew Emerton on

Is there a typo on line 5 from the bottom? It says $f.f.(R)$, but the $f.f.$ doesn't seem to mean anything, so maybe it is just supposed to $R$?

Comment #1406 by sdf on

I believe this notation means field of fractions. See e.g. lemma 10.113.1 tag 00P0

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