The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

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 , but the doesn't seem to mean anything, so maybe it is just supposed to ?

Comment #1406 by sdf on

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


Post a comment

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.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 07PV. Beware of the difference between the letter 'O' and the digit '0'.