## Tag `0BRA`

Chapter 10: Commutative Algebra > Section 10.45: Universal homeomorphisms

Lemma 10.45.6. Let $\varphi : R \to S$ be a ring map. Let $p$ be a prime number. Assume

- (a) $S$ is generated as an $R$-algebra by elements $x$ such that there exists an $n > 0$ with $x^{p^n} \in \varphi(R)$ and $p^nx \in \varphi(R)$, and
- (b) $\mathop{\rm Ker}(\varphi)$ is locally nilpotent,
Then $\varphi$ induces a homeomorphism of spectra and induces residue field extensions satisfying the equivalent conditions of Lemma 10.45.5. For any ring map $R \to R'$ the ring map $R' \to R' \otimes_R S$ also satisfies (a) and (b).

Proof.Assume (a) and (b). Note that (b) is equivalent to condition (2) of Lemma 10.45.3. Let $T \subset S$ be the set of elements $x \in S$ such that there exists an integer $n > 0$ such that $x^{p^n} , p^n x \in \varphi(R)$. We claim that $T = S$. This will prove that condition (1) of Lemma 10.45.3 holds and hence $\varphi$ induces a homeomorphism on spectra. By assumption (a) it suffices to show that $T \subset S$ is an $R$-sub algebra. If $x \in T$ and $y \in R$, then it is clear that $yx \in T$. Suppose $x, y \in T$ and $n, m > 0$ such that $x^{p^n}, y^{p^m}, p^n x, p^m y \in \varphi(R)$. Then $(xy)^{p^{n + m}}, p^{n + m}xy \in \varphi(R)$ hence $xy \in T$. We have $x + y \in T$ by Lemma 10.45.4 and the claim is proved.Since $\varphi$ induces a homeomorphism on spectra, it is in particular surjective on spectra which is a property preserved under any base change, see Lemma 10.29.3. Therefore for any $R \to R'$ the kernel of the ring map $R' \to R' \otimes_R S$ consists of nilpotent elements, see Lemma 10.29.6, in other words (b) holds for $R' \to R' \otimes_R S$. It is clear that (a) is preserved under base change. Finally, the condition on residue fields follows from (a) as generators for $S$ as an $R$-algebra map to generators for the residue field extensions. $\square$

The code snippet corresponding to this tag is a part of the file `algebra.tex` and is located in lines 10402–10415 (see updates for more information).

```
\begin{lemma}
\label{lemma-p-ring-map}
Let $\varphi : R \to S$ be a ring map. Let $p$ be a prime number. Assume
\begin{enumerate}
\item[(a)] $S$ is generated as an $R$-algebra by elements $x$ such
that there exists an $n > 0$ with $x^{p^n} \in \varphi(R)$ and
$p^nx \in \varphi(R)$, and
\item[(b)] $\Ker(\varphi)$ is locally nilpotent,
\end{enumerate}
Then $\varphi$ induces a homeomorphism of spectra and induces
residue field extensions satisfying the equivalent conditions
of Lemma \ref{lemma-p-ring-map-field}. For any ring map $R \to R'$
the ring map $R' \to R' \otimes_R S$ also satisfies (a) and (b).
\end{lemma}
\begin{proof}
Assume (a) and (b). Note that (b) is equivalent to condition (2)
of Lemma \ref{lemma-powers}. Let $T \subset S$ be the set of
elements $x \in S$ such that there exists an
integer $n > 0$ such that $x^{p^n} , p^n x \in \varphi(R)$.
We claim that $T = S$. This will prove that condition (1) of
Lemma \ref{lemma-powers} holds and hence $\varphi$ induces
a homeomorphism on spectra.
By assumption (a) it suffices to show that $T \subset S$ is an $R$-sub algebra.
If $x \in T$ and $y \in R$, then it is clear that $yx \in T$.
Suppose $x, y \in T$ and $n, m > 0$ such that
$x^{p^n}, y^{p^m}, p^n x, p^m y \in \varphi(R)$.
Then $(xy)^{p^{n + m}}, p^{n + m}xy \in \varphi(R)$
hence $xy \in T$. We have $x + y \in T$ by Lemma \ref{lemma-help-with-powers}
and the claim is proved.
\medskip\noindent
Since $\varphi$ induces a homeomorphism on spectra, it is in particular
surjective on spectra which is a property preserved under any base change, see
Lemma \ref{lemma-surjective-spec-radical-ideal}.
Therefore for any $R \to R'$ the kernel of the ring map
$R' \to R' \otimes_R S$ consists of nilpotent elements, see
Lemma \ref{lemma-image-dense-generic-points},
in other words (b) holds for $R' \to R' \otimes_R S$.
It is clear that (a) is preserved under base change.
Finally, the condition on residue fields follows from (a)
as generators for $S$ as an $R$-algebra map to generators for
the residue field extensions.
\end{proof}
```

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