## Tag `00HY`

Chapter 10: Commutative Algebra > Section 10.40: Going up and going down

Lemma 10.40.5. Let $R \to S$ be a ring map. Let $T \subset \mathop{\rm Spec}(R)$ be the image of $\mathop{\rm Spec}(S)$. If $T$ is stable under specialization, then $T$ is closed.

Proof.We give two proofs.First proof. Let $\mathfrak p \subset R$ be a prime ideal such that the corresponding point of $\mathop{\rm Spec}(R)$ is in the closure of $T$. This means that for every $f \in R$, $f \not \in \mathfrak p$ we have $D(f) \cap T \not = \emptyset$. Note that $D(f) \cap T$ is the image of $\mathop{\rm Spec}(S_f)$ in $\mathop{\rm Spec}(R)$. Hence we conclude that $S_f \not = 0$. In other words, $1 \not = 0$ in the ring $S_f$. Since $S_{\mathfrak p}$ is the directed limit of the rings $S_f$ we conclude that $1 \not = 0$ in $S_{\mathfrak p}$. In other words, $S_{\mathfrak p} \not = 0$ and considering the image of $\mathop{\rm Spec}(S_{\mathfrak p}) \to \mathop{\rm Spec}(S) \to \mathop{\rm Spec}(R)$ we see there exists a $\mathfrak p' \in T$ with $\mathfrak p' \subset \mathfrak p$. As we assumed $T$ closed under specialization we conclude $\mathfrak p$ is a point of $T$ as desired.

Second proof. Let $I = \mathop{\rm Ker}(R \to S)$. We may replace $R$ by $R/I$. In this case the ring map $R \to S$ is injective. By Lemma 10.29.5 all the minimal primes of $R$ are contained in the image $T$. Hence if $T$ is stable under specialization then it contains all primes. $\square$

The code snippet corresponding to this tag is a part of the file `algebra.tex` and is located in lines 9276–9281 (see updates for more information).

```
\begin{lemma}
\label{lemma-image-stable-specialization-closed}
Let $R \to S$ be a ring map. Let $T \subset \Spec(R)$
be the image of $\Spec(S)$. If $T$ is stable under specialization,
then $T$ is closed.
\end{lemma}
\begin{proof}
We give two proofs.
\medskip\noindent
First proof. Let $\mathfrak p \subset R$ be a prime ideal such that
the corresponding point of $\Spec(R)$ is in the closure
of $T$. This means that for every $f \in R$, $f \not \in \mathfrak p$
we have $D(f) \cap T \not = \emptyset$. Note that $D(f) \cap T$
is the image of $\Spec(S_f)$ in $\Spec(R)$. Hence
we conclude that $S_f \not = 0$. In other words, $1 \not = 0$ in
the ring $S_f$. Since $S_{\mathfrak p}$ is the directed limit
of the rings $S_f$ we conclude that $1 \not = 0$ in
$S_{\mathfrak p}$. In other words, $S_{\mathfrak p} \not = 0$ and
considering the image of $\Spec(S_{\mathfrak p})
\to \Spec(S) \to \Spec(R)$ we see there exists
a $\mathfrak p' \in T$ with $\mathfrak p' \subset \mathfrak p$.
As we assumed $T$ closed under specialization we conclude $\mathfrak p$
is a point of $T$ as desired.
\medskip\noindent
Second proof. Let $I = \Ker(R \to S)$. We may replace $R$ by $R/I$.
In this case the ring map $R \to S$ is injective.
By Lemma \ref{lemma-injective-minimal-primes-in-image}
all the minimal primes of $R$ are contained in the image $T$. Hence
if $T$ is stable under specialization then it contains all primes.
\end{proof}
```

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