Lemma 10.30.3. Let $\varphi : R \to S$ be a ring map. The following are equivalent:

The map $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ is surjective.

For any ideal $I \subset R$ the inverse image of $\sqrt{IS}$ in $R$ is equal to $\sqrt{I}$.

For any radical ideal $I \subset R$ the inverse image of $IS$ in $R$ is equal to $I$.

For every prime $\mathfrak p$ of $R$ the inverse image of $\mathfrak p S$ in $R$ is $\mathfrak p$.

In this case the same is true after any base change: Given a ring map $R \to R'$ the ring map $R' \to R' \otimes _ R S$ has the equivalent properties (1), (2), (3) as well.

**Proof.**
If $J \subset S$ is an ideal, then $\sqrt{\varphi ^{-1}(J)} = \varphi ^{-1}(\sqrt{J})$. This shows that (2) and (3) are equivalent. The implication (3) $\Rightarrow $ (4) is immediate. If $I \subset R$ is a radical ideal, then Lemma 10.17.2 guarantees that $I = \bigcap _{I \subset \mathfrak p} \mathfrak p$. Hence (4) $\Rightarrow $ (2). By Lemma 10.17.9 we have $\mathfrak p = \varphi ^{-1}(\mathfrak p S)$ if and only if $\mathfrak p$ is in the image. Hence (1) $\Leftrightarrow $ (4). Thus (1), (2), (3), and (4) are equivalent.

Assume (1) holds. Let $R \to R'$ be a ring map. Let $\mathfrak p' \subset R'$ be a prime ideal lying over the prime $\mathfrak p$ of $R$. To see that $\mathfrak p'$ is in the image of $\mathop{\mathrm{Spec}}(R' \otimes _ R S) \to \mathop{\mathrm{Spec}}(R')$ we have to show that $(R' \otimes _ R S) \otimes _{R'} \kappa (\mathfrak p')$ is not zero, see Lemma 10.17.9. But we have

\[ (R' \otimes _ R S) \otimes _{R'} \kappa (\mathfrak p') = S \otimes _ R \kappa (\mathfrak p) \otimes _{\kappa (\mathfrak p)} \kappa (\mathfrak p') \]

which is not zero as $S \otimes _ R \kappa (\mathfrak p)$ is not zero by assumption and $\kappa (\mathfrak p) \to \kappa (\mathfrak p')$ is an extension of fields.
$\square$

## Comments (2)

Comment #4244 by Kazuki Masugi on

Comment #4419 by Johan on