## 10.29 More on images

In this section we collect a few additional lemmas concerning the image on $\mathop{\mathrm{Spec}}$ for ring maps. See also Section 10.40 for example.

Lemma 10.29.1. Let $R \subset S$ be an inclusion of domains. Assume that $R \to S$ is of finite type. There exists a nonzero $f \in R$, and a nonzero $g \in S$ such that $R_ f \to S_{fg}$ is of finite presentation.

Proof. By induction on the number of generators of $S$ over $R$. During the proof we may replace $R$ by $R_ f$ and $S$ by $S_ f$ for some nonzero $f \in R$.

Suppose that $S$ is generated by a single element over $R$. Then $S = R[x]/\mathfrak q$ for some prime ideal $\mathfrak q \subset R[x]$. If $\mathfrak q = (0)$ there is nothing to prove. If $\mathfrak q \not= (0)$, then let $h \in \mathfrak q$ be a nonzero element with minimal degree in $x$. Write $g = f x^ d + a_{d - 1} x^{d - 1} + \ldots + a_0$ with $a_ i \in R$ and $f \not= 0$. After inverting $f$ in $R$ and $S$ we may assume that $h$ is monic. We obtain a surjective $R$-algebra map $R[x]/(h) \to S$. We have $R[x]/(h) = R \oplus Rx \oplus \ldots \oplus Rx^{d - 1}$ as an $R$-module and by minimality of $d$ we see that $R[x]/(h)$ maps injectively into $S$. Thus $R[x]/(h) \cong S$ is finitely presented over $R$.

Suppose that $S$ is generated by $n > 1$ elements over $R$. Say $x_1, \ldots , x_ n \in S$ generate $S$. Denote $S' \subset S$ the subring generated by $x_1, \ldots , x_{n-1}$. By induction hypothesis we see that there exist $f\in R$ and $g \in S'$ nonzero such that $R_ f \to S'_{fg}$ is of finite presentation. Next we apply the induction hypothesis to $S'_{fg} \to S_{fg}$ to see that there exist $f' \in S'_{fg}$ and $g' \in S_{fg}$ such that $S'_{fgf'} \to S_{fgf'g'}$ is of finite presentation. We leave it to the reader to conclude. $\square$

Lemma 10.29.2. Let $R \to S$ be a finite type ring map. Denote $X = \mathop{\mathrm{Spec}}(R)$ and $Y = \mathop{\mathrm{Spec}}(S)$. Write $f : Y \to X$ the induced map of spectra. Let $E \subset Y = \mathop{\mathrm{Spec}}(S)$ be a constructible set. If a point $\xi \in X$ is in $f(E)$, then $\overline{\{ \xi \} } \cap f(E)$ contains an open dense subset of $\overline{\{ \xi \} }$.

Proof. Let $\xi \in X$ be a point of $f(E)$. Choose a point $\eta \in E$ mapping to $\xi$. Let $\mathfrak p \subset R$ be the prime corresponding to $\xi$ and let $\mathfrak q \subset S$ be the prime corresponding to $\eta$. Consider the diagram

$\xymatrix{ \eta \ar[r] \ar@{|->}[d] & E \cap Y' \ar[r] \ar[d] & Y' = \mathop{\mathrm{Spec}}(S/\mathfrak q) \ar[r] \ar[d] & Y \ar[d] \\ \xi \ar[r] & f(E) \cap X' \ar[r] & X' = \mathop{\mathrm{Spec}}(R/\mathfrak p) \ar[r] & X }$

By Lemma 10.28.2 the set $E \cap Y'$ is constructible in $Y'$. It follows that we may replace $X$ by $X'$ and $Y$ by $Y'$. Hence we may assume that $R \subset S$ is an inclusion of domains, $\xi$ is the generic point of $X$, and $\eta$ is the generic point of $Y$. By Lemma 10.29.1 combined with Chevalley's theorem (Theorem 10.28.9) we see that there exist dense opens $U \subset X$, $V \subset Y$ such that $f(V) \subset U$ and such that $f : V \to U$ maps constructible sets to constructible sets. Note that $E \cap V$ is constructible in $V$, see Topology, Lemma 5.15.4. Hence $f(E \cap V)$ is constructible in $U$ and contains $\xi$. By Topology, Lemma 5.15.15 we see that $f(E \cap V)$ contains a dense open $U' \subset U$. $\square$

At the end of this section we present a few more results on images of maps on Spectra that have nothing to do with constructible sets.

Lemma 10.29.3. Let $\varphi : R \to S$ be a ring map. The following are equivalent:

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

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

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

4. 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 10.16.2 guarantees that $I = \bigcap _{I \subset \mathfrak p} \mathfrak p$. Hence (3) $\Rightarrow$ (2). By Lemma 10.16.9 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 10.16.9. 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. $\square$

Lemma 10.29.4. Let $R$ be a domain. Let $\varphi : R \to S$ be a ring map. The following are equivalent:

1. The ring map $R \to S$ is injective.

2. The image $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ contains a dense set of points.

3. There exists a prime ideal $\mathfrak q \subset S$ whose inverse image in $R$ is $(0)$.

Proof. Let $K$ be the field of fractions of the domain $R$. Assume that $R \to S$ is injective. Since localization is exact we see that $K \to S \otimes _ R K$ is injective. Hence there is a prime mapping to $(0)$ by Lemma 10.16.9.

Note that $(0)$ is dense in $\mathop{\mathrm{Spec}}(R)$, so that the last condition implies the second.

Suppose the second condition holds. Let $f \in R$, $f \not= 0$. As $R$ is a domain we see that $V(f)$ is a proper closed subset of $R$. By assumption there exists a prime $\mathfrak q$ of $S$ such that $\varphi (f) \not\in \mathfrak q$. Hence $\varphi (f) \not= 0$. Hence $R \to S$ is injective. $\square$

Lemma 10.29.5. Let $R \subset S$ be an injective ring map. Then $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ hits all the minimal primes of $\mathop{\mathrm{Spec}}(R)$.

Proof. Let $\mathfrak p \subset R$ be a minimal prime. In this case $R_{\mathfrak p}$ has a unique prime ideal. Hence it suffices to show that $S_{\mathfrak p}$ is not zero. And this follows from the fact that localization is exact, see Proposition 10.9.12. $\square$

Lemma 10.29.6. Let $R \to S$ be a ring map. The following are equivalent:

1. The kernel of $R \to S$ consists of nilpotent elements.

2. The minimal primes of $R$ are in the image of $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$.

3. The image of $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ is dense in $\mathop{\mathrm{Spec}}(R)$.

Proof. Let $I = \mathop{\mathrm{Ker}}(R \to S)$. Note that $\sqrt{(0)} = \bigcap _{\mathfrak q \subset S} \mathfrak q$, see Lemma 10.16.2. Hence $\sqrt{I} = \bigcap _{\mathfrak q \subset S} R \cap \mathfrak q$. Thus $V(I) = V(\sqrt{I})$ is the closure of the image of $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$. This shows that (1) is equivalent to (3). It is clear that (2) implies (3). Finally, assume (1). We may replace $R$ by $R/I$ and $S$ by $S/IS$ without affecting the topology of the spectra and the map. Hence the implication (1) $\Rightarrow$ (2) follows from Lemma 10.29.5. $\square$

Lemma 10.29.7. Let $R \to S$ be a ring map. If a minimal prime $\mathfrak p \subset R$ is in the image of $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$, then it is the image of a minimal prime.

Proof. Say $\mathfrak p = \mathfrak q \cap R$. Then choose a minimal prime $\mathfrak r \subset S$ with $\mathfrak r \subset \mathfrak q$, see Lemma 10.16.2. By minimality of $\mathfrak p$ we see that $\mathfrak p = \mathfrak r \cap R$. $\square$

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