## 10.29 Images of ring maps of finite presentation

In this section we prove some results on the topology of maps $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ induced by ring maps $R \to S$, mainly Chevalley's Theorem. In order to do this we will use the notions of constructible sets, quasi-compact sets, retrocompact sets, and so on which are defined in Topology, Section 5.15.

Lemma 10.29.1. Let $U \subset \mathop{\mathrm{Spec}}(R)$ be open. The following are equivalent:

1. $U$ is retrocompact in $\mathop{\mathrm{Spec}}(R)$,

2. $U$ is quasi-compact,

3. $U$ is a finite union of standard opens, and

4. there exists a finitely generated ideal $I \subset R$ such that $X \setminus V(I) = U$.

Proof. We have (1) $\Rightarrow$ (2) because $\mathop{\mathrm{Spec}}(R)$ is quasi-compact, see Lemma 10.17.10. We have (2) $\Rightarrow$ (3) because standard opens form a basis for the topology. Proof of (3) $\Rightarrow$ (1). Let $U = \bigcup _{i = 1\ldots n} D(f_ i)$. To show that $U$ is retrocompact in $\mathop{\mathrm{Spec}}(R)$ it suffices to show that $U \cap V$ is quasi-compact for any quasi-compact open $V$ of $\mathop{\mathrm{Spec}}(R)$. Write $V = \bigcup _{j = 1\ldots m} D(g_ j)$ which is possible by (2) $\Rightarrow$ (3). Each standard open is homeomorphic to the spectrum of a ring and hence quasi-compact, see Lemmas 10.17.6 and 10.17.10. Thus $U \cap V = (\bigcup _{i = 1\ldots n} D(f_ i)) \cap (\bigcup _{j = 1\ldots m} D(g_ j)) = \bigcup _{i, j} D(f_ i g_ j)$ is a finite union of quasi-compact opens hence quasi-compact. To finish the proof note that (4) is equivalent to (3) by Lemma 10.17.2. $\square$

Lemma 10.29.2. Let $\varphi : R \to S$ be a ring map. The induced continuous map $f : \mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ is quasi-compact. For any constructible set $E \subset \mathop{\mathrm{Spec}}(R)$ the inverse image $f^{-1}(E)$ is constructible in $\mathop{\mathrm{Spec}}(S)$.

Proof. We first show that the inverse image of any quasi-compact open $U \subset \mathop{\mathrm{Spec}}(R)$ is quasi-compact. By Lemma 10.29.1 we may write $U$ as a finite open of standard opens. Thus by Lemma 10.17.4 we see that $f^{-1}(U)$ is a finite union of standard opens. Hence $f^{-1}(U)$ is quasi-compact by Lemma 10.29.1 again. The second assertion now follows from Topology, Lemma 5.15.3. $\square$

Lemma 10.29.3. Let $R$ be a ring. A subset of $\mathop{\mathrm{Spec}}(R)$ is constructible if and only if it can be written as a finite union of subsets of the form $D(f) \cap V(g_1, \ldots , g_ m)$ for $f, g_1, \ldots , g_ m \in R$.

Proof. By Lemma 10.29.1 the subset $D(f)$ and the complement of $V(g_1, \ldots , g_ m)$ are retro-compact open. Hence $D(f) \cap V(g_1, \ldots , g_ m)$ is a constructible subset and so is any finite union of such. Conversely, let $T \subset \mathop{\mathrm{Spec}}(R)$ be constructible. By Topology, Definition 5.15.1, we may assume that $T = U \cap V^ c$, where $U, V \subset \mathop{\mathrm{Spec}}(R)$ are retrocompact open. By Lemma 10.29.1 we may write $U = \bigcup _{i = 1, \ldots , n} D(f_ i)$ and $V = \bigcup _{j = 1, \ldots , m} D(g_ j)$. Then $T = \bigcup _{i = 1, \ldots , n} \big (D(f_ i) \cap V(g_1, \ldots , g_ m)\big )$. $\square$

Lemma 10.29.4. Let $R$ be a ring and let $T \subset \mathop{\mathrm{Spec}}(R)$ be constructible. Then there exists a ring map $R \to S$ of finite presentation such that $T$ is the image of $\mathop{\mathrm{Spec}}(S)$ in $\mathop{\mathrm{Spec}}(R)$.

Proof. The spectrum of a finite product of rings is the disjoint union of the spectra, see Lemma 10.21.2. Hence if $T = T_1 \cup T_2$ and the result holds for $T_1$ and $T_2$, then the result holds for $T$. By Lemma 10.29.3 we may assume that $T = D(f) \cap V(g_1, \ldots , g_ m)$. In this case $T$ is the image of the map $\mathop{\mathrm{Spec}}((R/(g_1, \ldots , g_ m))_ f) \to \mathop{\mathrm{Spec}}(R)$, see Lemmas 10.17.6 and 10.17.7. $\square$

Lemma 10.29.5. Let $R$ be a ring. Let $f$ be an element of $R$. Let $S = R_ f$. Then the image of a constructible subset of $\mathop{\mathrm{Spec}}(S)$ is constructible in $\mathop{\mathrm{Spec}}(R)$.

Proof. We repeatedly use Lemma 10.29.1 without mention. Let $U, V$ be quasi-compact open in $\mathop{\mathrm{Spec}}(S)$. We will show that the image of $U \cap V^ c$ is constructible. Under the identification $\mathop{\mathrm{Spec}}(S) = D(f)$ of Lemma 10.17.6 the sets $U, V$ correspond to quasi-compact opens $U', V'$ of $\mathop{\mathrm{Spec}}(R)$. Hence it suffices to show that $U' \cap (V')^ c$ is constructible in $\mathop{\mathrm{Spec}}(R)$ which is clear. $\square$

Lemma 10.29.6. Let $R$ be a ring. Let $I$ be a finitely generated ideal of $R$. Let $S = R/I$. Then the image of a constructible subset of $\mathop{\mathrm{Spec}}(S)$ is constructible in $\mathop{\mathrm{Spec}}(R)$.

Proof. If $I = (f_1, \ldots , f_ m)$, then we see that $V(I)$ is the complement of $\bigcup D(f_ i)$, see Lemma 10.17.2. Hence it is constructible, by Lemma 10.29.1. Denote the map $R \to S$ by $f \mapsto \overline{f}$. We have to show that if $\overline{U}, \overline{V}$ are retrocompact opens of $\mathop{\mathrm{Spec}}(S)$, then the image of $\overline{U} \cap \overline{V}^ c$ in $\mathop{\mathrm{Spec}}(R)$ is constructible. By Lemma 10.29.1 we may write $\overline{U} = \bigcup D(\overline{g_ i})$. Setting $U = \bigcup D({g_ i})$ we see $\overline{U}$ has image $U \cap V(I)$ which is constructible in $\mathop{\mathrm{Spec}}(R)$. Similarly the image of $\overline{V}$ equals $V \cap V(I)$ for some retrocompact open $V$ of $\mathop{\mathrm{Spec}}(R)$. Hence the image of $\overline{U} \cap \overline{V}^ c$ equals $U \cap V(I) \cap V^ c$ as desired. $\square$

Lemma 10.29.7. Let $R$ be a ring. The map $\mathop{\mathrm{Spec}}(R[x]) \to \mathop{\mathrm{Spec}}(R)$ is open, and the image of any standard open is a quasi-compact open.

Proof. It suffices to show that the image of a standard open $D(f)$, $f\in R[x]$ is quasi-compact open. The image of $D(f)$ is the image of $\mathop{\mathrm{Spec}}(R[x]_ f) \to \mathop{\mathrm{Spec}}(R)$. Let $\mathfrak p \subset R$ be a prime ideal. Let $\overline{f}$ be the image of $f$ in $\kappa (\mathfrak p)[x]$. Recall, see Lemma 10.17.9, that $\mathfrak p$ is in the image if and only if $R[x]_ f \otimes _ R \kappa (\mathfrak p) = \kappa (\mathfrak p)[x]_{\overline{f}}$ is not the zero ring. This is exactly the condition that $f$ does not map to zero in $\kappa (\mathfrak p)[x]$, in other words, that some coefficient of $f$ is not in $\mathfrak p$. Hence we see: if $f = a_ d x^ d + \ldots + a_0$, then the image of $D(f)$ is $D(a_ d) \cup \ldots \cup D(a_0)$. $\square$

We prove a property of characteristic polynomials which will be used below.

Lemma 10.29.8. Let $R \to A$ be a ring homomorphism. Assume $A \cong R^{\oplus n}$ as an $R$-module. Let $f \in A$. The multiplication map $m_ f: A \to A$ is $R$-linear and hence has a characteristic polynomial $P(T) = T^ n + r_{n-1}T^{n-1} + \ldots + r_0 \in R[T]$. For any prime $\mathfrak {p} \in \mathop{\mathrm{Spec}}(R)$, $f$ acts nilpotently on $A \otimes _ R \kappa (\mathfrak {p})$ if and only if $\mathfrak p \in V(r_0, \ldots , r_{n-1})$.

Proof. This follows quite easily once we prove that the characteristic polynomial $\bar P(T) \in \kappa (\mathfrak p)[T]$ of the multiplication map $m_{\bar f}: A \otimes _ R \kappa (\mathfrak p) \to A \otimes _ R \kappa (\mathfrak p)$ which multiplies elements of $A \otimes _ R \kappa (\mathfrak p)$ by $\bar f$, the image of $f$ viewed in $\kappa (\mathfrak p)$, is just the image of $P(T)$ in $\kappa (\mathfrak p)[T]$. Let $(a_{ij})$ be the matrix of the map $m_ f$ with entries in $R$, using a basis $e_1, \ldots , e_ n$ of $A$ as an $R$-module. Then, $A \otimes _ R \kappa (\mathfrak p) \cong (R \otimes _ R \kappa (\mathfrak p))^{\oplus n} = \kappa (\mathfrak p)^ n$, which is an $n$-dimensional vector space over $\kappa (\mathfrak p)$ with basis $e_1 \otimes 1, \ldots , e_ n \otimes 1$. The image $\bar f = f \otimes 1$, and so the multiplication map $m_{\bar f}$ has matrix $(a_{ij} \otimes 1)$. Thus, the characteristic polynomial is precisely the image of $P(T)$.

From linear algebra, we know that a linear transformation acts nilpotently on an $n$-dimensional vector space if and only if the characteristic polynomial is $T^ n$ (since the characteristic polynomial divides some power of the minimal polynomial). Hence, $f$ acts nilpotently on $A \otimes _ R \kappa (\mathfrak p)$ if and only if $\bar P(T) = T^ n$. This occurs if and only if $r_ i \in \mathfrak p$ for all $0 \leq i \leq n - 1$, that is when $\mathfrak p \in V(r_0, \ldots , r_{n - 1}).$ $\square$

Lemma 10.29.9. Let $R$ be a ring. Let $f, g \in R[x]$ be polynomials. Assume the leading coefficient of $g$ is a unit of $R$. There exists elements $r_ i\in R$, $i = 1\ldots , n$ such that the image of $D(f) \cap V(g)$ in $\mathop{\mathrm{Spec}}(R)$ is $\bigcup _{i = 1, \ldots , n} D(r_ i)$.

Proof. Write $g = ux^ d + a_{d-1}x^{d-1} + \ldots + a_0$, where $d$ is the degree of $g$, and hence $u \in R^*$. Consider the ring $A = R[x]/(g)$. It is, as an $R$-module, finite free with basis the images of $1, x, \ldots , x^{d-1}$. Consider multiplication by (the image of) $f$ on $A$. This is an $R$-module map. Hence we can let $P(T) \in R[T]$ be the characteristic polynomial of this map. Write $P(T) = T^ d + r_{d-1} T^{d-1} + \ldots + r_0$. We claim that $r_0, \ldots , r_{d-1}$ have the desired property. We will use below the property of characteristic polynomials that

$\mathfrak p \in V(r_0, \ldots , r_{d-1}) \Leftrightarrow \text{multiplication by }f\text{ is nilpotent on } A \otimes _ R \kappa (\mathfrak p).$

This was proved in Lemma 10.29.8.

Suppose $\mathfrak q\in D(f) \cap V(g)$, and let $\mathfrak p = \mathfrak q \cap R$. Then there is a nonzero map $A \otimes _ R \kappa (\mathfrak p) \to \kappa (\mathfrak q)$ which is compatible with multiplication by $f$. And $f$ acts as a unit on $\kappa (\mathfrak q)$. Thus we conclude $\mathfrak p \not\in V(r_0, \ldots , r_{d-1})$.

On the other hand, suppose that $r_ i \not\in \mathfrak p$ for some prime $\mathfrak p$ of $R$ and some $0 \leq i \leq d - 1$. Then multiplication by $f$ is not nilpotent on the algebra $A \otimes _ R \kappa (\mathfrak p)$. Hence there exists a prime ideal $\overline{\mathfrak q} \subset A \otimes _ R \kappa (\mathfrak p)$ not containing the image of $f$. The inverse image of $\overline{\mathfrak q}$ in $R[x]$ is an element of $D(f) \cap V(g)$ mapping to $\mathfrak p$. $\square$

Theorem 10.29.10 (Chevalley's Theorem). Suppose that $R \to S$ is of finite presentation. The image of a constructible subset of $\mathop{\mathrm{Spec}}(S)$ in $\mathop{\mathrm{Spec}}(R)$ is constructible.

Proof. Write $S = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m)$. We may factor $R \to S$ as $R \to R[x_1] \to R[x_1, x_2] \to \ldots \to R[x_1, \ldots , x_{n-1}] \to S$. Hence we may assume that $S = R[x]/(f_1, \ldots , f_ m)$. In this case we factor the map as $R \to R[x] \to S$, and by Lemma 10.29.6 we reduce to the case $S = R[x]$. By Lemma 10.29.1 suffices to show that if $T = (\bigcup _{i = 1\ldots n} D(f_ i)) \cap V(g_1, \ldots , g_ m)$ for $f_ i , g_ j \in R[x]$ then the image in $\mathop{\mathrm{Spec}}(R)$ is constructible. Since finite unions of constructible sets are constructible, it suffices to deal with the case $n = 1$, i.e., when $T = D(f) \cap V(g_1, \ldots , g_ m)$.

Note that if $c \in R$, then we have

$\mathop{\mathrm{Spec}}(R) = V(c) \amalg D(c) = \mathop{\mathrm{Spec}}(R/(c)) \amalg \mathop{\mathrm{Spec}}(R_ c),$

and correspondingly $\mathop{\mathrm{Spec}}(R[x]) = V(c) \amalg D(c) = \mathop{\mathrm{Spec}}(R/(c)[x]) \amalg \mathop{\mathrm{Spec}}(R_ c[x])$. The intersection of $T = D(f) \cap V(g_1, \ldots , g_ m)$ with each part still has the same shape, with $f$, $g_ i$ replaced by their images in $R/(c)[x]$, respectively $R_ c[x]$. Note that the image of $T$ in $\mathop{\mathrm{Spec}}(R)$ is the union of the image of $T \cap V(c)$ and $T \cap D(c)$. Using Lemmas 10.29.5 and 10.29.6 it suffices to prove the images of both parts are constructible in $\mathop{\mathrm{Spec}}(R/(c))$, respectively $\mathop{\mathrm{Spec}}(R_ c)$.

Let us assume we have $T = D(f) \cap V(g_1, \ldots , g_ m)$ as above, with $\deg (g_1) \leq \deg (g_2) \leq \ldots \leq \deg (g_ m)$. We are going to use induction on $m$, and on the degrees of the $g_ i$. Let $d = \deg (g_1)$, i.e., $g_1 = c x^{d_1} + l.o.t$ with $c \in R$ not zero. Cutting $R$ up into the pieces $R/(c)$ and $R_ c$ we either lower the degree of $g_1$ (and this is covered by induction) or we reduce to the case where $c$ is invertible. If $c$ is invertible, and $m > 1$, then write $g_2 = c' x^{d_2} + l.o.t$. In this case consider $g_2' = g_2 - (c'/c) x^{d_2 - d_1} g_1$. Since the ideals $(g_1, g_2, \ldots , g_ m)$ and $(g_1, g_2', g_3, \ldots , g_ m)$ are equal we see that $T = D(f) \cap V(g_1, g_2', g_3\ldots , g_ m)$. But here the degree of $g_2'$ is strictly less than the degree of $g_2$ and hence this case is covered by induction.

The bases case for the induction above are the cases (a) $T = D(f) \cap V(g)$ where the leading coefficient of $g$ is invertible, and (b) $T = D(f)$. These two cases are dealt with in Lemmas 10.29.9 and 10.29.7. $\square$

Comment #2602 by fanjun meng on

Two ) should not appear in the nineth line and tenth line of the proof of theorem 10.28.9.

Comment #6454 by Jonas Ehrhard on

Constructible sets are defined in 04ZC, not 0059. Maybe add this to the text in the introduction?

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).