## 10.28 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 5.12.

Lemma 10.28.1. Let $U \subset \mathop{\mathrm{Spec}}(R)$ be open. The following are equivalent:

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

$U$ is quasi-compact,

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

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 10.16.10. 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 10.16.6 and 10.16.10. 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 10.16.2.
$\square$

Lemma 10.28.2. 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 10.28.1 we may write $U$ as a finite open of standard opens. Thus by Lemma 10.16.4 we see that $f^{-1}(U)$ is a finite union of standard opens. Hence $f^{-1}(U)$ is quasi-compact by Lemma 10.28.1 again. The second assertion now follows from Topology, Lemma 5.15.3.
$\square$

Lemma 10.28.3. 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.**
Let $T \subset \mathop{\mathrm{Spec}}(R)$ be constructible. The spectrum of a finite product of rings is the disjoint union of the spectra, see Lemma 10.20.2. Hence if $T = T_1 \cup T_2$ and the result holds for $T_1$ and $T_2$, then the result holds for $T$. In particular we may assume that $T = U \cap V^ c$, where $U, V \subset \mathop{\mathrm{Spec}}(R)$ are retrocompact open. By Lemma 10.28.1 we may write $T = (\bigcup D(f_ i)) \cap (\bigcup D(g_ j))^ c = \bigcup \big (D(f_ i) \cap V(g_1, \ldots , g_ m)\big )$. In fact we may assume that $T = D(f) \cap V(g_1, \ldots , g_ m)$ (by the argument on unions above). In this case $T$ is the image of the map $R \to (R/(g_1, \ldots , g_ m))_ f$, see Lemmas 10.16.6 and 10.16.7.
$\square$

Lemma 10.28.4. 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 10.28.1 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 10.16.6 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 10.28.5. Let $R$ be a ring. Let $I$ be a finitely generated ideal of $R$. Let $S = R/I$. Then the image of a constructible 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 10.16.2. Hence it is constructible, by Lemma 10.28.1. 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 10.28.1 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 10.28.6. 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 10.16.9, 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 10.28.7. 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 10.28.8. 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 10.28.7.

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 10.28.9 (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 10.28.5 we reduce to the case $S = R[x]$. By Lemma 10.28.1 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 10.28.4 and 10.28.5 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 descending 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 10.28.8 and 10.28.6.
$\square$

## Comments (2)

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