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10.28. Images of ring maps of finite presentation

In this section we prove some results on the topology of maps $\mathop{\rm Spec}(S) \to \mathop{\rm Spec}(R)$ 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 $U \subset \mathop{\rm Spec}(R)$ be open. The following are equivalent:

  1. $U$ is retrocompact in $\mathop{\rm Spec}(R)$,
  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 $\mathop{\rm Spec}(R)$ 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 $\mathop{\rm Spec}(R)$ it suffices to show that $U \cap V$ is quasi-compact for any quasi-compact open $V$ of $\mathop{\rm Spec}(R)$. 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 $f : \mathop{\rm Spec}(S) \to \mathop{\rm Spec}(R)$ is quasi-compact. For any constructible set $E \subset \mathop{\rm Spec}(R)$ the inverse image $f^{-1}(E)$ is constructible in $\mathop{\rm Spec}(S)$.

Proof. We first show that the inverse image of any quasi-compact open $U \subset \mathop{\rm Spec}(R)$ 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 $T \subset \mathop{\rm Spec}(R)$ be constructible. Then there exists a ring map $R \to S$ of finite presentation such that $T$ is the image of $\mathop{\rm Spec}(S)$ in $\mathop{\rm Spec}(R)$.

Proof. Let $T \subset \mathop{\rm Spec}(R)$ 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 $U, V \subset \mathop{\rm Spec}(R)$ 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 $\mathop{\rm Spec}(S)$ is constructible in $\mathop{\rm Spec}(R)$.

Proof. We repeatedly use Lemma 10.28.1 without mention. Let $U, V$ be quasi-compact open in $\mathop{\rm Spec}(S)$. We will show that the image of $U \cap V^c$ is constructible. Under the identification $\mathop{\rm Spec}(S) = D(f)$ of Lemma 10.16.6 the sets $U, V$ correspond to quasi-compact opens $U', V'$ of $\mathop{\rm Spec}(R)$. Hence it suffices to show that $U' \cap (V')^c$ is constructible in $\mathop{\rm Spec}(R)$ 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 $\mathop{\rm Spec}(S)$ is constructible in $\mathop{\rm Spec}(R)$.

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 $\mathop{\rm Spec}(S)$, then the image of $\overline{U} \cap \overline{V}^c$ in $\mathop{\rm Spec}(R)$ 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 $\mathop{\rm Spec}(R)$. Similarly the image of $\overline{V}$ equals $V \cap V(I)$ for some retrocompact open $V$ of $\mathop{\rm Spec}(R)$. 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 $\mathop{\rm Spec}(R[x]) \to \mathop{\rm Spec}(R)$ 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 $\mathop{\rm Spec}(R[x]_f) \to \mathop{\rm Spec}(R)$. 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 $\mathfrak{p} \in \mathop{\rm Spec}(R)$, $f$ 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 $\mathop{\rm Spec}(R)$ 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 $\mathop{\rm Spec}(S)$ in $\mathop{\rm Spec}(R)$ 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 $\mathop{\rm Spec}(R)$ 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 $$ \mathop{\rm Spec}(R) = V(c) \amalg D(c) = \mathop{\rm Spec}(R/(c)) \amalg \mathop{\rm Spec}(R_c)), $$ and correspondingly $\mathop{\rm Spec}(R[x]) = V(c) \amalg D(c) = \mathop{\rm Spec}(R/(c)[x]) \amalg \mathop{\rm Spec}(R_c[x])$. 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 $\mathop{\rm Spec}(R)$ 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 $\mathop{\rm Spec}(R/(c))$, respectively $\mathop{\rm Spec}(R_c)$.

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. $\square$

    The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 5095–5429 (see updates for more information).

    \section{Images of ring maps of finite presentation}
    \label{section-images-finite-presentation}
    
    \noindent
    In this section we prove some results on the
    topology of maps $\Spec(S) \to \Spec(R)$
    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 \ref{topology-section-quasi-compact}.
    
    \begin{lemma}
    \label{lemma-qc-open}
    Let $U \subset \Spec(R)$ be open. The following
    are equivalent:
    \begin{enumerate}
    \item $U$ is retrocompact in $\Spec(R)$,
    \item $U$ is quasi-compact,
    \item $U$ is a finite union of standard opens, and
    \item there exists a finitely generated ideal $I \subset R$ such
    that $X \setminus V(I) = U$.
    \end{enumerate}
    \end{lemma}
    
    \begin{proof}
    We have (1) $\Rightarrow$ (2) because $\Spec(R)$ is quasi-compact, see
    Lemma \ref{lemma-quasi-compact}. 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 $\Spec(R)$ it suffices to show that $U \cap V$ is quasi-compact for any
    quasi-compact open $V$ of $\Spec(R)$. 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 \ref{lemma-standard-open} and
    \ref{lemma-quasi-compact}. 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 \ref{lemma-Zariski-topology}.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-affine-map-quasi-compact}
    Let $\varphi : R \to S$ be a ring map.
    The induced continuous map $f : \Spec(S) \to \Spec(R)$
    is quasi-compact. For any constructible set $E \subset \Spec(R)$
    the inverse image $f^{-1}(E)$ is constructible in $\Spec(S)$.
    \end{lemma}
    
    \begin{proof}
    We first show that the inverse image of any quasi-compact
    open $U \subset \Spec(R)$ is quasi-compact. By
    Lemma \ref{lemma-qc-open} we may write $U$ as a finite
    open of standard opens. Thus by Lemma \ref{lemma-spec-functorial}
    we see that $f^{-1}(U)$ is a finite union of standard opens.
    Hence $f^{-1}(U)$ is quasi-compact by Lemma \ref{lemma-qc-open} again.
    The second assertion now follows from Topology, Lemma
    \ref{topology-lemma-inverse-images-constructibles}.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-constructible-is-image}
    Let $R$ be a ring and let $T \subset \Spec(R)$
    be constructible. Then there exists a ring map $R \to S$ of
    finite presentation such that $T$ is the image of
    $\Spec(S)$ in $\Spec(R)$.
    \end{lemma}
    
    \begin{proof}
    Let $T \subset \Spec(R)$ be constructible.
    The spectrum of a finite product of rings
    is the disjoint union of the spectra, see
    Lemma \ref{lemma-spec-product}. 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 $U, V \subset \Spec(R)$
    are retrocompact open. By Lemma \ref{lemma-qc-open} 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
    \ref{lemma-standard-open} and \ref{lemma-spec-closed}.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-open-fp}
    Let $R$ be a ring.
    Let $f$ be an element of $R$.
    Let $S = R_f$.
    Then the image of a constructible subset of $\Spec(S)$
    is constructible in $\Spec(R)$.
    \end{lemma}
    
    \begin{proof}
    We repeatedly use Lemma \ref{lemma-qc-open} without mention.
    Let $U, V$ be quasi-compact open in $\Spec(S)$.
    We will show that the image of $U \cap V^c$ is constructible.
    Under the identification
    $\Spec(S) = D(f)$ of Lemma \ref{lemma-standard-open}
    the sets $U, V$ correspond to quasi-compact opens
    $U', V'$ of $\Spec(R)$.
    Hence it suffices to show that $U' \cap (V')^c$
    is constructible in $\Spec(R)$ which is clear.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-closed-fp}
    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 $\Spec(S)$
    is constructible in $\Spec(R)$.
    \end{lemma}
    
    \begin{proof}
    If $I = (f_1, \ldots, f_m)$, then we see that
    $V(I)$ is the complement of $\bigcup D(f_i)$,
    see Lemma \ref{lemma-Zariski-topology}.
    Hence it is constructible, by Lemma \ref{lemma-qc-open}.
    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 $\Spec(S)$, then the
    image of $\overline{U} \cap \overline{V}^c$
    in $\Spec(R)$ is constructible.
    By Lemma \ref{lemma-qc-open} 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
    $\Spec(R)$. Similarly the image of $\overline{V}$ equals
    $V \cap V(I)$ for some retrocompact open $V$ of $\Spec(R)$.
    Hence the image of $\overline{U} \cap \overline{V}^c$
    equals $U \cap V(I) \cap V^c$ as desired.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-affineline-open}
    Let $R$ be a ring. The map $\Spec(R[x]) \to \Spec(R)$
    is open, and the image of any standard open is a quasi-compact
    open.
    \end{lemma}
    
    \begin{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
    $\Spec(R[x]_f) \to \Spec(R)$.
    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 \ref{lemma-in-image},
    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 $D(a_d) \cup \ldots \cup D(a_0)$.
    \end{proof}
    
    \noindent
    We prove a property of characteristic polynomials which
    will be used below.
    
    \begin{lemma}
    \label{lemma-characteristic-polynomial-prime}
    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
    $\mathfrak{p} \in \Spec(R)$, $f$ acts nilpotently on $A
    \otimes_R \kappa(\mathfrak{p})$ if and only if $\mathfrak p \in
    V(r_0, \ldots, r_{n-1})$.
    \end{lemma}
    
    \begin{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 $P(T)$.
    
    \medskip\noindent
    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 $\mathfrak p \in
    V(r_0, \ldots, r_{n - 1}).$
    \end{proof}
    
    \begin{lemma}
    \label{lemma-affineline-special}
    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 $\Spec(R)$ is
    $\bigcup_{i = 1, \ldots, n} D(r_i)$.
    \end{lemma}
    
    \begin{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 \ref{lemma-characteristic-polynomial-prime}.
    
    \medskip\noindent
    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 $\mathfrak p \not \in  V(r_0, \ldots, r_{d-1})$.
    
    \medskip\noindent
    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 $\mathfrak p$.
    \end{proof}
    
    \begin{theorem}[Chevalley's Theorem]
    \label{theorem-chevalley}
    Suppose that $R \to S$ is of finite presentation.
    The image of a constructible subset of
    $\Spec(S)$ in $\Spec(R)$ is constructible.
    \end{theorem}
    
    \begin{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 \ref{lemma-closed-fp} we reduce to
    the case $S = R[x]$. By Lemma \ref{lemma-qc-open} 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 $\Spec(R)$ is
    constructible. Since finite unions of constructible sets
    are constructible, it suffices to deal with the case $n = 1$,
    i.e., when $T = D(f) \cap V(g_1, \ldots, g_m)$.
    
    \medskip\noindent
    Note that if $c \in R$, then we have
    $$
    \Spec(R) =
    V(c) \amalg D(c) =
    \Spec(R/(c)) \amalg \Spec(R_c)),
    $$
    and correspondingly $\Spec(R[x]) =
    V(c) \amalg D(c) = \Spec(R/(c)[x]) \amalg
    \Spec(R_c[x])$. 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 $\Spec(R)$ is the union of the image of
    $T \cap V(c)$ and $T \cap D(c)$. Using Lemmas \ref{lemma-open-fp}
    and \ref{lemma-closed-fp} it suffices to prove the images of both
    parts are constructible in $\Spec(R/(c))$, respectively
    $\Spec(R_c)$.
    
    \medskip\noindent
    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.
    
    \medskip\noindent
    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 \ref{lemma-affineline-special}
    and \ref{lemma-affineline-open}.
    \end{proof}

    Comments (2)

    Comment #2602 by fanjun meng on June 10, 2017 a 12:24 pm UTC

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

    Comment #2627 by Johan (site) on July 7, 2017 a 12:40 pm UTC

    Thanks, fixed here.

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