The Stacks project

Proof. We have (1) $\Rightarrow $ (2) because $\mathop{\mathrm{Spec}}(R)$ is quasi-compact, see Lemma 10.17.8. 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 10.17.8. 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. $\square$

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. $\square$

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 $T = \bigcup _{i = 1, \ldots , n} \big (D(f_ i) \cap V(g_1, \ldots , g_ m)\big )$. $\square$

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. $\square$

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. $\square$

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. $\square$

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 10.18.6, 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 $D(a_ d) \cup \ldots \cup D(a_0)$. $\square$

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 $\mathfrak p \in V(r_0, \ldots , r_{n - 1}).$ $\square$

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

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 $\mathfrak p$. $\square$

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

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 $d_1 = \deg (g_1)$, 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. $\square$