Constructible images for finite-presentation affine maps

By [F3] and [F4] write $B\cong P\mathrm{/}I$ with $P=A[x_1,\dots ,x_n]$ and $I=(h_1,\dots ,h_m)$ a finitely generated ideal, and choose $f\in P$ mapping to $b$. Let $T=D(f)\cap V(h_1,\dots ,h_m)\subseteq \mathrm{Spec}P$. Then $T$ is constructible in $\mathrm{Spec}P$ by [F2], because $D(f)$ is retrocompact open and $V(h_1,\dots ,h_m)$ is the complement of the retrocompact open $D(h_1)\cup \cdots \cup D(h_m)$. Let $\pi :\mathrm{Spec}P\to \mathrm{Spec}A$ be the induced map.

We prove the following claim by induction on $k\ge 0$: for every commutative ring $R$ and every constructible subset $E\subseteq \mathrm{Spec}R[x_1,\dots ,x_k]$, the image of $E$ under the structure map $\mathrm{Spec}R[x_1,\dots ,x_k]\to \mathrm{Spec}R$ is constructible. For $k=0$ the structure map is the identity, so the claim is immediate.

One-variable claim: for every commutative ring $R$, every $f\in R[x]$ and all $g_1,\dots ,g_m\in R[x]$, the image of $D(f)\cap V(g_1,\dots ,g_m)$ under $\mathrm{Spec}R[x]\to \mathrm{Spec}R$ is constructible. We prove this by well-founded induction on the pair $\mu =(m;{\delta }_1\le \cdots \le {\delta }_m)$, where ${\delta }_1,\dots ,{\delta }_m$ are the degrees of the nonzero polynomials among $g_1,\dots ,g_m$ arranged in increasing order, ordered lexicographically with the empty tuple smallest. Dropping a zero polynomial decreases $m$, and replacing a polynomial by one of strictly smaller degree decreases the tuple of degrees with $m$ fixed, so $\mu$ strictly decreases in the reductions below.

Base case $m=0$: write $f={\sum }_{i=0}^N{a}_i{x}^i$. For $\mathfrak{p}\in \mathrm{Spec}R$ the fibre of $\mathrm{Spec}R[x]\to \mathrm{Spec}R$ over $\mathfrak{p}$ is $\mathrm{Spec}\kappa (\mathfrak{p})[x]$, and the fibre of $D(f)$ is $D(\bar{f})$, which is nonempty exactly when some coefficient $a_i$ has nonzero image in $\kappa (\mathfrak{p})$, that is, when $a_i\mathrm{\notin }\mathfrak{p}$. Hence the image of $D(f)$ is ${\bigcup }_{i=0}^{N}D(a_i)$, a constructible set; for $f=0$ all the basic opens are empty and the image is empty.

Fix $c\in R$ and a constructible set $T\subseteq D(f)\cap V(g_1,\dots ,g_m)$ built from $f$ and the $g_i$. The space $\mathrm{Spec}R$ is the disjoint union of the closed subspace $V(c)=\mathrm{Spec}R\mathrm{/}(c)$ and the open subspace $D(c)=\mathrm{Spec}{R}_c$. Accordingly the image of $T$ is the union of its parts over $V(c)$ and over $D(c)$: over $V(c)$ one computes in $R\mathrm{/}(c)[x]$, over $D(c)$ in $R_c[x]$, and the images of these two parts are the corresponding subsets of $\mathrm{Spec}R$ because $\mathrm{Spec}R\mathrm{/}(c)\to \mathrm{Spec}R$ is a closed immersion onto $V(c)$ and $\mathrm{Spec}{R}_c\to \mathrm{Spec}R$ is an open immersion onto $D(c)$. A subset of $V(c)$ of the form ${\bigcup }_{j}D({\bar{f}}_j)\cap V({\bar{g}}_{j1},\dots )$ corresponds in $\mathrm{Spec}R$ to ${\bigcup }_{j}D(f_j)\cap V(g_{j1},\dots ,c)$, and similarly on $D(c)$ the images are themselves constructible subsets of $\mathrm{Spec}R$ described by the same expressions read in $R[x]$.

Base case of the induction: $m=1$, $g\in R[x]$ with invertible leading coefficient $u$, and $f\in R[x]$. Multiplying $g$ by $u^{-1}$ we may assume $g$ monic of degree $d\ge 0$. Let $S=R[x]\mathrm{/}(g)$, a free $R$-module with basis the classes of $1,x,\dots ,x^{d-1}$ by [F6]; let $M$ be the matrix of multiplication by $f$ on $S$ in this basis, and put $P(T)=\det (T{I}_d-M)={\sum }_{i=0}^d{r}_i{T}^i\in R[T]$, a monic polynomial of degree $d$ with $r_d=1$, with the convention $P=1$ when $d=0$.

The quotient map $P\to B$ induces a closed immersion $\mathrm{Spec}B\to \mathrm{Spec}P$ whose image is exactly $V(I)$, and for the reduction $\bar{f}$ of $f$ the preimage of $D(\bar{f})\subseteq \mathrm{Spec}B$ is $D(f)\cap V(I)=T$. Hence the image of $D(b)$ under $\mathrm{Spec}B\to \mathrm{Spec}A$ equals $\pi (T)$.

For the induction step from $k-1$ to $k$, put $R^{\mathrm{\prime }}=R[x_1,\dots ,x_{k-1}]$, so that $R[x_1,\dots ,x_k]=R^{\mathrm{\prime }}[x_k]$. By the one-variable claim of steps 1.3-3.1, applied over the ring $R^{\mathrm{\prime }}$, the image $E^{\mathrm{\prime }}$ of $E$ under $\mathrm{Spec}{R}^{\mathrm{\prime }}[x_k]\to \mathrm{Spec}{R}^{\mathrm{\prime }}$ is a constructible subset of $\mathrm{Spec}{R}^{\mathrm{\prime }}$. By the induction hypothesis for $k-1$ applied to the ring $R$ and the constructible subset $E^{\mathrm{\prime }}$, the image of $E^{\mathrm{\prime }}$ in $\mathrm{Spec}R$ is constructible. Since the image of $E$ in $\mathrm{Spec}R$ is the image of $E^{\mathrm{\prime }}$, this proves the claim for $k$.

For the induction step assume $m\ge 1$ and that the claim is known for all smaller $\mu$. If some $g_j=0$ then $V(g_1,\dots ,g_m)=V(\text{the remaining }{g}_i)$, the pair's first entry drops, and the claim follows from the induction hypothesis. Hence assume all $g_i\mathrm{\ne }0$, and reorder the $g_i$ so that $g_1$ has minimal degree $d$, with leading coefficient $c\in R$, $c\mathrm{\ne }0$.

Over $V(c)$: in $R\mathrm{/}(c)[x]$ the image of $g_1$ has degree $<d$ or is zero, while the images of $g_2,\dots ,g_m$ have degrees at most their original degrees. Hence the pair ${\mu }^{\mathrm{\prime }}$ of the tuple in $R\mathrm{/}(c)[x]$ is strictly smaller than $\mu$: either a polynomial has disappeared, or the smallest of the degrees strictly decreased. By the induction hypothesis applied over the ring $R\mathrm{/}(c)$, the image of the part of $T$ over $V(c)$ inside $\mathrm{Spec}(R\mathrm{/}(c))[x]$ is constructible, and by step 1.5 its image in $\mathrm{Spec}R$ is constructible.

Over $D(c)$: the element $c$ becomes a unit, so after multiplying $g_1$ by the unit $c^{-1}$ we may assume $g_1$ is monic of degree $d$; this does not change the ideal $(g_1)$ nor its zero set. If $m=1$ the claim over $D(c)$ is the one-equation case treated below, whose constructibility is then moved back to $\mathrm{Spec}R$ by step 1.5. If $m\ge 2$ and $d=0$, then $g_1=c$ is a unit in $R_c[x]$, so $V(g_1)=\mathrm{\varnothing }$ over $D(c)$ and the part of the image over $D(c)$ is empty.

Over $D(c)$ with $m\ge 2$ and $d\ge 1$, apply [F6] to each $g_i$, $i\ge 2$, and $g_1$: there are $q_i,r_i\in R_c[x]$ with $g_i=q_i{g}_1+r_i$ and $r_i=0$ or $\mathrm{deg}{r}_i<d$. Then $(g_1,g_2,\dots ,g_m)=(g_1,r_2,\dots ,r_m)$ as ideals, hence $V(g_1,\dots ,g_m)=V(g_1,r_2,\dots ,r_m)$ over $D(c)$. Every nonzero $r_i$ has degree $<d$, so the new pair ${\mu }^{\mathrm{\prime }\mathrm{\prime }}$ is strictly smaller than $\mu$: its smallest degree is $<d$ or the tuple is shorter. The induction hypothesis over the ring $R_c$ makes the image of this part constructible in $\mathrm{Spec}{R}_c$, and step 1.5 moves it to a constructible subset of $\mathrm{Spec}R$.

Claim: for $\mathfrak{p}\in \mathrm{Spec}R$, the element $f$ is nilpotent in $S{\otimes }_R\kappa (\mathfrak{p})$ if and only if $\mathfrak{p}\supseteq (r_0,\dots ,r_{d-1})$. Indeed $S{\otimes }_R\kappa (\mathfrak{p})\cong \kappa (\mathfrak{p})[x]\mathrm{/}(\bar{g})$ has $\kappa (\mathfrak{p})$-basis the classes of $1,x,\dots ,x^{d-1}$, and multiplication by the image of $f$ has matrix $\bar{M}$ obtained by reducing the entries of $M$; since the determinant is given by the Leibniz formula [F7], the characteristic polynomial of that multiplication is the reduction $\bar{P}(T)=T^d+{\bar{r}}_{d-1}{T}^{d-1}+\cdots +{\bar{r}}_0$ of $P$ over $\kappa (\mathfrak{p})$, in the sense of [F8]. When $d\ge 1$ and the fibre algebra is nonzero, [F9] says that this multiplication is nilpotent exactly when $\bar{P}=T^d$, that is, exactly when all ${\bar{r}}_i=0$, i.e. $\mathfrak{p}\supseteq (r_0,\dots ,r_{d-1})$; when $d=0$ the multiplication acts on the zero ring, is nilpotent, and the condition $\mathfrak{p}\supseteq \mathrm{\varnothing }$ holds.

Claim: the image of $D(f)\cap V(g)$ in $\mathrm{Spec}R$ is ${\bigcup }_{i=0}^{d-1}D(r_i)$. For the inclusion $\subseteq$, let $\mathfrak{q}\in D(f)\cap V(g)$ lie over $\mathfrak{p}$. Then $\mathfrak{q}$ is a prime of the fibre algebra $S{\otimes }_R\kappa (\mathfrak{p})$ containing the image of $g$ and avoiding the image of $f$, so multiplication by $f$ on that fibre algebra is not nilpotent: a nilpotent element lies in every prime ideal of its ring. By step 2.7 some $r_i\mathrm{\notin }\mathfrak{p}$, that is, $\mathfrak{p}\in D(r_i)$ for some $i$. Conversely, if $r_i\mathrm{\notin }\mathfrak{p}$ for some $i<d$, then by step 2.7 $f$ is not nilpotent in the fibre algebra $S{\otimes }_R\kappa (\mathfrak{p})$. Fact [F10] supplies a prime ideal of that fibre algebra avoiding $f$. Its preimage under $R[x]\to S{\otimes }_R\kappa (\mathfrak{p})$ is a prime of $R[x]$ lying in $D(f)\cap V(g)$ over $\mathfrak{p}$.

This completes the induction of step 1.3 and hence the one-variable claim, which supplies step 2.2 and proves the induction claim of step 1.2 for every $k$. Applying step 1.2 with $R=A$, $k=n$ and the constructible set $T$ of step 1.1, the image $\pi (T)$ is constructible in $\mathrm{Spec}A$. By step 2.1 the image of $D(b)$ under $\mathrm{Spec}B\to \mathrm{Spec}A$ equals $\pi (T)$, so it is constructible. The Axiom of Choice [F5] is used only in step 3.1, to choose a prime ideal avoiding a non-nilpotent element; all other selections are finite.