Support dimension under field extension

Choose a finite affine open cover $U_1,\dots ,U_r$ of $X$, $U_i=\mathrm{Spec}{A}_i$ with $A_i$ a finitely generated $k$-algebra; this is possible by [F2] because $X$ is quasi-compact. Refining this cover if necessary, [F3] lets us assume that on each $U_i$ there is a finitely generated $A_i$-module $M_i$ with $\mathcal{F}{\mathrm{\mid }}_{U_i}\cong \widetilde{M_i}$; the refinement may be taken finite and inside the charts supplied by [F3], and its coordinate rings are again finitely generated $k$-algebras by [F2]. The same opens give affine charts $\mathrm{Spec}(A_i{\otimes }_{k}K)$ of $X_K$ over them by [F1].

Chartwise support before base change. Since restriction preserves stalks, $\mathrm{Supp}(\mathcal{F})\cap U_i=\mathrm{Supp}(\mathcal{F}{\mathrm{\mid }}_{U_i})=\mathrm{Supp}(\widetilde{M_i})=V({\mathrm{Ann}}_{A_i}{M}_i)$ as a subset of $U_i=\mathrm{Spec}{A}_i$, by [F4]; here ${\mathrm{Ann}}_{A_i}{M}_i=A_i$ exactly when $M_i=0$, and then the chart contributes empty support.

Chartwise support after base change. Let $B_i=A_i{\otimes }_{k}K$ and let ${\pi }_i:\mathrm{Spec}{B}_i\to \mathrm{Spec}{A}_i$ be the chart morphism of [F1]. The restriction of ${\mathcal{F}}_K$ to this chart is ${\pi }_i^{\ast }(\mathcal{F}{\mathrm{\mid }}_{U_i})$, which by [F5] is $\widetilde{(B_i{\otimes }_{A_i}{M}_i)}$; hence, again by [F4], the part of $\mathrm{Supp}({\mathcal{F}}_K)$ inside $\mathrm{Spec}{B}_i$ is $V({\mathrm{Ann}}_{B_i}(B_i{\otimes }_{A_i}{M}_i))$.

Annihilator under base change. Let $A$ be a commutative ring, $M$ a finitely generated $A$-module and $B$ a flat $A$-algebra, for instance $B=A{\otimes }_{k}K$ with the flatness of [F9]. Choose a presentation $A^n\mathrm{/}N\cong M$ with $N\subseteq A^n$ (possible by choosing finitely many generators of $M$), so that ${\mathrm{Ann}}_A(M)=(N:{}_A{A}^n)$. Tensoring $0\to N\to A^n\to M\to 0$ with $B$ and using flatness gives $0\to NB\to B^n\to M{\otimes }_{A}B\to 0$, so the image of $N$ generates $NB$ and ${\mathrm{Ann}}_B(M{\otimes }_{A}B)=(NB:{}_B{B}^n)$. The colon identity $(N:{}_A{A}^n)B=(NB:{}_B{B}^n)$ holds: the inclusion $\subseteq$ is immediate by multiplying, and for the converse the scalar-action map $A\to {\mathrm{Hom}}_A(A^n,A^n\mathrm{/}N)$, $a\mapsto (v\mapsto av<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>N)$, has kernel $(N:{}_A{A}^n)$. Tensoring its kernel sequence with the flat $B$ and using that $A^n$ is finite free identifies ${\mathrm{Hom}}_A(A^n,A^n\mathrm{/}N){\otimes }_{A}B$ with ${\mathrm{Hom}}_B(B^n,B^n\mathrm{/}NB)$; the resulting scalar-action map from $B$ has kernel $(NB:{}_B{B}^n)$. Hence $(N:{}_A{A}^n)B=(NB:{}_B{B}^n)$ and ${\mathrm{Ann}}_B(M{\otimes }_{A}B)=({\mathrm{Ann}}_{A}M)B$.

Field extension preserves dimension of a finite-type algebra. Let $C$ be a nonzero finite-type $k$-algebra and $C\to C{\otimes }_{k}K$ the base change. By [F8] there are algebraically independent $z_1,\dots ,z_d\in C$ such that $C$ is module-finite over $P=k[z_1,\dots ,z_d]\subseteq C$; the inclusion $P\subseteq C$ is injective and integral, so $\dim C=\dim P=d$ by [F8]. Tensoring with $K$ over the flat $k$-module $K$ of [F9] preserves the injection $P\to C$ and the module finiteness, so $K[z_1,\dots ,z_d]=P{\otimes }_{k}K\to C{\otimes }_{k}K$ is an injective integral extension of nonzero rings; hence $\dim (C{\otimes }_{k}K)=\dim K[z_1,\dots ,z_d]=d$ by [F8]. Therefore $\dim C=d=\dim (C{\otimes }_{k}K)$.

Reduction to dimension of a finite-type algebra. With the notation of steps 1.2 and 1.3 and [F7], the closed subset $V({\mathrm{Ann}}_{A_i}{M}_i)$ of $\mathrm{Spec}{A}_i$ is homeomorphic to $\mathrm{Spec}(A_i\mathrm{/}{\mathrm{Ann}}_{A_i}{M}_i)$ by the contraction bijection, so its dimension equals the dimension of that spectrum, and likewise for the base-changed chart with the ideal $({\mathrm{Ann}}_{A_i}{M}_i)B_i$; by step 1.4 that ideal is ${\mathrm{Ann}}_{B_i}(B_i{\otimes }_{A_i}{M}_i)$. Since $B_i\mathrm{/}({\mathrm{Ann}}_{A_i}{M}_i)B_i\cong (A_i\mathrm{/}{\mathrm{Ann}}_{A_i}{M}_i){\otimes }_{k}K$, the desired chartwise equality of dimensions is exactly the equality $\dim C=\dim (C{\otimes }_{k}K)$ for the finite-type $k$-algebra $C=A_i\mathrm{/}{\mathrm{Ann}}_{A_i}{M}_i$, which is a finitely generated $k$-algebra because $A_i$ is one. If $C=0$, then $M_i=0$ and both chart supports are empty; otherwise $C$ is nonzero and the next step applies.

Global comparison. The open sets $U_i$ cover $X$ and the charts ${\pi }^{-1}(U_i)=\mathrm{Spec}{B}_i$ cover $X_K$ by [F1]; their intersections with the respective supports give finite open covers of those supports. For either support $T$ and a strict chain $Z_0⊊\cdots ⊊Z_d$ of nonempty irreducible closed subsets of $T$, choose a chart open $V$ meeting $Z_0$. Each $Z_j\cap V$ is a nonempty irreducible closed subset of $T\cap V$, and it is dense in $Z_j$ because it is a nonempty open subset of that irreducible space. The intersections remain strict: equality $Z_{j-1}\cap V=Z_j\cap V$ would put the dense subset $Z_j\cap V$ inside the closed proper subset $Z_{j-1}$ of $Z_j$. Thus $d\le \dim (T\cap V)$, while every chain in a chart is a chain in $T$ after taking closures in $T$ (the closures remain irreducible and strict, since their intersections with the chart recover the original chain). Hence $\dim T$ is the maximum of the dimensions of its finitely many chart intersections, also when $T=\mathrm{\varnothing }$. Steps 1.2, 1.3, 1.4, 2.1 and 1.5 give equality of these chartwise dimensions for every $i$, including $-\mathrm{\infty }$ for empty chart supports, so the two global dimensions coincide.

Boundary bookkeeping. If $\mathcal{F}=0$ then every $M_i=0$, both supports are empty and both sides are $-\mathrm{\infty }$; if $\mathcal{F}$ is nonzero on some chart then the corresponding algebra $C$ of step 2.1 is nonzero and steps 2.1 and 1.5 apply there. The Axiom of Choice is used through [F1] (choice of the glued base change), [F2] and [F3] (finite affine covers and associated sheaves), [F4] (associated-sheaf presentation) and the pullback supplier [F5]; no other selection is made.