The Stacks project

Proof. Let us explain how to prove (1) in a more elementary way and part (2) using previously developed general theory.

Proof of (1). Since formation of $\text{Tor}$ commutes with localization we may assume $X$ is affine. Hence $X = \mathop{\mathrm{Spec}}(A)$ for some Noetherian ring $A$ and $\mathcal{F}$, $\mathcal{G}$ correspond to finite $A$-modules $M$ and $N$ (Cohomology of Schemes, Lemma 30.9.1). By Derived Categories of Schemes, Lemma 36.3.9 we may compute the $\text{Tor}$'s by first computing the $\text{Tor}$'s of $M$ and $N$ over $A$, and then taking the associated $\mathcal{O}_ X$-module. Since the modules $\text{Tor}_ p^ A(M, N)$ are finite by Algebra, Lemma 10.75.7 we conclude.

By Derived Categories of Schemes, Lemma 36.10.3 the assumption is equivalent to asking $L$ and $K$ to be (locally) pseudo-coherent. Then $L \otimes _{\mathcal{O}_ X}^\mathbf {L} K$ is pseudo-coherent by Cohomology, Lemma 20.47.5. $\square$

Proof. The result on stalks was discussed above and it implies the support condition. The $\text{Tor}$'s are coherent by Lemma 43.14.1. The vanishing of negative $\text{Tor}$'s is immediate from the construction. The vanishing of $\text{Tor}_ p$ for $p > \dim (X)$ can be seen as follows: the local rings $\mathcal{O}_{X, x}$ are regular (as $X$ is nonsingular) of dimension $\leq \dim (X)$ (Algebra, Lemma 10.116.1), hence $\mathcal{O}_{X, x}$ has finite global dimension $\leq \dim (X)$ (Algebra, Lemma 10.110.8) which implies that $\text{Tor}$-groups of modules vanish beyond the dimension (More on Algebra, Lemma 15.66.19). $\square$

Proof. Let $(A, \mathfrak m, \kappa ) = (\mathcal{O}_{X, \xi }, \mathfrak m_\xi , \kappa (\xi ))$ where $\xi \in Z$ is the generic point. Then $\dim (A) = \dim (X) - \dim (Z)$, see Varieties, Lemma 33.20.3. Let $I, J \subset A$ cut out the trace of $V$ and $W$ in $\mathop{\mathrm{Spec}}(A)$. Set $\overline{I} = I + \mathfrak m^2/\mathfrak m^2$. Then $\dim _\kappa \overline{I} \leq \dim (X) - \dim (V)$ with equality if and only if $A/I$ is regular (this follows from the lemma cited above and the definition of regular rings, see Algebra, Definition 10.60.10 and the discussion preceding it). Similarly for $\overline{J}$. If the multiplicity is $1$, then $\text{length}_ A(A/I + J) = 1$, hence $I + J = \mathfrak m$, hence $\overline{I} + \overline{J} = \mathfrak m/\mathfrak m^2$. Then we get equality everywhere (because the intersection is proper). Hence we find $f_1, \ldots , f_ a \in I$ and $g_1, \ldots g_ b \in J$ such that $\overline{f}_1, \ldots , \overline{g}_ b$ is a basis for $\mathfrak m/\mathfrak m^2$. Then $f_1, \ldots , g_ b$ is a regular system of parameters and a regular sequence (Algebra, Lemma 10.106.3). The same lemma shows $A/(f_1, \ldots , f_ a)$ is a regular local ring of dimension $\dim (X) - \dim (V)$, hence $A/(f_1, \ldots , f_ a) \to A/I$ is an isomorphism (if the kernel is nonzero, then the dimension of $A/I$ is strictly less, see Algebra, Lemmas 10.106.2 and 10.60.13). We conclude $I = (f_1, \ldots , f_ a)$ and $J = (g_1, \ldots , g_ b)$ by symmetry. Thus the Koszul complex $K_\bullet (A, f_1, \ldots , f_ a)$ on $f_1, \ldots , f_ a$ is a resolution of $A/I$, see More on Algebra, Lemma 15.30.2. Hence

Since we've seen above that $f_1 \bmod J, \ldots , f_ a \bmod J$ is a regular system of parameters in the regular local ring $A/J$ we conclude that there is only one cohomology group, namely $H_0 = A/(I + J) = \kappa$. This finishes the proof. $\square$