In Nitsure's Part 2. Construction of Hilbert and Quot schemes in Fundamental Algebraic Geometry, there is the following Lemma
Lemma 5.21. (3) Let $S$ be a noetherian scheme, and let $f: X \to S$ and $g: Y \to S$ be finite type flat morphisms. Let $\pi: Y \to X$ be any morphism such that $g = f \circ \pi$. Let $y \in Y$, let $x = \pi(y)$, and let $s = g(y) = f(x)$. If the restricted morphism $\pi_s: Y_s \to X_s$ between the fibers over $s$ is flat at $y \in Y_s$, then $\pi$ is flat at $y \in Y$.
He notes that this is a consequence of the local criterion for flatness. So one version of the local criterion for flatness is the following (from Matsumura, Commutative ring theory, Thm 22.3)
If $A$ is a noetherian ring and $M$ is an $I$-adically ideal-separated module, then the following conditions are equivalent.
- $M$ is flat over $A$;
- $M/IM$ is flat over $A/I$ and $\operatorname{Tor}_1^A(A/I, M) = 0$;
$I$-adically ideal-separated means that for any ideal $\mathfrak a \subset A$, the module $\mathfrak a \otimes_A M$ is $I$-adically separated. For example if $M$ is a finite $A$-module this is always satisfied, right?
Since we want to know if $\mathcal O_{X, x} \to \mathcal O_{Y, y}$ is flat, I think we want to apply the local criterion for flatness to $A = \mathcal O_{X, x}$ and $M = \mathcal O_{Y, y}$. Since we already know that $Y_s \to X_s$ is flat, we know that $\mathcal O_{X,x} / \mathfrak m_s \mathcal O_{X, x} \to \mathcal O_{Y, y} / \mathfrak m_s \mathcal O_{Y, y}$ is flat, so I guess $I = \mathfrak m_s \mathcal O_{X, x}$ is the correct choice.
Then $\mathcal O_{Y, y}$ is $I$-adically ideal-separated, because any ideal $\mathfrak a \subset \mathcal O_{X, x}$ is finitely generated, so $\mathfrak a \otimes_A \mathcal O_{Y, y}$ is a quotient of finitely many copies of $\mathcal O_{Y, y}$ and as $\mathcal O_{Y, y}$ is noetherian, the quotient is also $I$-adically separated.
But I don't know how to show $$\operatorname{Tor}^A_1(A/I, M) = 0.$$
I'm aware that I didn't yet use the flatness of $X \to S$ and $Y \to S$, maybe that is important here?
Any help would be appreciated 🙂
Best Answer
Claim. For all $p \geq 1$ we have $\operatorname{Tor}_p^{\mathcal O_{x}}(\mathcal O_{x} / \mathfrak m_s \mathcal O_{x}, \mathcal O_{y}) = 0$.
Proof. Choose a free resolution $$F_\bullet \to \mathcal O_{s} / \mathfrak m_s \to 0$$ of $\mathcal O_{s}$-modules. As $\mathcal O_{s} \to \mathcal O_{x}$ is flat, we obtain a free resolution $$F_\bullet \otimes_{\mathcal O_s}\mathcal O_x \to \mathcal O_x / \mathfrak m_s \mathcal O_x \to 0$$ by tensorizing, and we may use this resolution to compute $\operatorname{Tor}_p^{\mathcal O_x}(\mathcal O_x / \mathfrak m_s \mathcal O_x, \mathcal O_y)$. To do this, we have to tensorize with $\mathcal O_{y}$, so observe $$(F_\bullet \otimes_{\mathcal O_s}\mathcal O_x) \otimes_{\mathcal O_x} \mathcal O_y = F_\bullet \otimes_{\mathcal O_s} \mathcal O_y.$$ But $\mathcal O_s \to \mathcal O_y$ is flat, so this complex is still a resolution, hence the $\operatorname{Tor}$-groups (which are higher homology groups) vanish.