Let $X$ be an affine, complex variety, $A$ be a $\mathbb{C}$-algebra (not necessarily noetherian) and $F_A$ is a coherent sheaf over $X \times \mbox{Spec}(A)$, flat over $\mbox{Spec}(A)$. Denote by $Y \subset X \times \mbox{Spec}(A)$ the scheme-theoretic support of $F_A$. Then, does there exist a non-empty open subset $U \subset \mbox{Spec}(A)$ such that $Y \cap (X \times U)$ is flat over $U$? This is true if $A$ does not have any nilpotent elements.


1 Answer 1


Let $A$ be a non-reduced, Artinian ring with maximal ideal $\mathfrak{m}$. The underlying topological space of $\text{Spec}\ A$ is a one-point space. Thus, the unique nonempty open is the entire space $\text{Spec}\ A$. Thus, the problem asks, for every $A$-flat coherent sheaf on a finite type $A$-scheme, whether the scheme-theoretic support of the coherent sheaf is also $A$-flat. That is not true.

Consider the $A$-algebra $\widetilde{B}:= A\epsilon \times A\eta $ that is free of rank $2$ as an $A$-module, i.e., $$\widetilde{B}=A[\epsilon,\eta]/\langle\ \epsilon+\eta-1,\ \epsilon\eta\ \rangle= A\epsilon \oplus A\eta.$$ For every ideal $H\subset A$, consider the $A$-subalgebra of $\widetilde{B}$, $$B_H =\{\ a\epsilon + b\eta\in \widetilde{B} \ |\ a-b\in H\ \}.$$ Finally, consider the following $B_H$-module, $$M := B_H/\left( \langle \epsilon \rangle \cap B_H \right) \oplus B_H/\left( \langle \epsilon \rangle \cap B_H \right).$$

1. For every nonzero ideal $H\subset \mathfrak{m}$, the $A$-module $B_H$ is not flat.
2. The $A$-module $M$ is flat of rank $2$.
3. Also, the $B_H$-annihilator of $M$ equals the zero ideal.
Thus, on the scheme $Y=\text{Spec}\ B_H$, the scheme-theoretic support of the $A$-flat coherent sheaf $\widetilde{M}$ equals $Y$, and $Y$ is not $A$-flat.

Proof. The $A$-module $B_H$ equals $A\cdot 1 \oplus H\cdot \epsilon$. Since $H$ is not $A$-flat, also $B_H$ is not $A$-flat.

As an $A$-algebra, each of the following quotient $\widetilde{B}$-algebras is isomorphic to $A$ itself, $$\widetilde{B}/\langle \epsilon \rangle, \ \ \widetilde{B}/\langle \eta \rangle.$$ Since the $A$-subalgebra $A\cdot 1$ is contained in $B_H$, also the quotient $B_H$-algebras also equal $A$, $$B_H/\langle \epsilon \rangle \cap B_H = A = B_H/\langle \eta \rangle \cap B_H.$$ Thus, the $A$-module $M$ is free of rank $2$.

Finally, the common intersection $\langle \epsilon \rangle \cap \langle \eta \rangle$ in $B$ is the zero ideal. Thus, the common intersection in $B_H$ is also the zero ideal. Therefore the annihilator ideal of $M$ is the zero ideal. Thus, the scheme-theoretic support of $M$ equals all of $\text{Spec}\ B_H$. QED


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