# Is the support of a flat module generically flat?

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.

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, $$M := B_H/\left( \langle \epsilon \rangle \cap B_H \right) \oplus B_H/\left( \langle \epsilon \rangle \cap B_H \right).$$

Proposition.
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