# Effective Cartier divisor is an open property

Let $$X$$ be a regular affine $$\mathbb{C}$$-scheme, $$A$$ a (finitely-generated) $$\mathbb{C}$$-algebra. Let $$Y \subset X \times \mathrm{Spec}(A)$$ be a closed subscheme of codimension $$1$$ such that for each $$t \in \mbox{Spec}(A)$$, the fiber $$Y_t$$ is an effective Cartier divisor of $$X \times \{t\}$$. Is $$Y$$ flat over $$\mbox{Spec}(A)$$? In other words, does fiberwise Cartier imply globally Cartier?

• Laurent Moret-Bailly already answered your question. I am just repeating his second counterexample, since you seem not to have understood his point. Let $X$ be $\mathbb{A}^1_k$ with coordinate $s$, let $A$ be $k[t]$, and let $Y$ be the closed subscheme with defining ideal $\langle s^2,st \rangle$. The fiber $Y_t$ is an effective Cartier divisor in $X\times\{t\}$ for every $t$ in $\text{Spec}\ A$. However, $Y$ is not flat over $\text{Spec}\ A$. Jan 17, 2019 at 19:40
• @JasonStarr Thank you. I understand now. Jan 17, 2019 at 20:01

I assume that "scheme" means $$\mathbb{C}$$-scheme, and that the product is over $$\mathbb{C}$$. I also assume that $$Y$$ is a closed subscheme.
First (trivial) counterexample: take for $$Y$$ an effective Cartier divisor in $$X\times t$$ ($$t\in \mathrm{Spec}(A)(\mathbb{C})$$), viewed as a subscheme of $$X\times \mathrm{Spec}(A)$$.
Second counterexample: if $$X$$ is a regular curve, every proper closed subscheme of $$X$$ is a Cartier divisor in $$X$$. So, take for $$Y$$ any closed subscheme not containing any fiber component, and not flat over $$\mathrm{Spec}(A)$$.