# Flatness preserved under Cartesian product

Let $X_1, X_2, X_3, Y_1, Y_2$ and $Y_3$ be projective schemes. Let $f_1:X_1 \to Y_1, f_2:X_2 \to Y_2$ and $f_3:X_3 \to Y_3$ be flat morphisms. Suppose there are morphism $g_1:X_1 \to X_2$, $g_2:X_3 \to X_2$, $h_1:Y_1 \to Y_2$ and $h_2:Y_3 \to Y_2$ such that $f_2 \circ g_1=h_1 \circ f_1$ and $f_2 \circ g_2=h_2 \circ f_3$. Is it then true that the induced morphism from $X_1 \times_{X_2} X_3$ to $Y_1 \times_{Y_2} Y_3$ is flat?

• The title doesn't fit to the question. Better choose "flatness preserved by fiber products" or something like that. – Martin Brandenburg Sep 10 '13 at 15:47

## 1 Answer

Let's work with arbitrary schemes first. Recall that flat morphisms are stable under base change and composition. Now we factor the morphism as

$X_1 \times_{X_2} X_3 \to X_1 \times_{Y_2} X_3 \to Y_1 \times_{Y_2} X_3 \to Y_1 \times_{Y_2} Y_3.$

The first arrow is a base change of $\Delta : X_2 \to X_2 \times_{Y_2} X_2$. The second morphism is flat because it is a base change of $X_1 \to Y_1$, and the third morphism is flat because it is a base change of $X_3 \to Y_3$.

This shows: We are OK when $\Delta$ is flat, and conversely flatness of $\Delta$ is a special case of the claim (take $X_1=X_2=X_3=Y_1=Y_3$).

But when $X_2 \to Y_2$ is separated (which happens, for example, when $X_2$ and $Y_2$ are separated), the diagonal is a closed immersion, hence almost never flat. For example, in the affine case, $R/I$ can only be a flat $R$-module when $I=I^2$. If $I$ is finitely generated, this even implies that $I$ is generated by an idempotent element.

• @Brandenburg: Thanks for the answer. Since my scheme is projective in most cases we will have that $X_2$ and $Y_2$ are separated in most cases. Could you suggest any other condition? – Chen Sep 10 '13 at 18:31
• You obviously haven't understood my answer. It is basically "No, the morphism is not flat, except for pathological cases." – Martin Brandenburg Sep 10 '13 at 20:24