# Closed image of a product of morphisms

This may be pretty trivial, but I can't figure it out. Suppose that $S$ is any scheme, and $f_1:X_1\to Y_1$ and $f_2:X_2\to Y_2$ are two morphisms of $S$-schemes, such that the closed image of each one exists. That is, there is a smallest closed subscheme $Z_i$ of $Y_i$ over which $f_i$ factorizes (for $i=1,2$).

Is it true then that the closed image of $f_1\times_S f_2:X_1\times_S X_2\to Y_1\times_S Y_2$ exists? In this case, is it equal to $Z_1\times_S Z_2$ (with its natural closed immersion into $Y_1\times_S Y_2$)?

• the closed image always exists. – Martin Brandenburg Apr 22 '10 at 23:24
• To quote chapter and verse, this is Hartshorne II, Ex. 3.11(d) – Mariano Suárez-Álvarez Apr 23 '10 at 1:11
• Ok, this is true, sorry. In EGA I (9.5) everything is stated as if the closed image does not always exist. One sufficient condition for this is that the direct image of the structure sheaf is quasi-coherent, which is not always true. But in any case, one can take the smallest ideal containing it which is quasi-coherent and that works. – unknown Apr 23 '10 at 8:04

The closed image of $Spec(B) \to Spec(A)$ is the spectrum of $A/K$, where $K$ is the kernel of $A \to B$. Let $A',B',K$ be analogously defined. The closed image of $Spec(B \otimes_R B') \to Spec(A \otimes_R A')$ is the spectrum of $(A \otimes_R A')/L$, where $L$ is the kernel of $A \otimes_R A' \to B \otimes_R B'$. We have to compare this ring with $A/K \otimes_R A'/K = (A \otimes_R A')/\langle K,K' \rangle$. If $K=K'=0$, this asks if the tensor product of two injective rings maps is injective, which is false in general.
This discussion shows that everything is fine if, in your notation, $X_1,Y_2$ or $X_2,Y_1$ are flat over $S$.