Lemma 29.6.1. Let $f : X \to Y$ be a morphism of schemes. There exists a closed subscheme $Z \subset Y$ such that $f$ factors through $Z$ and such that for any other closed subscheme $Z' \subset Y$ such that $f$ factors through $Z'$ we have $Z \subset Z'$.

Proof. Let $\mathcal{I} = \mathop{\mathrm{Ker}}(\mathcal{O}_ Y \to f_*\mathcal{O}_ X)$. If $\mathcal{I}$ is quasi-coherent then we just take $Z$ to be the closed subscheme determined by $\mathcal{I}$, see Lemma 29.2.3. This works by Schemes, Lemma 26.4.6. In general the same lemma requires us to show that there exists a largest quasi-coherent sheaf of ideals $\mathcal{I}'$ contained in $\mathcal{I}$. This follows from Lemma 29.4.2. $\square$

Comment #3251 by Laurent Moret-Bailly on

I don't think Lemma 28.4.2 (tag 01QZ) is needed. Just use the fact that the set of closed subschemes of $Y$ has arbitrary intersections, which are products in the category of all $Y$-schemes. (Of course this generalizes to affine morphisms).

Comment #3347 by on

Yes, that is right, but is your argument really different? You'd still have to prove the thing about intersections of closed subschemes (the material on products and more generally limits of schemes comes a bit later in the Stacks project, namely in Chapter 32, so we can't use it here). The problem with lemmas like this is that they are obvious to those who have mastered the material, but they aren't immediate from the definitions. Anyway, since the proof is correct, I am going to leave it as is.

There are also:

• 8 comment(s) on Section 29.6: Scheme theoretic image

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).