# The proof concerning morphism between varieties in T. A. Springer's book

The Theorem 1.9.5 in T. A. Springer's Linear Algebraic Groups (page 19) states that:

Let $$\phi: X \rightarrow Y$$ be a morphism of varieties. Then $$\phi X$$ contains a non-empty open subset of its closure $$\overline{\phi X}$$.

(Here, a variety is defined as a quasi-compact ringed space to have an open cover by affine subsets along with being separated.)

The first line of Springer's proof wrotes:

Using a covering of $$Y$$ by affine open sets, we reduce the proof to the case that $$Y$$ is affine.

I wonder how such reduction proceeds, particularly whether the preimage of a affine open subset of $$Y$$ is a subvariety of $$X$$. (Initially I wrongly claim that the preimage of a affine open subset of $$Y$$ is not necessarily quasi-compact.)

Hope for an answer, thanks in advance!

Actually, the map $$\phi:X\to Y$$ is quasi-compact, so the preimage of any quasi-compact set (like an affine open set of $$Y$$) is again quasi-compact. The key result is the following:

Lemma (Stacks Project 03GI): Let $$f:X\to Y$$ and $$g:Y\to Z$$ be morphisms of schemes. If $$g\circ f$$ is quasi-compact and $$g$$ is quasi-separated, then $$f$$ is quasi-compact.

In our situation, we take $$Z=\operatorname{Spec} k$$, $$g$$ the structure morphism, $$f=\phi$$, and then we have that $$g\circ\phi$$ is equal to the structure morphism $$X\to \operatorname{Spec} k$$ by assumption that $$\phi$$ is a map of $$k$$-schemes. As a separated morphism is quasi-separated, we have that $$g$$ is quasi-separated, and because $$X$$ is quasi-compact and $$\operatorname{Spec} k$$ is a point, we have that $$g\circ f$$ is quasi-compact. Thus your situation satisfies the assumptions of the lemma and we may conclude that $$\phi$$ is quasi-compact.

(I do not have a copy of this book in front of me right now, but I remember that some of these algebraic groups texts are notorious for not doing things correctly from a modern algebraic-geometry perspective. If it's doing some wacky thing like only taking closed points, do not fear: these geometric objects embed as subspaces with the induced topology of the schemes in question, and the maps between the not-quite-schemey versions are exactly the restrictions of the scheme maps to these subspaces and everything is okay.)

• May I ask that utilizing the method above, whether we can prove the preimage of a subvariety is still a subvariety? – Wembley Inter Oct 23 '20 at 8:32
• @WembleyInter What do you mean? What parts of that claim are tripping you up? This shows quasi-compactness, and any locally closed subscheme of a separated scheme is again separated, so that would seem to do it by your definitions. – KReiser Oct 23 '20 at 9:39
• Ah... Now I get it. Thanks! – Wembley Inter Oct 23 '20 at 11:01
• @WembleyInter if this resolves your issues, please consider accepting the answer. If not, is there something you feel is missing from the answer? – KReiser Oct 23 '20 at 22:12

Actually, $$\phi$$ is a quasi-compact map.

Indeed, let $$B$$ be an affine base scheme, so that $$X,Y$$ are separated quasicompact over $$B$$. Then $$\phi$$ is the composition of $$\phi_1: X= X \times_Y Y \rightarrow X \times_B Y$$ and $$\phi_2: X \times_B Y \rightarrow Y$$.

$$Y \rightarrow B$$ is separated so $$\phi_1$$ is a closed immersion so is quasi-compact. But $$\phi_2$$ is a base change of $$X \rightarrow B$$ which is a quasi-compact map so is quasi-compact. So the composition $$\phi$$ of $$\phi_1$$ and $$\phi_2$$ is quasi-compact.