# Base change and irreducibility/reducedness/connectedness in Qing Liu's book (3.2.7 and 3.2.11 using 3.2.6)

This is from Qing Liu's Algebraic Geometry and Arithmetic Curves (page 89). When he proves (c), he first prove the result when $$K/k$$ is finite, and then says that the general case immediately follows from the finite case and lemma 2.6 (posted below). I have no clue of how the lemma can be use to reduce the general case to the finite case.

Moreover, there is a remark pictured below that also uses the lemma 2.6 and once again I'm confused about how to use it. How can I complete these details?

Let us explain the reduction in the purely inseparable $$\Rightarrow$$ homeomorphism claim first. In order to reduce to the finite case, we need to show that if $$X_K\to X$$ isn't a homeomorphism, then there is some finite subextension $$k\subset F\subset K$$ so that $$X_F$$ isn't a homeomorphism. As a homeomorphism is equivalent to a closed bijection, we'll tick off the three conditions of surjective, closed, and injective.

We know $$\operatorname{Spec} K\to \operatorname{Spec} k$$ is surjective for any field extension $$k\subset K$$, since both are just a single closed point. By exercise 3.1.8 (page 86), any base change of this morphism is surjective. So $$X_K\to X$$ is surjective for any field extension $$k\subset K$$, as it is a base change of $$\operatorname{Spec} K\to\operatorname{Spec} k$$ by $$X\to\operatorname{Spec} k$$. So we need not worry about surjectivity, since it's automatically satisfied.

Now we want to show that if $$k\subset K$$ is algebraic and $$X_K\to X$$ isn't closed, then there is a finite subextension $$F$$ so that $$X_F\to X$$ is not closed. Suppose $$X_K \to X$$ isn't closed. Then there is some reduced closed subscheme $$Z\subset X_K$$ so that the image of $$Z$$ is not closed. We apply lemma 2.6 to get an $$F$$ and a reduced closed subscheme $$W\subset X_F$$ with $$Z=W_K$$ - then the image of $$Z$$ in $$X$$ is the same as the image of $$W$$ in $$X$$, and so $$X_F\to X$$ is not closed.

It remains to show that if $$X_K\to X$$ isn't injective, then $$X_F\to X$$ isn't injective for some finite subextension $$k\subset F\subset K$$. If we can find two closed points $$y,y'\in X_K$$ so that their images are the same point $$x$$ in $$X$$, the we win: via two applications of the lemma, we get finite extensions $$k\subset F,F'$$ and reduced closed subschemes $$Y\subset X_F$$ and $$Y'\subset X_{F'}$$ so that $$y=Y_K,y'=Y'_K$$. Next, via considering a finite extension $$\widehat{F}$$ which contains both $$F,F'$$ we get that $$Y_\widehat{F}$$ and $$Y_\widehat{F}'$$ are distinct single points of $$X_\widehat{F}$$ which both map to $$x$$, which proves the claim modulo the statement about closed points.

In order to produce two distinct closed points from two distinct arbitrary points having the same image, suppose we have two points $$y,z\in X_K$$ which have image $$x\in X$$. By base-changing along the closed immersion $$\overline{\{x\}}\to X$$ where we put the reduced induced structure on the source, we may assume that both $$y,z$$ map to the generic point of $$X$$. Now I claim that neither is in the closure of the other. If this were so, say $$z$$ were in the closure of $$y$$, then because $$X_K\to X$$ is surjective, we would have that $$\dim \overline{\{z\}}$$ is at least the dimension of $$X$$, and as $$\overline{\{z\}}$$ is properly contained in $$\overline{\{y\}}$$, this would imply $$\dim \overline{\{y\}} > \dim X$$, which contradicts part (a) of proposition 3.2.7. Now we have two distinct irreducible components $$Y,Z$$ of $$X_K$$ which surject on to $$X$$ and are of the same dimension as $$X$$. So $$\dim Y\cap Z < \dim Y=\dim Z$$, and the image of $$Y\cap Z$$ cannot be all of $$X$$ because it is contained inside a proper closed subscheme of strictly smaller dimension. Pick a closed point $$p\in X$$ in the complement of the image of $$Y\cap Z$$, and then the fiber of $$Y$$ and $$Z$$ both have a closed point and by the choice of $$p$$, these are distinct. So we have obtained our two distinct closed points.

For the other portion of the question, we will show that if $$k\subset K$$ is an algebraic field extension, then $$X_K$$ has the property reduced, irreducible, or connected iff $$X_F$$ has that property for every finite extension $$k\subset F\subset K$$. In particular, this shows that if $$X_\overline{k}$$ is reduced/irreducible/connected, then $$X_K$$ is for any algebraic extension via two applications of the previous statement.

Let $$k\subset K$$ be an algebraic field extension. Suppose $$X_K$$ is not reduced. Then $$X_K^{red}$$, the reduction, is a closed reduced subvariety. Applying lemma 2.6, we can find an intermediate field $$k\subset K'\subset K$$ and a reduced closed subvariety $$Z\subset X_{K'}$$ so that $$X_K^{red}=Z_K$$. We note that such a $$Z$$ cannot be equal to $$X_{K'}$$, as $$(X_{K'})_K=X_K\neq X_K^{red}$$. On the other hand, if there is a finite subextension $$k\subset F\subset K$$ so that $$X_F$$ is non-reduced, then $$(X_F^{red})_K$$ gives closed subscheme of $$X_K$$ which contains all the points of $$X_K$$ but is not equal to $$X_K$$, and thus $$X_K$$ is not reduced. This shows that $$X_K$$ is reduced iff $$X_F$$ is reduced for every finite extension $$k\subset F \subset K$$, and we are done as outlined in the first paragraph of this section.

Irreducible is similar. If $$X_K=Y\cup Y'$$ is reducible as a topological space with $$Y,Y'$$ closed, then by putting the reduced induced structure on $$Y,Y'$$ we get two closed reduced subschemes $$Y,Y'\subset X_K$$ and by two applications of the lemma, $$k\subset F$$ and $$k\subset F'$$ both finite extensions with $$Z\subset X_F$$ and $$Z'\subset X_{F'}$$ so that $$Y=Z_F$$ and $$Y'=Z'_{F'}$$. Now let $$k\subset\widehat{F}\subset K$$ be a finite subextension containing $$F,F'$$ - then $$Z_\widehat{F}$$ and $$Z_\widehat{F}'$$ are reduced closed subvarieties of $$X_\widehat{F}$$ which do not contain eachother and have union $$X_\widehat{F}$$ (as sets). So $$X_K$$ reducible implies that there is a finite extension $$k\subset \widehat{K}$$ so that $$X_\widehat{K}$$ is reducible. On the other hand, we immediately verify that if $$X_F=X_1\cup X_2$$ is reducible with $$X_1\not\subset X_2$$ nor $$X_2\not\subset X_1$$, $$X_K$$ is reducible: the closed subschemes $$X_1$$ and $$X_2$$ produce closed subschemes $$(X_1)_K$$ and $$(X_2)_K$$ of $$X_K$$ which are closed, have union (as sets) $$X_\overline{k}$$, and do not contain eachother. We are done as in the first paragraph.

Connected is the same proof as irreducible, except instead of just $$X_1,X_2$$ or $$Y,Y'$$ not containing each other, we require that they do not intersect at all.

• the part of your proof that that if $k \subset K$ is an algebraic field extension, then $X_K$ is reduced iff $X_F$ is reduced for every finite extension $k \subset F \subset K$ I not completely understand. You use both times contraposition:
– user784905
May 7, 2020 at 3:06
• On first direction: You assume that $X_K$ is not reduced. $X_K^{red}$ is by def reduced, then by lemma 2.6 we find tower $k\subset K'\subset K$ and reduced closed $Z\subset X_{K'}$ with $X_K^{red}=Z_K$. Well, the lemma tells us also that $Z$ is unique with this property $X_K^{red}=Z_K$ and you also conclude correctly that $Z \neq X_{K'}$. But why does this imply that $X_{K‘}$ is not reduced? There was nowhere a claim that $Z$ is the "biggest" reduced subvariety of $X_K$,right? So I not see why $X_K^{red}=Z_K$ and $Z \neq X_{K'}$ imply that $X_F$ is not reduced.
– user784905
May 8, 2020 at 0:45
• The "$\Rightarrow$" direction confuses me as well: we assume that we have again a tower of extensions $k\subset F\subset K$ with $X_F$ non reduced. What do we know about $(X_F^{red})_K$? Topologically it coinsides with $X_K$, that is it contains all points. But is $(X_F^{red})_K$ reduced? In addition, why $(X_F^{red})_K \neq X_K$ and if we know the later, why this imply that $X_K$ is not reduced?
– user784905
May 8, 2020 at 0:50
• @FShrike That probably works too - it's been 4 years since I wrote this, I can't remember any specific reasons why I wrote it this way besides it seemed natural. Anyways, if $Y\subset X_F$ is such that $y=Y_K$, then $Y$ must also be a single point: consider the base change diagram with rows $Y_K \to Y_{\widehat{F}} \to Y$ and $\operatorname{Spec} K\to \operatorname{Spec}\widehat{F}\to \operatorname{Spec}F$ and note that since the bottom maps are all surjective, the top maps are too, and $Y_K$ is a single point, implying the other schemes in the top are all one point. Feb 4 at 2:25
• @FShrike a closed immersion is of finite type and a composition of finite type morphisms is again of finite type. Both of these are results you should meet early on in your study of algebraic geometry, and they're both pretty approachable. Feb 4 at 17:26