Proof of rigidity lemma

Question

I am trying to understand in full details the proof of rigidity lemma as proved here http://staff.science.uva.nl/~bmoonen/boek/DefBasEx.pdf [Lemma 1.11, Pag. 12].

The statement in this reference is

Lemma Let $X$, $Y$ and $Z$ be algebraic varieties over a field $k$. Suppose that $X$ is complete. If $f : X × Y \rightarrow Z$ is a morphism with the property that, for some $y \in Y (k)$, the fibre $X \times_k \{y\}$ is mapped to a point $z \in Z(k)$ then $f$ factors through the projection $pr_Y : X \times_k Y → Y$.

I will express what I am not understanding of the proof in the following questions. (Please, read the proof as in the link to understand the notation).

1. Why we can reduce to the case $k=\bar{k}$?
2. Why can we choose $x_0\in X(k)$, that is, why $X(k)$ is non-empty?
3. What does it mean "since $X\times_k Y$ is reduced it sufficies to prove it for $k$-rational points"? Why is this possible?
4. Why $f$ is constant on $X\times_k \{P\}$?
5. At the end they say: $f=g \circ pr_Y$ on the non-empy open set $X\times_k (Y-V)$. So $f=g \circ pr_Y$ everywhere because $X\times_k Y$ is irreducible. Why?

For question (5) I would argue like this: since $X\times_k Y$ is irreducible, $X\times_k (Y-V)$ is a dense subset. Then by continuity $f=g \circ pr_Y$ everywhere. Is this argument correct?

Any help is appreciated, thank you.

Answer 1

(1) That's a diagram chase: by base change you get a morphism $X \times Y \times \overline{k} \to Z \times \overline{k}$ and if this one factors through the projection to $Y \times k$, you can factor the original through $Y$.

(2) If $k$ is an algebraically closed field then every variety $X$ has a $k$-rational point. A $k$-rational point is the same as a morphism from $\text{Spec}k$ to $X$. If $X=\text{Spec}A$ is affine then this is the same as a homomorphism from $A$ to $k$. By Hilbert's Nullstellensatz every maximal ideal in $A$ gives you such a homomorphism.

(3) In general, a pair of morphisms of schemes that induce the same map on the set of closed points may not be the same (there is much more data in a morphism of schemes then just what it does to the closed points. Arbitrary schemes need not even have closed points!)

But for varieties (reduced, separated, of finite type over an algebraically closed field, say), this is true: the morphism is determined by what it does to the closed points, set-theoretically. So they prove that the maps $f$ and $g \circ \text{pr}_{Y}$ are the same on the closed points.

(4) This variety is isomorphic to $X$ and hence complete. Every morphism from a complete irreducible variety to an affine variety is constant.

The image of the morphism must be closed and irreducible and hence affine. Now an affine variety of positive dimension has non-constant global sections. By pullback, so does $X$. But this is impossible: on a complete variety, all global sections are constant (well, assuming of course that we are working over an algebraically closed field and that the variety is reduced)

(5) This is close, but not true: you are using a sort of "Hausdorffness" argument, that if two continuous maps agree on a dense subset, then they are the same. However, a variety is not Hausdorff. It is true, however, that for separated schemes (and this includes varieties), the locus where two morphisms agree is closed. If in addition the scheme is reduced, then it has no proper closed subschemes with the same underlying topological space, and then two morphisms that agree on a dense subset are the same. Note that we are using here again the fact that a morphism is determined by what it does to the closed points set theoretically.