Irreducibility of $X\times_k\overline{k}$ (Hartshorne 3.15 (a))

Question

I'm working on Hartshorne's exercise II.3.15 (a), namely:

Let $X$ be a scheme of finite type over a field $k$ (not necessarily algebraically closed). Show that the following three conditions are equivalent:

(i) $X\times_k\overline{k}$ is irreducible.

(ii) $X\times_kk_s$ is irreducible (where $k_s$ is the separable closure of $k$)

(iii) $X\times_kK$ is irreducible for every field extension $K|k$.

I've found the following solution on the internet, which I'm having a hard time trying to understand (I've selected part (ii) $\Rightarrow$ (i), which seems to be the essential one).

My questions are:

1. Why can we restrict ourselves to purely inseparable extensions $K|k$? What about the others?

2. Why can we immediatly reduce to the case $X=\text{Spec}(A)$?

3. Why is it relevant to consider that $\text{Spec}(A)$ is homeomorphic to $\text{Spec}(A_{\text{red}})$? (I know the homeomorphism exists, I just can't see why it is important here)

4. He says that "$A\otimes_k K$ having a zero-divisor is equivalent to a system of equations with coefficients in $k$ having a solution over $K$". I have no idea what this means. An element in $A\otimes_k K$ is something of the form $\sum_ia_i\otimes b_i$ and the product of things like that looks just like a big mess. How do we find this system he is talking about?

Thank you!

Answer 1

Noting that we are trying to prove that $X\times_k k_s$ irreducible implies $X\times_k \overline{k}$ irreducible resolves point 1: in the proof (ii) implies (i), we have that $\overline{k}/k_s$ is a purely inseparable extension. Therefore it is enough to show that purely inseparable extensions preserve irreducibility.

For point 2, there is something more going on here. The following lemma is useful:

Lemma. A scheme $S$ is irreducible if and only if there exists an affine open covering $S=\bigcup_{i\in I} U_i$ so that $I$ is nonempty, $U_i$ is irreducible for all $i\in I$, and $U_i\cap U_j\neq\emptyset$ for all $i,j\in I$.

Proof. The forward direction is clear: any open subset of an irreducible topological space is irreducible, and the generic point is in every open subset, so any affine open cover suffices. To show the reverse direction, suppose $S=Z_1\cup Z_2$ is a union of two closed subsets. For every $i$, we see that $U_i\subset Z_1$ or $U_i\subset Z_2$, so pick some $i$ and assume (up to possibly renumbering the $Z$) that $U_i\in Z_1$. For any $j\in I$, the open subset $U_i\cap U_j$ is dense in $U_j$ and contained in the closed subset $Z_1\cap U_j$. So $U_j\subset Z_1$ as well, and thus $X=Z_1$. $\blacksquare$

This implies we can reduce to the affine case: if $\{U_i\}$ is an affine open cover of $X$ and $(U_i)_K$ is again irreducible, then $\{(U_i)_K)\}$ fulfills the conditions of the lemma and $X_K$ will be irreducible.

For point 3, the goal is to reduce from proving "$\operatorname{Spec} A$ irreducible implies $\operatorname{Spec} A\otimes_k K$ irreducible" to the claim "$A$ a finitely generated domain over $k$ implies $A\otimes_k K$ a finitely generated domain over $K$". To explain why this works, the map $X_{red} \to X$ is a universal homeomorphism, so the base change of $X_{red}\to X$ by $X_K\to X$ is again a universal homeomorphism, so $(X_{red})_K$ is irreducible iff $X_K$ is. Really, the underlying message here is that since irreducibility is a topological property, we should be free to choose a nice scheme structure to work with. The nicest choice we can make for this part of the problem is $X_{red}$.

Point 4 is a fun trick. Write $\{a_i\}$ for a basis of $A$ as a $k$-vector space; then the $a_i\otimes 1$ (which by abuse of notation I will also call $a_i$) form a basis for $A\otimes_k K$ as a $K$-vector space. These have a multiplication table: $a_ia_j=\sum\Gamma_{ij\ell} a_\ell$ for $\Gamma_{ij\ell}\in k$, so if we have two elements of $A$ or $A\otimes_kK$, say $z_1=\sum_i c_{1i}a_i$ and $z_2=\sum_i c_{2i} a_i$, we can write down their product as $$z_1z_2= \sum_\ell\left( \sum_{i,j} c_{1i}c_{2j}\Gamma_{ij\ell} \right)a_\ell.$$ Now let $z_1,z_2\in A\otimes_kK$ be two nonzero elements such that $z_1z_2=0$. Then this gives us a list of equations: we know that $\sum_{i,j} c_{1i}c_{2j}\Gamma_{ij\ell}=0$ for all $\ell$. As only finitely many of $c_{1i}$ and $c_{2j}$ are nonzero, we get a finite list of equations in finitely many variables (the $c_{1i}$ and $c_{2j}$) with coefficients in $k$ which have a solution over $K$.