Irreducibility is preserved under base extension

Question

I want to prove that if $A$ is a finitely generated $k$-algebra ($k$ is a field) with prime nilradical then for any field extension $k\rightarrow K$, the $K$-algebra $A\otimes_kK$ has also prime nilradical, that is, if for some $f,g\in A\otimes_kK$ we have $(fg)^n=0$ for some $n$ then we must conclude that $f^n=0$ or $g^n=0$ for some $n$.

I am considering the case $K=\overline{k}$. First I take an algebraic extension $K$ which is purely inseparable over $k$. If $f\in A\otimes_kK$ then it is easy to show that $f^{p^n}\in A$ for some $n$ ($p=Char\ k$) and with this observation it is easy to conclude that $A\otimes_kK$ has prime nilradical in this case. Now if I could conclude the proof in the case where $K$ is a finite separable extension then I am done because taking $K=k_s$, where $k_s$ is the separable closure of $k$ then it is not difficult to conclude that $A\otimes_kk_s$ has prime nilradical and finally I can use the observation that $\overline{k}$ is a purely inseparable extension of $k_s$.

A final observation is that if $k\rightarrow F\rightarrow K$ are extensions and if $A\otimes_kK$ has prime nilradical then $A\otimes_kF$ has prime nilradical. So the problem can be reduced to:

If $A$ is a finitely generated $k$-algebra with prime nilradical and if $K$ is a finite Galois extension of $k$ then $A\otimes_kK$ has prime nilradical.

Any help with this problem will be strongly appreciated.

Answer 1

The $\mathbb R$-algebra $\mathbb R[x]/(x^2 + 1)$ has prime nilradical but $\mathbb C[x]/(x^2 + 1)$ does not.

Answer 2

This is true for example when $k$ is algebraically closed (and this one the main steps in the exercise in Hartshorne's book mentioned in the comments). In fact, the following is true:

If $k$ is an algebraically closed field and $X,Y$ are irreducible (integral, reduced) $k$-schemes, then the same is true for $X \times_k Y$.

Proof sketch: For "integral", first do the affine case: Let $A,B$ be integral $k$-algebras, we have to show that $A \otimes_k B$ is integral. By a colimit argument, we may assume that $A,B$ are finitely generated. By Hilbert's Nullstellensatz, $A \to \prod_{\mathfrak{m}} A/\mathfrak{m}$ is injective, where $\mathfrak{m}$ runs through the maximal ideals of $A$, and in each case we have $A/\mathfrak{m}=k$. The same for $B$. Now if $x,y \in A \otimes_k B$ with $xy=0$ and $y \neq 0$, choose some maximal ideal $\mathfrak{n} \subseteq B$ with $0 \neq y \bmod \mathfrak{n}$. For all $\mathfrak{m} \subseteq A$ it follows that $0 = (x \bmod \mathfrak{m})(y \bmod \mathfrak{n})$ in $A/\mathfrak{m} \otimes_k B/\mathfrak{n}=k \otimes_k k=k$, hence $0 = x \bmod \mathfrak{m}$. This shows $x=0$.

For "reduced", reduce to the affine case, and then a similar argument as above works, see here.

For "irreducible", first do the affine case: Let $A,B$ be $k$-algebras such that $A_{red},B_{red}$ are integral. We already know that $A_{red} \otimes_k B_{red}$ is integral, in particular irreducible. Now apply the Lemma below twice to conclude that $A \otimes_k B$ is irreducible. For the general case of two $k$-schemes $X,Y$, it is not hard to show that the generic points of the affine charts coincide, by considering their intersection.

Finally, integral = reduced + irreducible. $\square$

Lemma: For $S$-schemes $X,Y$ the canonical morphism $(X \times_S Y)_{red} \to X_{red} \times_S Y$ is a homeomorphism.

In fact, one can check (for example in affine charts) that this morphism is a closed immersion whose ideal is nil.