Is a product of two Noetherian schemes over Spec $\mathbb Z$ a Noetherian scheme?

Question

In Hartshorne's proof of Proposition 6.6 in Chapter 2, he says that if $X$ being Noetherian implies $X\times\mathbf A^1$ is "clearly" Noetherian.

I assume this is because $X$ can be covered by affine open sets $U_i=\text{Spec}\ A_i$ with Noetherian $A_i$ and then $X\times\mathbf A^1$ is covered by the affine open sets $U_i\times_{\mathbb Z}\mathbf A^1$, which are $\text{Spec}\ \big(A_i\otimes_{\mathbb Z}\mathbb Z[x]\big)=\text{Spec}\ A_i[x]$, and $A_i[x]$ is Noetherian from Hilbert's basis theorem. Is there a simpler way to see this?

More generally if $X,Y$ are schemes over $\mathbb Z$, is $X\times_{\text{Spec}\ \mathbb Z}Y$ always Noetherian? With a similar argument as above this is the same as asking whether $A\otimes_{\mathbb Z}B$ is a Noetherian ring if $A$ and $B$ are, correct? I guess my confusion arises from "Noetherianness" as modules vs. "Noetherianess" as a ring.

Over what schemes $S$ (instead of $\text{Spec}\ \mathbb Z$) is it true that products of two Noetherian schemes over $S$ is a Noetherian scheme?

Answer 1

Here is an example of an extension field $k\to K$ such that $K\otimes_kK$ is not noetherian.
It immediately implies that the product $\text{Spec}(K)\times_k \text{Spec}(K)$ of two copies of the noetherian affine scheme $\text{Spec}(K)$ is not noetherian .

Take for $k$ any non perfect field of characteristic $p\gt 0$ (the rational function field $k=F(T)$ of any field $F$ of characteristic $p$ for example) and consider its perfect closure $K=k^{p^{-\infty}}$.
If $a\in K$ is of level $n$ (i.e. $a^{p^n}\in k$ but $a^{p^{n-1}}\notin k$), then $b=a\otimes 1-1\otimes a\in K\otimes_k K$ is nilpotent of order $p^n$ (i.e. $b^{p^n}=0$ but $b^{p^n-1}\neq 0$)
Since there exist elements $a\in K$ of any level $n\geq 1$ the nilpotent radical $Nil (K\otimes_k K)$ of the ring $K\otimes_k K$ contains nilpotents $b$ of arbitrary high order $p^n$, which is not possible in a noetherian ring .
[Indeed given a noetherian ring $A$, if $Nil(A)=\langle n_1,...,n_k\rangle $ and if the nilpotency order of all $n_i$'s is $\leq N$ then for every $b=\sum _{i=1}^k a_in_i$ we have $b^{k(N-1)+1}=0$, so that all elements of $Nil(A)$ have order of nilpotence $\leq k(N-1)+1$]