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?