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?

  • 4
    $\begingroup$ In general, the tensor product is not Noetherian: According to Theorem 3 in mat.uniroma3.it/scuola_orientamento/alumni/laureati/cittadini/… $F\otimes_k F$ is Noetherian if and only if $F/k$ is finitely generated. Hence, for example ${\mathbb Q}(X_1,X_2,...)\otimes_{\mathbb Q} {\mathbb Q}(X_1,X_2,...)\cong{\mathbb Q}(X_1,X_2,...)\otimes_{\mathbb Z} {\mathbb Q}(X_1,X_2,...)$ is not Noetherian. $\endgroup$
    – Hanno
    Feb 18, 2015 at 9:54
  • $\begingroup$ Thank you. Your example implies that fibre product over $\mathbb Z$ even of 1-point schemes ($\text{Spec}$s of fields) need not be a Noetherian scheme, right? Interesting. So $X\times\mathbf A^1$ being Noetherian is a consequence of this specific setup. $\endgroup$
    – aytio
    Feb 18, 2015 at 10:19
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    $\begingroup$ The reason $X\times_\mathbf{Z}\mathbf{A}^1=:\mathbf{A}^1_X$ is ``clearly Noetherian" if $X$ is is that morphisms of finite type are stable under base change. The morphism $\mathbf{A}^1=\mathbf{A}^1_\mathbf{Z}\to\mathrm{Spec}(\mathbf{Z})$ is of finite type, so the base change $\mathbf{A}_X^1=X\times_{\mathbf{Z}}\mathbf{A}^1\to X$ is of finite type. But as $X$ is Noetherian, any $X$-scheme of finite type is itself Noetherian. $\endgroup$ Feb 18, 2015 at 14:24
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    $\begingroup$ That $X\times A^1$ is noetherian is more or less equivalent to Hilbert's Basis theorem! $\endgroup$ Feb 18, 2015 at 15:16
  • $\begingroup$ Dear @GeorgesElencwajg, thank you for your answer below. I do have access to the article by Vámos (pdf), but I am not sure how I could share it with you via stackexchange. $\endgroup$
    – aytio
    Feb 19, 2015 at 10:09

1 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 $\leq k(N-1)+1$]


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