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Let $p, q$ be prime numbers which may or may not be distinct. Let $\mathbb{Z}_p$ be the $p$-adic completion of $\mathbb{Z}$. Similarly for $\mathbb{Z}_q$. Is $\mathbb{Z}_p\otimes_{\mathbb{Z}} \mathbb{Z}_q$ noetherian?

Makoto Kato
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2 Answers2

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No, since otherwise also the localization $p^{-1} q^{-1} \mathbb{Z}_p \otimes_{\mathbb{Z}} \mathbb{Z}_q = \mathbb{Q}_p \otimes_{\mathbb{Q}} \mathbb{Q}_q$ is noetherian, which is not the case, since the transcendence degree of $\mathbb{Q}_p$ over $\mathbb{Q}$ is infinite (see my answer here).

For $p=q$, I can give you an explicit example of an ideal which is not generated: The kernel $I$ of the multiplication $\mathbb{Z}_p \otimes_\mathbb{Z} \mathbb{Z}_p \to \mathbb{Z}_p$, i.e. $I = \langle a \otimes 1 - 1 \otimes a : a \in \mathbb{Z}_p \rangle$. I don't know a direct argument, but one can use $$I/I^2 \cong \Omega^1_{\mathbb{Z}_p / \mathbb{Z}} \cong \Omega^1_{\mathbb{Z}[[X]]/(X-p) ~/\mathbb{Z}} \cong \Omega^1_{\mathbb{Z}[[X]]/\mathbb{Z}} /\langle X-p,d(X-p) \rangle$$ and MO/21189.

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Martin Brandenburg
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  • Could you explain what exactly $p^{-1} q^{-1} \mathbb{Z}p \otimes{\mathbb{Z}} \mathbb{Z}_q$ is? – Makoto Kato Feb 21 '14 at 09:58
  • If $A$ is a ring and $f \in A$, then $f^{-1} A := A_f$ is the localization of $A$ at (the multiplicative subset generated by) $f$. – Martin Brandenburg Feb 21 '14 at 17:29
  • So I suppose you mean $p^{-1} q^{-1} \mathbb{Z}p \otimes{\mathbb{Z}} \mathbb{Z}q = p^{-1}(q^{-1}(\mathbb{Z}_p \otimes{\mathbb{Z}} \mathbb{Z}q))$, right? Then could you explain how you prove that $p^{-1} q^{-1} \mathbb{Z}_p \otimes{\mathbb{Z}} \mathbb{Z}q = \mathbb{Q}_p \otimes{\mathbb{Q}} \mathbb{Q}_q$? – Makoto Kato Feb 22 '14 at 00:15
  • Use $p^{-1} Z_p = Q_p$ and general nonsense. – Martin Brandenburg Feb 22 '14 at 00:35
  • I understand $p^{-1} \mathbb{Z}p = \mathbb{Q}_p$. Could you elaborate a bit on $p^{-1} q^{-1} \mathbb{Z}_p \otimes{\mathbb{Z}} \mathbb{Z}q = \mathbb{Q}_p \otimes{\mathbb{Q}} \mathbb{Q}_q$? – Makoto Kato Feb 22 '14 at 01:40
  • Use that $p^{-1} \mathbb{Z}p = \mathbb{Q}_p$ and that localizations commute with tensor products, and that $\mathbb{Q}$-linear is the same as $\mathbb{Z}$-linear, so that you can replace $\mathbb{Q}_p \otimes{\mathbb{Z}} \mathbb{Q}q$ by $\mathbb{Q}_p \otimes{\mathbb{Q}} \mathbb{Q}_q$. – Martin Brandenburg Feb 27 '14 at 22:00
  • "$Q$-linear is the same as $Z$-linear, so that you can replace $\mathbb{Q}p\otimes{\mathbb{Z}} \mathbb{Q}q$ by $\mathbb{Q}_p\otimes{\mathbb{Q}} \mathbb{Q}_q$." I'm afraid I don't understand. Could you please explain this? – Makoto Kato Feb 28 '14 at 06:00
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    Lemma: If $M,N$ are $S^{-1} R$-modules, then $M \otimes_{S^{-1} R} N = M \otimes_R N$. Proof: By the Yoneda Lemma, it suffices to prove that every $R$-bilinear map $M \times N \to T$ (with $T$ an $S^{-1} R$-module) is already $S^{-1} R$-bilinear. This is easy to check. – Martin Brandenburg Feb 28 '14 at 15:57
  • I have another question on the issue. Let $S = {p^nq^m \ |\ n, m = 0, 1, 2,\cdots }$. Then $S^{-1}(\mathbb{Z}p \otimes{\mathbb{Z}} \mathbb{Z}q) = \mathbb{Q}_p \otimes{S^{-1}\mathbb{Z}} \mathbb{Q}q$. Are you saying that $\mathbb{Q}_p \otimes{S^{-1}\mathbb{Z}} \mathbb{Q}q = \mathbb{Q}_p \otimes{\mathbb{Z}} \mathbb{Q}_q$? – Makoto Kato Mar 01 '14 at 23:34
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We assume all rings are commutative and have multiplicative identity elements.

Notation Let $A$ be a ring, $B$ an $A$-algebra, $M$ an $A$-module. Then we denote $M\otimes_A B$ by $M_B$.

Lemma 1 Let $A$ be a ring, $B$ an $A$-algebra. Let $M$ and $N$ be $A$-modules. Then $(M\otimes_A N)_B \cong M_B \otimes_B N_B$.

Proof: $(M\otimes_A N)_B \cong (M \otimes_A N) \otimes_A B \cong M \otimes_A N_B \cong M \otimes_A (B\otimes_B N_B) \cong (M \otimes_A B) \otimes_B N_B) \cong M_B \otimes_B N_B$.

Corollary Let $A$ be a ring, $S$ a multiplicative subset of $A$. Let $M$ and $N$ be $A$-modules. Then $S^{-1}(M\otimes_A N) \cong S^{-1}M \otimes_{S^{-1}A} S^{-1}N$.

Proof: This is clear because $S^{-1}M \cong M\otimes_A S^{-1}A$, $S^{-1}N \cong N\otimes_A S^{-1}A$.

Lemma 2 Let $p$ be a prime number. Let $S = \mathbb{Z} - \{0\}$. Then $S^{-1} \mathbb{Z}_p = \mathbb{Q}_p$.

Proof: Let $x$ be a nonzero element of $\mathbb{Q}_p$. There exists an integer $n$ such that $p^n x \in \mathbb{Z}_p$. Hence $S^{-1} \mathbb{Z}_p = \mathbb{Q}_p$.

Lemma 3 Let $p, q$ be prime numbers which may or may not be distinct. Let $S = \mathbb{Z} - \{0\}$. Then $S^{-1} (\mathbb{Z}_p \otimes_{\mathbb{Z}} \mathbb{Z}_q) \cong \mathbb{Q}_p \otimes_{\mathbb{Q}} \mathbb{Q}_q$.

Proof: Clear by Lemma 2 and the corollary of Lemma 1.

Proposition Let $p, q$ be prime numbers which may or may not be distinct. Then $\mathbb{Z}_p \otimes_{\mathbb{Z}} \mathbb{Z}_q$ is not noetherian.

Proof: Otherwise $\mathbb{Q}_p \otimes_{\mathbb{Q}} \mathbb{Q}_q$ is noetherian by Lemma 3. However, this is not the case by this question.

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Makoto Kato
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