Is $\mathbb{Z}_p\otimes_{\mathbb{Z}} \mathbb{Z}_q$ noetherian? 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?
 A: 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.
A: 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.
