Let $\mathbb{Q}_p$ denote the field of fractions of $\mathbb{Z}_p$. By the answers to this quesition the tensor product $\mathbb{Q}_p \otimes_{\mathbb{Q}} \mathbb{Q}_p$ cannot be a Noetherian ring (alternatively, this follows because the transcendence degree of $\mathbb{Q} \to \mathbb{Q}_p$ is infinite). One could instead hope for the weaker result that this ring is coherent. Is this true?


You can use the following:

Lemma. Let $A = \operatorname{colim}_i A_i$ be a filtered colimit of coherent rings such that $A$ is flat over each $A_i$. Then $A$ is coherent.

For example, this is true if all the transition maps $A_i \to A_j$ are flat.

Proof. Let $I \subseteq A$ be a finitely generated ideal. Then $I = AI_i$ for some finitely generated ideal $I_i \subseteq A_i$ for some $i$. By assumption, $I_i$ is finitely presented as $A_i$-module, i.e. there is an exact sequence $$A_i^m \to A_i^n \to A_i \to A_i/I_i \to 0.$$ By flatness of $A_i \to A$, the sequence $$A^m \to A^n \to A \to A/I \to 0$$ is exact as well, i.e. $I$ is finitely presented. $\square$

Example 1. Let $K \subseteq L$ and $K \subseteq M$ be field extensions. Then $A = L \otimes_K M$ is coherent. Indeed, it can be written as a colimit $$A = \underset{\substack{\longrightarrow \\ K \subseteq L_i \subseteq L \\ K \subseteq M_j \subseteq M}}{\operatorname{colim}} L_i \underset K\otimes M_j,$$ where the colimit runs over all finitely generated subextensions $K \subseteq L_i \subseteq L$ and $K \subseteq M_j \subseteq M$. Each $L_i \otimes_K M_j$ is Noetherian, so in particular coherent, and the transition maps are flat because both $L_i \to L_{i'}$ and $M_j \to M_{j'}$ are.

Example 2. The algebraic integers $\bar{\mathbf Z}$ are coherent as the colimit of all $\mathcal O_K$ for $\mathbf Q \subseteq K$ finite. The transition maps $\mathcal O_K \to \mathcal O_L$ are flat because $\mathcal O_K$ is a Dedekind domain and $\mathcal O_L$ is torsion-free.

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  • $\begingroup$ Thank you, that is very nice! I have to admit, I expected the answer to be negative, and as such didn't ask what I was most interested in - do you happen to know if $\mathbb{Z}_p \otimes_{\mathbb{Z}} \mathbb{Z}_p$ is coherent? $\endgroup$ – Drew Heard Jul 2 at 20:45
  • $\begingroup$ At the end, "$\mathcal O_K$ is DVR" should be "is a Dedekind domain", or say that its localizations at primes are DVRs $\endgroup$ – Wojowu Jul 2 at 21:38
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    $\begingroup$ Dear Remy, thank you for the update. But I don't quite understand the example - it seems you claim that the maps in the filtered colimit factor through the localizations? But this does not appear to be the case in general, already for $\mathbb{Z}_{(p)} \to \mathbb{Z}_p$ $\endgroup$ – Drew Heard Jul 4 at 9:27
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    $\begingroup$ Right, that didn't work at all. In fact it's somehow the opposite: the valuations on the intermediate rings (induced by the valuation on $\mathbf Z_p$) are basically never divisorial, because that would give a nontrivial extension of residue fields (for dimension reasons). In particular, $\mathbf Z_p$ cannot be written as a colimit of DVRs that are essentially of finite type. I think it's fair to ask the $\mathbf Z_p$-version as a separate question ― I don't see any way to apply my flatness argument. $\endgroup$ – R. van Dobben de Bruyn Jul 5 at 7:03

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