# Is $\mathbb{Q}_p \otimes_{\mathbb{Q}}\mathbb{Q}_p$ coherent?

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.

• 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? – Drew Heard Jul 2 at 20:45
• At the end, "$\mathcal O_K$ is DVR" should be "is a Dedekind domain", or say that its localizations at primes are DVRs – Wojowu Jul 2 at 21:38
• 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 $\mathbb{Z}_{(p)} \to \mathbb{Z}_p$ – Drew Heard Jul 4 at 9:27
• 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. – R. van Dobben de Bruyn Jul 5 at 7:03