13
$\begingroup$

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?

$\endgroup$
15
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.