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?
 A: 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.
