Let $\mathbb{Q}\_p$ denote the field of fractions of $\mathbb{Z}\_p$. By the answers to [this quesition](https://math.stackexchange.com/questions/19426/when-is-a-tensor-product-of-two-commutative-rings-noetherian) 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?



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


