Suppose that $k$ is an algebraically closed field. Let $F/k$ be a (possibly nonfinitely generated) field extension. Is
$$ k[[x]] \otimes_{k} F $$
noetherian?
If not, is the natural map $k[[x]] \otimes_{k} F \to F[[x]]$ injective?
Suppose that $k$ is an algebraically closed field. Let $F/k$ be a (possibly nonfinitely generated) field extension. Is $$ k[[x]] \otimes_{k} F $$ noetherian? If not, is the natural map $k[[x]] \otimes_{k} F \to F[[x]]$ injective? 


The answer is no: for example, $k[[x]]\otimes_k k((x))$ is not noetherian. Given an extension of fields $K/F$ the tensor product $K\otimes_F K$ is noetherian if and only if $K$ is finitely generated as a field over $F$. Full confession 


I was emailed the following argument: We prove that $k[[x]]\otimes_{k} k((x))$ is not noetherian by showing directly that $k((x)) \otimes k((x))$ is not noetherian (as suggested by Georges Elencwajg). I will just handle the case $k=\bar{\mathbb{Q}}$ and then make a remark about the general case at the end. The field $k((x))$ has only countably many finite separable extensions because every such extension is obtained by adjoining a root of $x$. On the other hand, the transcendence degree of $k((x))$ over $k$ must be uncountable because $k((x))$ is uncountable and $k$ is countable. Fix a transcendence basis $( t_i )_{i \in I}$ for $k((x))$ over $k$. The algebraic extension $k((x))$ of $K := k((t_i))$ is algebraic with infinite separable degree. Indeed, if the separable degree was finite, then $K$ would admit at most countably many finite separable extensions as this is true for $k((x))$. This is absurd because $( t_i )$ is uncountable. Because the separable degree of $k((x))$ over $k((t_i))$ is infinite, $(k((x)) \otimes_{K} k((x)))_{\text{red}}$ has infinitely many idempotents and so $$k((x)) \otimes_{K} k((x))$$ and hence $$k((x)) \otimes_{k} k((x))$$ are nonnoetherian. This completes the proof. With work, this proof can be modified to hold when $k$ is a finite field $\mathbb{F}$. In this case, one must argue more carefully to show that $k((x))$ has only countably many finite separable extensions. (The email indicated that one should use local compactness together with Krasner's Lemma.) Finally, one can deduce the case of a more general field $k$ from the case $k=\bar{\mathbb{Q}}$ or $\mathbb{F}$ by using a faithfully flat descent argument. 

