Is the tensor product of a power series ring and a field noetherian? Suppose that $k$ is an algebraically closed field.  Let $F/k$ be a (possibly non-finitely 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?
 A: 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 non-noetherian.  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.
A: The answer is no: for example, $k[[x]]\otimes_k k((x))$ is not noetherian.
Indeed, if it were, so would be $ k((x))\otimes_k k((x))$.
But this would contradict the following interesting general theorem of Vámos:    
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 have only  read an abstract of Vámos's  article because I have no access to it. Anyway, here is the reference:
P. Vámos, On the minimal prime ideal of a tensor product of two fields, Math. Proc.
Cambridge Philos. Soc. 84 (1978), no. 1, p.25-35.
