I am studying these notes and I am trying to generalize a bit the setting of the Section 3, because there doesn’t seem to be a fundamental reason to only study $p$-adic fields. So all the fields considered in this question are complete non-Archimedean valued fields with a non-trivial discrete valuation.
For most of the notes, the only change is to add some separability hypotheses for the extensions and their residue fields.
And then I hit a roadblock at the unproven (because trivial in the author’s setting) Corollary 3.21.
The reason is the following:
Original form of the question: let $K/F$ be a finite separable field extension with separable residue field extension. Is there a Galois extension $L/F$ containing $K$ such that the residue field extension is still separable?
The only candidate to check would be the Galois closure of $K/F$. After all, it is generated by the conjugates of $K$, all of which have separable residue fields extensions.
But it isn’t obvious that the residue field of the composition of two separable field extensions is generated by the residue fields of the extensions. More formally,
First reformulation (sufficient to answer ”yes” to the original question): let $K/F,L/F$ be subextensions of a finite separable extension $M/F$, such that the residue fields of $L$ and $K$ are separable over that of $F$ and $L\cdot K=M$. Is the residue field of $M$ separable over that of $F$?
This reformulation begs the natural stronger following questions: with the same hypotheses, are $k_K \otimes_{k_F} k_L \rightarrow k_M$ or even $O_K \otimes_{O_F} O_L \rightarrow O_M$ surjective (where $k_{\cdot}$ is the residue field)?
The proof scheme for this statement seemed classical enough: consider the isomorphism of $K \otimes_F L$ to a product of field extensions of $L$ (corresponding to the factors in $L$ of the minimal polynomial of a primitive element of $K$), and identify $O_K \otimes O_L$ as the subalgebra of $z$ such that $(z^n)_n$ is bounded.
There is an obvious inclusion but the reverse one is significantly less obvious – nothing I tried (mostly tracking dimensions and factors of the minimal polynomial to prove inequalities, as in several similar-looking proofs) seemed to work – basically (and unsurprisingly) we lack control of the residue field of $k_M$ – and I now doubt it’s true.
The best I was able to find out is that if $z^n \rightarrow 0$ and $z \in O_K \otimes O_L$, then $z$ must be in the ideal of generated by the maximal ideals of the valuation rings.
Less ambitiously, in the same setting, we can consider the maximal unramified subextension $F’ \subset M$ of $M/F$, and define $K’=M’K,L’=M’L$. Then $K’/K,L’/L$ are unramified so have separable residue field extensions, and thus the residue fields of $K’,L’$ are separable over that of $F$ and contain the residue field of $F’$, so that $F’,L’,K’$ have the same residue fields (equal to the separable closure of the residue field of $F$ in the residue field of $M$). So the “first reformulation” is implied by the following statement:
Second reformulation: let $M/F$ be a separable extension such that the extension of residue fields is purely inseparable. Let $K/F,L/F$ be subextensions of $M/F$ with trivial residue field extension and assume $K\cdot L=M$. Then is the residue field extension of $M/F$ trivial?
In this case, $O_L,O_K$ are generated over $O_F$ by uniformizers whose minimal polynomial is Eisenstein, so it should be easier to do, but I can’t find either an argument or a counter-example. One reason for this is that unlike in the unramified case, Eisenstein polynomials quickly aren’t Eisenstein any more when changing fields.
So I’m quite stuck and I would appreciate some other ideas. Thank you for your help!
Best Answer
Yes. The $2$-adic valuation on $$F=\Bbb{Q}_2(x)$$ (non-complete but it will stay the same over the completions) is given by
$$v_2(f)=\inf_j v_2(c_j),\qquad f=\sum_{j=0}^J c_j x^j \in \Bbb{Q}_2[x]$$ The valuation ring $O_F$ has uniformizer $2$ and residue field $\Bbb{F}_2(x)$.
It extends naturally to a valuation on $$K=\Bbb{Q}_2(2^{1/4}x^{1/4}),\qquad v_2(\sum_{j=0}^J c_j (2^{1/4}x^{1/4} )^j)=\inf_j v_2(c_j 2^{j/4}), \qquad c_j\in \Bbb{Q}_2$$ The valuation ring $O_K$ has uniformizer $2^{1/4}x^{1/4}$ and residue field $\Bbb{F}_2(x)$.
The Galois closure of $K/F$ is $$L=\Bbb{Q}_2(2^{1/4}x^{1/4}, i)$$ and it contains $$\frac{2^{1/2} x^{1/2}}{1+i}= \zeta_8^{-1} x^{1/2}$$ Whence the residue field of $O_L$ contains a $4$-th root of $-x^2$ ie. it contains $\Bbb{F}_2(x^{1/2})$.