# Does $\mathbb{F}_p((X))$ has only finitely many extension of a given degree?

We know that $\mathbb{Q}_p$ has only finitely many extensions of a given degree in its algebraic closure. Is it the same for $\mathbb{F}_p((X))$?

• So now we have answers saying Yes and No respectively, both of which have received upvotes. Which one should be trusted? – Dilip Sarwate Apr 8 '13 at 12:13
• The local class field theory makes it clear that there are infinitely many cyclic extensions of degree $p$. – Lubin Apr 8 '13 at 12:22
• @Dilip: don't trust. Think for yourself about the underlying math. – KCd Apr 8 '13 at 12:23

No. There are infinitely many extensions of degree $p$. Let $K = {\mathbf F}_p((X))$ and $\wp(x) = x^p - x$ (Artin--Schreier operator, which is additive in characteristic $p$). For $c \in K - \wp(K)$, the polynomial $T^p - T - c$ is irreducible in $K[T]$ and its splitting field over $K$ has degree $p$. For two $c$ and $c'$ in $K - \wp(K)$, the splitting fields of $T^p - T - c$ and $T^p - T - c'$ in an alg. closure of $K$ are equal iff $c = c'$ in $K/\wp(K)$. This quotient group is infinite. In particular, the reciprocal powers $1/X^n$ for $n$ positive and not divisible by $p$ are distinct in $K/\wp(K)$. This is discussed in section 1 of http://math.stanford.edu/~conrad/248APage/handouts/weirdfield.pdf.

There are only finitely many extensions of each degree relatively prime to $p$ over ${\mathbf F}_p((X))$ in an algebraic closure, by the same proof as over ${\mathbf Q}_p$. When you deal with extensions of degree divisible by $p$ in characteristic $p$, watch out.

Given that I was initially lead astray I am trying to learn something by pinpointing (from the article in Ted's link) a step that does not generalize to our case. I show how I located what I think is the key point using an example. I utilize a subset of the infinitely many non-isomorphic degree $p$ extension from KCd's answer.

Let $n=n_\ell=p\ell-1, \ell$ a natural number, in KCd's answer. If $y$ is a root of $$T^p-T-\frac1{x^n}=0,\qquad(*)$$ then we do the usual tricks to put this into Eisenstein form, so let $v=x^\ell y$. Multiplying $(*)$ by $x^{p\ell}$ shows that $v$ is a root of the Eisenstein polynomial $$p_\ell(T):=T^p-x^{(p-1)\ell}T-x=0.$$ Clearly $v$ and $y$ generate the same extension field. But any neighborhood (after the mapping in the proof of 3.19) of $g(T)=T^p-x$ contains all the polynomials $p_\ell(T)$ with $\ell$ large enough. Now we hit a troublesome point. The polynomial $g(T)$ is inseparable. Therefore Corollary 3.17 cannot be applied, so we do not get a cover of the parameter space $U\times P^p$ by such neighborhoods that all the Eisenstein polynomials in a neighborhood would give the same extension field. The argument in the proof of 3.19 breaks down here (but does work in chararcteristic zero, as separability is not a problem).

As prof. Conrad pointed out, if we are looking at extensions of degree coprime to $p$, this problem (inseparability) does not arise.

Yes. The key properties are (1) the field is complete with respect to a nonarchimedean valuation, and (2) the residue field is finite. With those two properties in hand, then the same proof as for $\mathbb{Q}_p$ goes through. See (for example) this proof (Theorem 3.19).

EDIT: As others have pointed out, this isn't quite right; the proof fails for extensions of degree divisible by $p$. See other answers.

• Sorry, not right, 3.19 is for $\mathbb Q_p$ only. – Lubin Apr 8 '13 at 12:21