Finite extension of the field of formal Laurent series over a finite field

Question

Let $K=\mathbb{F}_q((T))$ and $L/K$ be a finite extension of degree n, then is it true that $L \cong \mathbb{F}_{q^n}((T))$ as fields? Thanks!

Answer 1

No. $\mathbb F_q (\!(T)\!) / \mathbb F_q (\!(T^n)\!)$ is a degree $n$ extension.

Since $\mathbb F_q (\!(T^n)\!) \cong \mathbb F_q (\!(T)\!)$, it’s indeed sufficient to look at $K =\mathbb F_q (\!(T^n)\!)$ instead and find a degree $n$ extension $L$ for it; here $L = \mathbb F_q (\!( T )\!)$. To see that $L \not\cong \mathbb F_{q^n} (\!(T)\!)$, just count the roots of unity.

It is true, however, that every unramified extenson $\mathbb F_q (\!(T)\!)$ of degree $n$ is isomorphic to $\mathbb F_{q^n} (\!(T)\!)$. I’ll add a reference as soon as I find one.

In any case, it’s implicit in the classification of local fields, see Wikipedia’s article on local fields:

• Any extension $L$ of $K$ of degree $n$ is again a local field, so $L \cong \mathbb F_{q'} (\!(T)\!)$ for some $q'$.
• If $L/K$ is unramified, their residue fields extension $\mathbb F_{q'} / \mathbb F_q$ is of degree $n$, hence $\mathbb F_{q'} = \mathbb F_{q^n}$.