Let $f(x)\in K((x))$ be an algebraic formal Laurent series and let $P(x,y)\in K(x)[y]$ be its minimal polynomial. Is $P(x,y)$
always separable? An example of non separable polynomial comes
from Puiseux series: the polynomial $y^2-x$ has a double root
$y=\sqrt{x}$. So I wonder what happen if we restrict to Laurent series.


Thank you in advance.



Is the minimal polynomial of an algebraic formal Laurent series always separable?
Yes. Let $p=char(K)$ and $\alpha\in \overline{K(x)}\cap K((x))$ assumed to be inseparable over $K(x)$.


* Let $L= K^{1/p^\infty}$ which is perfect. If $\alpha$ is inseparable over $L(x)$ then $\alpha$'s monic $L(x)$-minimal polynomial is $$f(y)=g(y^p)=h(y)^p$$ with $h(y)\in L(x)^{1/p}[y]=L(x^{1/p})[y]$ so $$[L(x^{1/p},\alpha):L(x)] = p \deg(h) =[L(x,\alpha):L(x)]$$ ie. $x^{1/p}\in L(x,\alpha)$.


This gives that $x^{1/p}\in L((x))[\alpha]= L((x))$ which is a contradiction.



> 
> Whence $\alpha$ is separable over $L(x)$ $\implies f(y)$ is separable.
> 
> 
> 


* There is a finite extension $E/K$ such that $f(y)\in E[y]$. Take a basis $E=\bigoplus\_{j=1}^q b\_j K$ with $b\_1=1$. We get that
$$E(x)=\bigoplus\_{j=1}^q b\_j K(x),\qquad
E((x))=\bigoplus\_{j=1}^q b\_j K((x))$$ $$f(y)=\sum\_{j=1}^q b\_j f\_j(y), \qquad f\_j(y)\in K(x)[y]$$


$f(\alpha)=\sum\_{j=1}^q b\_j f\_j(\alpha) = 0$ in $E((x))$ gives that $$f\_1(\alpha)=0$$
$f\_1(y)\in K(x)[y]$ being monic and of degree $= \deg f$ it must be that $f(y)=f\_1(y)$.


$f\_1$ is in $K(x)[y]$, separable, irreducible, which proves that $\alpha$ is in fact separable over $K(x)$.


