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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.

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    $\begingroup$ $y^2 (y-1) - x $ has a double root mod $x$ but also a single root. The single root lifts to an algebraic formal Laurent series by Hensel's lemma. Since $y^2 (y-1)-x$ is irreducible, its minimal polynomial has a double root. $\endgroup$
    – Will Sawin
    Commented Apr 22, 2022 at 23:49
  • $\begingroup$ @WillSawin This cubic polynomial is separable, its discriminant is $-27x^2 - 4x\ne 0 $ $\endgroup$
    – reuns
    Commented Apr 23, 2022 at 12:43
  • $\begingroup$ @reuns In view of the example $y^2 -x$, I thought we were considering separability mod $x$ (ignoring the claim that $y= \sqrt{x}$ is a double root, which suggests maybe instead we are working in characteristic $2$.) $\endgroup$
    – Will Sawin
    Commented Apr 23, 2022 at 12:45

1 Answer 1

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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 $K(x)$.

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  • $\begingroup$ Thank you for the answer! Could you tell me what is $K^{1/p^\infty}$? $\endgroup$
    – Jiu
    Commented Apr 24, 2022 at 1:09
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    $\begingroup$ @Jiu $\{ a\in \overline{K},\exists n, a^{p^n}\in K\}$ which is a field in characteristic $p$. But in my answer you can replace it by $\overline{K}$ it works the same way. $\endgroup$
    – reuns
    Commented Apr 24, 2022 at 1:59
  • $\begingroup$ And why is $f(y)=g(y^p)$? $\endgroup$
    – Jiu
    Commented Apr 24, 2022 at 3:19
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    $\begingroup$ If $f'\ne 0$ then $\gcd(f,f')$ divides $f$ $\endgroup$
    – reuns
    Commented Apr 24, 2022 at 3:30
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    $\begingroup$ One may argue using the following separability criterion (maybe it boils down to the same computations): $K(x)(f)/K(x)$ is separable if and only if $A := K(x)(f) \otimes_{K(x)} K(x)^{1/p}$ is reduced. We have $K(x)^{1/p}=K^{1/p}(x^{1/p})$ so that $A \hookrightarrow K((x)) \otimes_{K(x)} K^{1/p}(x^{1/p}) \hookrightarrow K^{1/p}((x^{1/p}))$. $\endgroup$
    – François Brunault
    Commented Apr 25, 2022 at 14:34

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