My class notes has as theorem (without proof): "Let $K/F$ be finite field extension, with $K=F(\alpha_1,\ldots,\alpha_n)$ and $\alpha_k$ is separable for all $k$. Then $K/F$ is separable".

My question is: in either finite field, or field characteristic $0$, will the $\alpha_k$ always be separable? (by "$\alpha_k$ separable" I mean that its minimal polynomial is separable). Also, could somebody give me an example for which an $\alpha_k$ is not separable?

  • $\begingroup$ also, as a side i have no idea how to prove the theorem haha $\endgroup$ – sh donny Apr 12 '16 at 19:40
  • 7
    $\begingroup$ $\mathbb{F}_p(X^{1/p})/\mathbb{F}_p(X)$ is the standard example. $\endgroup$ – Captain Lama Apr 12 '16 at 19:40

Separability is automatic in characteristic 0 and finite fields.

Let $f(x) \in \mathbb{F}[x]$. Let $f'(x)$ be its formal derivative. Then $f(x)$ has a repeated root (in some extension field) iff $f(x)$ and $f'(x)$ fail to be relatively prime.

If you take a field of characteristic $0$, non-constant polynomials have nonzero derivatives (of degree one less). Thus if $f(x)$ is irreducible, then $f'(x)$ must be relatively prime to $f'(x)$: If $g(x)$ divides $f(x)$ it's either $1$ or $f(x)$ up to associates. But $f'(x)$ cannot be divisible by $g(x)=f(x)$ -- it's degree is too small. Thus any common divisor $g(x)$ must be $1$.

This means that irreducible polynomials in fields of characteristic 0 cannot have repeated roots. Thus in characteristic 0 everything is separable (this explains why it's hard to think up inseparable stuff off the top of your head -- we tend to work in char 0 most of the time).

As for finite fields, every element in $\mathbb{F}_q$ (the field of order $q=p^n$ some prime $p$) is a root of $f(x) = x^q-x$. Thus the minimal polynomial of any element of a finite field must be a divisor of $f(x)$. Notice that $f'(x)=qx^{q-1}-1=-1$ (since $q=0$ in char $p$). The gcd of $f(x)$ and $f'(x)=-1$ is $1$ so $f(x)$ has no repeated roots. This means its factors cannot have repeated roots either. Thus every element in a finite field is separable.

Therefore, if you're looking for something that isn't separable, you'll need an infinite field of characteristic $p \not=0$. The canonical example is...

Consider $\mathbb{F}=\mathbb{Z}_p(y)$ (rational polynomials in $y$ with coefficients in $\mathbb{Z}_p$). Let $f(x) = x^p-y \in \mathbb{F}[x]$. [Notice that $f'(x)=px^{p-1}=0$.]

Now $f(x)$ is irreducible by Eisenstein's criterion: $y$ is prime in $\mathbb{F}$ since $\mathbb{F}/(y) \cong \mathbb{Z}_p$ (an integral domain) so $(y)$ is a prime ideal and so $y$ (being a nonzero generator of a prime ideal) is prime. Notice that $y$ divides all but the leading coefficients of $f(x)=x^p-y$ and $y^2$ does not divide $f(0)=y$ (the constant term).

Next, adjoin a root of $f(x)$. Let's call this root $\alpha$ (or in more suggestive notation we could say $\alpha=\sqrt[p]{y}$). This means that $f(\alpha)=\alpha^p-y=0$ so $\alpha^p=y$. Notice that $(x-\alpha)^p = x^p - \alpha^p$ (the middle binomial coefficients are divisible by $p$ and so are $=0$). Therefore, $f(x)=x^p-y = (x-\alpha)^p$. We have an irreducible polynomial with a single ($p$-times) repeated root. $y$ is inseparable!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.