# Example of finite field extension where root not separable

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?

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