Define a separable algebraic extension. Explain why algebraic extensions of characteristic-\(0\) fields and of finite fields are separable, and give an inseparable example.
Solution
An algebraic extension \(L/K\) is separable if the minimal polynomial over \(K\) of every \(\alpha\in L\) has distinct roots in an algebraic closure; equivalently, every irreducible polynomial over \(K\) having a root in \(L\) is separable.
1. Characteristic \(0\) fields are perfect.
If \(f\in K[x]\) is irreducible and \(\operatorname{char}K=0\), then \(f'\ne0\). Since \(f\) is irreducible, \(\gcd(f,f')=1\), so \(f\) has no repeated roots. Thus every algebraic extension is separable.
2. Finite fields are perfect.
Let \(K=\FF_q\). The Frobenius map \[ x\mapsto x^p \] is injective, hence surjective because \(K\) is finite. Therefore every element is a \(p\)th power. An irreducible polynomial with zero derivative would have the form \(g(x^p)\), and surjectivity of Frobenius would make it a \(p\)th power of a polynomial over \(K\), contradicting irreducibility unless it were linear. Hence every irreducible polynomial is separable.
Thus \(\QQ\) and every finite field have no inseparable algebraic extensions.
3. Inseparable example.
Let \[ K=\FF_p(t^p)\subset L=\FF_p(t). \] Then \(t\) satisfies \[ x^p-t^p\in K[x], \] which equals \((x-t)^p\) over \(L\). Its derivative is zero, and \(t\notin K\), so the minimal polynomial of \(t\) over \(K\) is inseparable. Hence \(L/K\) is not separable.