# Are all algebraic extensions of finite fields separable? What about fields of characteristic p in general?

I know that all algebraic extensions of fields of characteristic $$0$$ are separable, but what about a field of characteristic $$p$$, for example, $$\mathbb{F}_7$$?

I know that, for a finite field of characteristic $$p$$, all finite extensions are separable, but what happens with infinite algebraic extensions? What if the field itself is infinite?

• Finite fields are perfect, so any algebraic extension of them is separable. For infinite fields of positive characteristic, there are non-separable extensions. The standard example can be found here. May 22 at 15:44

What you are actually asking about is perfect fields. A field $$K$$ is called perfect if either the characteristic of $$K$$ is zero, or the characteristic is a prime $$p >0$$ and the Frobenius-map $$F:K \to K$$ associated to $$K$$, that sends $$K \ni \alpha \mapsto \alpha^p$$ is surjective (a field homomorphism $$K \to K$$ is always injective, so actually $$F$$ is an automorphism of $$K$$).
Using this definition, you can prove that a field $$K$$ is perfect iff every algebraic extension of $$K$$ is separable (over $$K$$). As you already state, for a finite field $$\mathbf{F}_{p^n}$$ of characteristic $$p>0$$ (prime) with $$n \in \mathbf{Z}_{>0}$$, all finite extensions will be separable. Fortunately, the Frobenius-map associated to $$\mathbf{F}_{p^n}$$ (sends $$\alpha \mapsto \alpha^p$$) is an automorphism of $$\mathbf{F}_{p^n}$$. Therefore, $$\mathbf{F}_{p^n}$$ is perfect.
However, if you look at the field $$\mathbf{F_p}(T)$$ where $$T$$ is a variable, then this variable is not a $$p$$-th power. Therefore, this is an imperfect field.
• So what would be an example of an algebraic extension over $\mathbf{F_p}(T)$ that is not separable? May 22 at 16:14
• @BraisRomero The link I gave you gives you one. The extension $\mathbf{F}_p(T^{1/p})$ over $\mathbf{F}_p(T)$ is not separable. May 22 at 16:29