An algebraic extension of a perfect field is a perfect field

I would like to show that an algebraic extension of a perfect field is a perfect field, using the following result:

Given a field $F$ and some family of perfect subfields $\{F_i\}_{i \in I}$ such that $F=\cup _{i\in I} F_i$, we have that $F$ is a perfect field.

EDIT: A perfect field is defined as follows: Any field of characteristic $0$ is perfect, and a field of characteristic $p$ is said to be perfect if any element in $F$ is a $p^{th}$ power of some element in $F$.

I've tried taking some element in the extended field and using the fact that it is algebraic over $F$ in order to construct a perfect subfield that contains the aforementioned element, however couldn't advance much.

Could anyone give me some direction toward the solution?

• It would help out others to know exactly which of several equivalent formulations of a perfect field you are taking as the definition. For instance, if I am allowed to define a perfect field to be one in which every algebraic extension is separable, then the question is almost trivial. – Pete L. Clark Apr 10 '11 at 21:50
• @Pete - You're right, I'll edit it. – Pandora Apr 11 '11 at 10:24

Let $K/F$ be an algebraic extension. We know that $K=\bigcup_{a\in K}F(a)$, so that it suffices to prove each $F(a)$ is perfect. For any algebraic extension $L/F(a)$, we have that $L/F$ is also algebraic, and hence separable because $F$ is perfect. Thus, for any $b\in L$, the polynomial $\text{Irr}(b,F)$ is separable (i.e. has no repeated roots). But the polynomial $\text{Irr}(b,F(a))$ is a factor of $\text{Irr}(b,F)$ (by the defining property of minimal polynomials), hence must also have no repeated roots, i.e. be separable. Thus any $b\in L$ is separable over $F(a)$, hence any algebraic $L/F(a)$ is separable, hence $F(a)$ is perfect.
To be honest, I feel like I should use somewhere that $[F(a):F]$ is finite, but at the moment the above argument seems ok.