Is every element in a finite extension of a field of characteristic 0 algebraic?

[futurebird]

Homework Statement

Show that any field of characteristic 0 is perfect.

2. The attempt at a solution

Let F be a field of characteristic 0.
Let K be a finite extension of F.
Let b be an element in K .

I need to show that b satisfies a polynomial over F having no multiple roots.

If f(x) is irreducible in F[x] then f(x) has no multiple roots.

I need to show that b satisfies a irreducible polynomial in F[x].

Well, suppose b can't satisfy any irreducible polynomial in F[x]. Can I get a contradiction? What kind of element could I have that didn't satisfy any irreducible polynomial?

Then how can b be in the finite extension...? A finite extension for a field of characteristic 0 is of the form F(a), it is generated by a single element. I'm stuck. I don't even know if what I've laid out so far is correct.

I'm having trouble connecting the arbitrary element b to a polynomial-- It's not obvious to me that b is the root of any polynomial in F[x].

 

[morphism]

Well K is a finite extension of F. So b is algebraic over F, and hence has a minimal polynomial in F[x].

If you don't know what a min poly is, think about what it means for K to be a finite extension of F. What does this say about the set {1, b, b^2, b^3, ...}?