If you have fields with algebraic over , show that,
is the minimal field containing and .
(Note: -a simple extension).
The map given by evaluation at is a ring homomorphism and hence its image, which you denote , is a subring of E. All you need to show is that it is closed under taking multiplicative inverses. Note for use in a moment that a subring of a field is an integral domain. The fact that is algebraic is equivalent to saying that is a finite-dimensional F-vector space. Let be a non-zero element of . Multiplication by is an F-linear endomorphism and has trivial kernel (integral domain), hence is surjective by finiteness of the dimension and the rank--nullity formula. So 1 is in the image, hence a multiple of , which is thus invertible.
I was thinking about this today, maybe you can do this?
You have with algebraic over . Let there exist a field with such as,
,
Any element can be expressed as,
with because every finite dimensional vector basis has a finite basis. Therefore thus, and thus, .
Am I missing something?
Any field that contains F and contains must contain any polynomial in with coefficients in F. The previous part of my argument showed that the ring comprised of polynomials in with coefficients in F is actually a field. Hence it is the minimal subfield of E containing both F and .