Thread: Subring is a field

Hello, I'm having difficulty with this problem:
Let F be an algebraic extension of K. Let R be an intermediate ring. Show that R is a field.

I know that R will contain the additive (0) and multiplicative (1) identities since it contains K, but I'm not sure how to show that it will be a division ring.

Thanks!

What are you assuming about F and K? That K is a ring? That F is a field? And what do you mean by "intermediate". Can we assume that neither F nor K itself would not be "intermediate"?

we're given from the get-go:

K < R < F.

since every element of F is algebraic over K, every element of R is algebraic over K (since every element of R is in F).

that means any non-zero element a of R satisfies some non-zero monic polynomial in K[x]. let m(x) be a monic polynomial for which m(a) = 0 of minimal degree (we know there is at least one since a is algebraic over K, so we can pick one of minimal degree).

i claim m(x) is irreducible over K. for suppose not, suppose that m(x) = f(x)g(x), where 1 ≤ deg(f),deg(g) < deg(m).

then 0 = m(a) = f(a)g(a), so either f(a) = 0, or g(a) = 0 (since the field F is certainly an integral domain). but that contradicts that m(x) is of minimal degree. so m(x) is irreducible (if it factors, one of f(x),g(x) must be a unit, that is, an element of K*).

so suppose

i claim that . for if so, m(x) is reducible over K (x is a factor). so we have:

, which means that:



which shows that any a in R* is a unit.

EDIT: note that this assumes K is a field. it is possible to have algebraic extensions of rings that are NOT fields, for example Z[i].

Thanks! Yes, forgot to mention that K is a field.