# Algebraicity of arbitrary extensions

• Feb 27th 2013, 08:49 AM
spudwish
Algebraicity of arbitrary extensions
Let F \subset K \subset E be field extensions. Assume that F \neq K and that there is x \in E s.t E=F(x). Show that E is algebraic over K.

I have no idea how to even begin with this one. I'm sitting with my books trying to find some quality to use, but all are about finite extensions, something I suppose I can't assume here. As you know, if E=F(x) is finite over K then it is also algebraic over K, and I'm thinking about using the contrapositive, but that doesn't seem to solve the problem in its entirety.

Thank you for any advice.
• Feb 27th 2013, 03:27 PM
spudwish
Re: Algebraicity of arbitrary extensions
I made some progress on my own, which I suppose was the easy part.

So, x is either algebraic or transcendental over F. If x is algebraic over F, then F(x)/F is finite => [F(x):F] = [F(x) : K][K : F] finite. The product if finite only if both factors are finite. In particular [F(x) : K] is finite, and hence algebraic. (I have one introductory algebra book stating the tower law (?) only for finite extensions, but Lang's Algebra states it for any extensions.)

As for the case x transcendental, then F(x)/F is infinite => either F(x)/K or K/F is infinite, at which point things get circular...
• Mar 1st 2013, 07:25 AM
spudwish
Re: Algebraicity of arbitrary extensions
Bump. Anyone?
• Mar 8th 2013, 10:52 AM
jakncoke
Re: Algebraicity of arbitrary extensions
So Assume, F $< K \leq M$ and $M = F(r_1)$. Assume $r_1$ was transcendental.
Now if there exist no intermediate field between F and M, then $F(r_1) = K$ and so M is algebraically closed over K.
If there does exist an intermediate field, then there exists an element $r_2 \in K$ which is not in F.
if $r_2$ was transcendental over F, then $K(r_2) \cong F(x) \cong F(r_1) \cong M$ a contradiction.
if $r_2$ was algebraic over F, kthen it implies, M is a transcendental extension of K, since it is a transcendental extension of F. There exists no polynomial in K[X], which makes $r_1$ 0. so $K(r_1) \cong K(x) < F(r_1) = M$. Since K is strictly bigger then F, this cannot be, thus with both contradictions, we see that M = $K$ thus again since no intermediate field exists $M=K$ is algebraically closed.

Now if $r_1 can algebraic$ there is a min deg monic polynomial in $F$ such that $f(r_1) = 0$. Then $[M:F] = n$ this implies, $[M:K][K:F] = [M:F]$ so $[M:K]$ is a finite extension, thus an algebraic extension.
if so.
• Mar 22nd 2013, 07:05 PM
spudwish
Re: Algebraicity of arbitrary extensions
jakncoke, thank you! I haven't been able to look through your explanation, but at least now I know where to go when I return to the subject :)