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.

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...

Re: Algebraicity of arbitrary extensions

Re: Algebraicity of arbitrary extensions

So Assume, F $\displaystyle < K \leq M $ and $\displaystyle M = F(r_1)$. Assume $\displaystyle r_1$ was transcendental.

Now if there exist no intermediate field between F and M, then $\displaystyle F(r_1) = K$ and so M is algebraically closed over K.

If there does exist an intermediate field, then there exists an element $\displaystyle r_2 \in K $ which is not in F.

if $\displaystyle r_2$ was transcendental over F, then $\displaystyle K(r_2) \cong F(x) \cong F(r_1) \cong M $ a contradiction.

if $\displaystyle r_2$ 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 $\displaystyle r_1$ 0. so $\displaystyle 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 = $\displaystyle K$ thus again since no intermediate field exists $\displaystyle M=K$ is algebraically closed.

Now if $\displaystyle r_1 can algebraic$ there is a min deg monic polynomial in $\displaystyle F$ such that $\displaystyle f(r_1) = 0$. Then $\displaystyle [M:F] = n$ this implies, $\displaystyle [M:K][K:F] = [M:F] $ so $\displaystyle [M:K]$ is a finite extension, thus an algebraic extension.

if so.

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 :)