Suppose $K$ is a field, and $K\subset E$ is an algebraic extension, and $K\subset F$ is any extension. Suppose that $L$ is some field which contains both $F$ and $E$ as subfields.

I want to show that $FE$ (the smallest subfield of $L$ containing both $F$ and $E$) is algebraic over $F$.

Lang mentions that the result is true because an element remains algebraic under lifting, but I can't prove the claim with this.

I interpret his comment to mean that if $x\in E$ is algebraic over $K$, then $x\in FE$ is algebraic over $F$. This is certainly true.

But this doesn't seem enough if I start with any element of $FE$.


First you must show that, given any extension $L/K$ of fields, the set of elements in $L$ which are algebraic over $K$ (the algebraic closure of $K$ in $L$) is itself a field (clearly containing $K$). This requires the following facts:

(1) An element $\alpha\in L$ is algebraic over $K$ if and only if $K(\alpha)$ is finite over $K$.

(2) A finite extension of $K$ is algebraic over $K$.

(3) If $K\subseteq L_1\subseteq L_2$ is a tower of extensions, $L_2/L_1$ finite and $L_1/K$ finite implies $L_2/K$ finite.

Now proceed as follows. Let $\alpha,\beta\in L$ be algebraic over $K$. Then $K(\alpha)$ is finite over $K$ by (1), and $K(\alpha,\beta)=K(\alpha)(\beta)$ is finite over $K(\alpha)$, also by (1) (since $\beta$ is algebraic over $K$, it is also algebraic over $K(\alpha)$). Thus $K(\alpha,\beta)$ is finite over $K$ by (3), so $K(\alpha,\beta)$ is algebraic over $K$ by (2). Since $\alpha\pm\beta,\alpha\beta,\alpha^{-1},\beta^{-1}\in K(\alpha,\beta)$, we see that all these elements are algebraic over $K$. Therefore the set of $K$-algebraic elements in $L$ is indeed a field.

Now we can get the result you want. We have $FE=F(E)$, i.e., $FE$ is generated as an extension of $F$ by $E$. As you point out in your post, every element of $E$ is algebraic over $F$. Now if $E^\prime$ is the set of elements of $EF$ algebraic over $F$, then it contains $E$. Since it also contains $F$, and is a field by the argument above, it contains $EF$, and hence equals it. Thus $EF/F$ is algebraic.


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