Is the class of algebraic extensions distinguished? In paragraph V.1 of Algebra proposition 1.7 Lang claims that the class of algebraic extensions is distinguished. I know that if $F/k$ and $E/F$ are algebraic extensions than so is the $E/k$ - that is easy to prove. However, the second required property is that if $E/k$ is algebraic and $F/k$ is arbitrary, then (assuming the compositum is defined) $EF/F$ is algebraic. Lang more or less skips the proof of this, only saying that "an element remains algebraic under lifting, and hence does the extension."
However, there are no finiteness conditions here and no element given in advance. Although $EF$ is generated over $F$ by a set of elements which are algebraic over $F$ (namely, all the elements of $E$), there are no facts known (up to this point in the book at least) about the infinitely generated extensions.
What am I missing?
Thank you.  
 A: In Lang's definition of distinguished, both conditions (2)-(3) require that $\,k\subset E,F\,$ and both $\,E,F\,$ are contained in some object $\,R\in\mathcal C\,$ , so $\,F/k\,$ cannot be "arbitrary": 
$\,\Bbb Q/\Bbb Q\,$ is algebraic and $\,\Bbb Q(\pi)/\Bbb Q\,$ is any extension, but $\,\Bbb Q\cdot\Bbb Q(\pi)/\Bbb Q=\Bbb Q(\pi)/\Bbb Q\,$ is not algebraic.
Added: The elements of $\,EF\,$ can be seen as rational functions in the elements of $\,E\,$ with coefficients of $\, F\, $ (and the other way around, of course). 
Let $\,x\in EF\,$. Then there exists a finite number of elements in $\,E\,$ , say $\,e_1,...,e_r\,$ which actually appear in the above rational expression of $\,x\,$, and thus $\,x\in F(e_1,...,e_n)\subset EF\,$.
Since clearly $\,F(e_1,...,e_n)/F\,$ is algebraic we're done.
A: If $E/F$ is an algebraic extension, $K/F$ is an extension. Prove that $EK/K$ is algebraic.
Assume $H=\{a\in EK|a \text{ is algebraic over } K\}$. It is easy to see that $H$ is a field and $K<H<EK$. $E/F$ is algebraic, so all the elements in $E$ are algebraic over $F$, also over $K$; so $E\subseteq H$; then we have $EK\subseteq H$, which means $EK=H$. $EK$ is an algebraic extension over $K$. ■
