TITLE: $M /K \land L /K$ algebraic $\implies ML/K$ algebraic?
QUESTION [1 upvotes]: Let $K \subset M$, $L\subset K'$, and let $ML$ denote the subfield of $K'$ generated by
$M$ and $L$.
Is the following true?
$M/K$ and $L/K$ algebraic $\implies ML/K$ algebraic?
Any hints proof or counter-example? Cheers

REPLY [0 votes]: This is just a lengthier version of Lubin's answer: Let $L$ and $M$ be algebraic extensions of a field $K$, and suppose that $L,M$ are contained in a larger extension $F$ of $K$, i.e., $K\subseteq L \subseteq F$ and $K\subseteq M \subseteq F$, so that $K\subseteq LM \subseteq F$. 
Now, let $\alpha$ be an element of $LM$. Then, $\alpha = \sum_{i=1}^n \ell_i m_i$, for some $\ell_i \in L$ and $m_i\in M$, for each $i=1,\ldots,n$. Hence, 
$$\alpha \in  K_\alpha:=K(\ell_1,\ldots,\ell_n,m_1,\ldots,m_n).$$
Since $L$ and $M$ are algebraic over $K$, each $\ell_i$ and $m_i$ is algebraic over $K$. Hence $K_\alpha$ is generated by (a finite number of) algebraic elements over $K$, and therefore $K_\alpha/K$ is algebraic. Therefore $\alpha$ is also algebraic over $K$. 
Since $\alpha$ was an arbitrary element of $LM$, and we have shown that $\alpha$ is algebraic over $K$, we conclude that $LM$ is an algebraic extension of $K$.
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