Let $L:K$ be a field extension. Let $A$ be a set of elements in $L$, all of which are algebraic over $K$. Construct the field extension $M=K(A)$. I have two questions:


[1] Is $M:K$ an algebraic field extension?


[2] Take $\beta\in M$ where $\beta$ is algebraic over $K$. Then $K(\beta):K$ is a finite extension. Can I assert that $\beta$ lies in a finite extension $K(a1,..., an)$ where $a1,..., an\in A$?


Both questions are trivial when $A$ is finite. So assume that $A$ is infinite; indeed assume that $[M:K]$ is infinite. Now what?



On algebraic field extensions
The question was a bit confusing because it was not clear what was meant as the definition of "the field extension $K(A)$". Nick says that "the smallest field containing" is the definition that he wanted, to make the question non-trivial. (But still routine, in my opinion.)


Let's agree then that both (1) and (2) are answered by this summary: The smallest field that contains $K$ and $A$ is the set of rational expressions in $A$ with coefficients in $K$. An element $\beta$ of this field $M$ can only use finitely many elements of $A$, so $\beta$ lies in the subfield generated by them. Moreover, $\beta$ is algebraic by the lemma that the sum or product of two algebraic numbers is algebraic, and the reciprocal of an algebraic number is algebraic.


I'm putting this summary here to hopefully take the question off of the unanswered stack.


