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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?

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    $\begingroup$ This is skirting kind of close to what could be homework for a Galois theory course, but could also be honest confusion. Both of these come down to remembering definitions. For (1), the definition of "algebraic" is that every element of beta of M obeys a polynomial with coefficients in K. For (2), what it the definition of K(A)? $\endgroup$
    – David E Speyer
    Commented Nov 12, 2009 at 14:59
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    $\begingroup$ I think what is missing in (1) is not the definition, but the lemma that in any field extension M of K, the sum and product of two elements algebraic over K are both algebraic over K. Reciprocals are also algebraic. Hence the algebraic subset of M is a subfield of M, and per (1) it can only be M itself. I agree that (2) is by definition. $\endgroup$
    – Greg Kuperberg
    Commented Nov 12, 2009 at 16:07
  • $\begingroup$ The lemma that Greg gave certainly answers (1). The definition of K(A) is that it is the smallest subfield of L that contains both K and A. Everyone I've asked here thought both questions were trivial, but couldn't quite dot the i's and t's (at least on the first parse). However after some thought I believe that question (1) is answered in the affirmative by recognising that K(A) is equal to the set of rational expressions with indeterminates in A, and coefficients in K. The point being that these rational expressions contain a FINITE set of elements of A. $\endgroup$
    – Nick Gill
    Commented Nov 12, 2009 at 16:37
  • $\begingroup$ Sorry I meant that "question (2) is answered in the affirmative" by the final observation. And you don't dot t's, you cross them. :-) $\endgroup$
    – Nick Gill
    Commented Nov 12, 2009 at 16:48
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    $\begingroup$ As you note, both questions are trivial when A is finite, therefore they are trivial in general because K(A) is the union of of fields K(B) with B a finite subset of A. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Nov 12, 2009 at 18:40

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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.

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