# Algebraic extension and infinite polynomial ring

Let $K$ a field and $F=K(\alpha_i : i\in I)$ an algebraic extension of $K$.

Is it true that for all $z\in F$ there exists $i_1,\dots,i_k\in I$ such that $z\in K(\alpha_{i_1},\dots,\alpha_{i_n})$ ?

Moreover, does $F$ equals to $K[\alpha_i : i\in I]$ ?

These results are true if $I$ is finite, but I don't know if they remain so if $I$ is infinite, since the proof doesn't seem straightforwardly generalisable. I haven't been able to find a book with this result either.

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By definition $F=K(\alpha_i : i\in I)$ is such that for all $z\in F$ there exist $i_1,\dots,i_k\in I$ such that $z\in K(\alpha_{i_1},\dots,\alpha_{i_n})$, so the answer to your first question is trivially yes. For such $i_1,\dots,i_k$ one has $K(\alpha_{i_1},\dots,\alpha_{i_n})=K[\alpha_{i_1},\dots,\alpha_{i_n}]$ so $F$ is also the smallest ring containg all these rings, which is by definition $K[\alpha_i : i\in I]$, so the answer to the second question is yes as well.
Hum, for me, $K(\alpha_i : i\in I)$ is defined to be the smallest extension of $K$ containing the $\alpha_i$, so my question for the first point would become "Why are these two definitions equivalent?" Thanks for the second one! –  Frank Apr 5 '12 at 10:41
One cannot define $F=K(\alpha_i:i\in I)$ to be the smallest extension such that ..., unless there is some set in which all extensions are embedded, which is not the case. However one can retain the following informal description: an element $z$ is only in $F$ if it has to, that is if any extension containing the $\alpha_i$ contains $z$. This only happens if $z$ is a rational function in finitely many $\alpha_i$, in which case it is also a polynomial in them. Indeed the union of all $K(\alpha_{i_1},\dots,\alpha_{i_n})$ is a field that contains only such elements. –  Marc van Leeuwen Apr 5 '12 at 11:44
Yes, indeed. I should have added the hypothesis that eveything happens in a big field $L$. Then how does one show that the two definitions coincide ? I still don't see why should every element of $F=\cap \dots$ be a rational function in finitely many $\alpha_i$. –  Frank Apr 6 '12 at 22:22
The union over all finite sets $\{\alpha_{i_1},\dots,\alpha_{i_n}\}$ of all fields $K[\alpha_{i_1},\dots,\alpha_{i_n}]$ is a field that contains precisely the elements that are polynomials in the $\alpha_i$ (which cannot be otherwise than in finitely many of them). (The main point to realise for this is that in spite of being a union, it is closed for field operations.) This particular field (containing all $\alpha_i$) therefore occurs in your intersection, which consequently cannot be bigger than it. It also clearly cannot be smaller. –  Marc van Leeuwen Apr 7 '12 at 9:31