1
$\begingroup$

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.

1 Answers 1

1

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.

  • 1
    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.2012-04-07