I have a conceptual problem regarding field extensions in Galois theory. My book (Basic Abstract Algebra, Bhattacharya) doesn't explicitly talk about it. Consider first the following statement from the book:
"If $\alpha \in \mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5},\ldots )$, then there exists a positive integer $r$ such that $\alpha \in \mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5},\ldots ,\sqrt{p_r})$" (p.291).
Why is this true? That is, why can't there be an element in this infinite extension that requires the root of every prime to be "constructed"?
In a similar manner,
If $F$ is a field and $E=F(S)$ is an extension field of $F$ generated by $F$ and an arbitrary collection $S=\{\alpha_i\}_{i\in \Lambda}$ (algebraic or not), is it then true that if $\beta \in E$, then $\beta \in F(S')$ for some finite $S'\subseteq S$?
Best Answer
Suppose $E=F(S)$.
Since $E$ is a field, it must contain the sum, product and difference of any two elements of $F\cup S$. Furthermore it must contain the sum, product and difference of any of the resulting expressions, etc.
It is clear that since this process of constucting elements using $+,-,\bullet$ started at $F\cup S$, and every element created in this process can be created in a finite number of steps, we have created every element in $F[S]$, the polynomial ring over $F$ in the commuting indeterminates $x \in S$.
But being a field, $E\hspace{1mm}$ must also contain the quotient of any two such expressions (so that the denominator is nonzero). The resulting structure, also denoted $F(S)$, is the field of rational functions/ expressions over $F$ in the commuting indeterminates $x \in S$. Since this is a field, we have by the minimality of $E$, that $E$ equals this field, that is: $$E=F(S)[\text{field generated by $F$ and $S$}]=F(S)\text{[field of rational functions]}$$
From this, it is clear that any element $\alpha \in E$ is constructed from only a finite number of elements of $S$, that is $\alpha \in E(S')$ for some finite $S' \subseteq S.$