# $K[X]=K(X)$ with X set of algebraic elements

Let $$F$$ an extension field of $$K$$ and $$X$$ a subset of algebraic elements of $$F$$ over $$K$$, $$K(X)$$ the intersections of all fields containing $$K$$ and $$X$$, and $$K[X]$$ the intersections of all rings containing $$K$$ and $$X$$.

Is it $$K(X)=K[X]$$?

In case $$X$$ is finite, I can prove it with induction starting from $$K(a)=K[a]$$ with $$a$$ algebraic (this result is in textbooks) then my question is for X infinite.

Yes, because every element of $$K(X)$$ belongs to $$K(Y)$$ where $$Y$$ is a finite subset of $$X$$: $$K(X) = \bigcup_{\substack{Y \subseteq X \\ Y \text{ finite}}} K(Y) = \bigcup_{\substack{Y \subseteq X \\ Y \text{ finite}}} K[Y] = K[X]$$
• I was thinking that your proof can be used also in finite and infinite case in this way: $$K(X) = \bigcup_{\substack{a \in X }} K(a) = \bigcup_{\substack{a \in X }} K[a] = K[X]$$ Is it correct? – asv Jul 3 at 13:49
• I was thinking that $K(X)=\{f(a)g(a)^{-1}|a \in X, f(x),g(x) \in K[x], g \neq 0\}$ but it is not. It occurs if X={a}. Then we must prove first the finite case and then the infinite case with your proof. If I'm not wrong – asv Jul 3 at 13:56