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 $X$: $$ K(X) = \bigcup_{\substack{Y \subseteq X \\ Y \text{ finite}}} K(Y) = \bigcup_{\substack{Y \subseteq X \\ Y \text{ finite}}} K[Y] = K[X] $$

  • $\begingroup$ Thank you very much for your answer $\endgroup$ – asv Jul 3 at 12:14
  • $\begingroup$ 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? $\endgroup$ – asv Jul 3 at 13:49
  • $\begingroup$ @asv, not directly, because your first equality is not clear. See en.wikipedia.org/wiki/Primitive_element_theorem $\endgroup$ – lhf Jul 3 at 13:54
  • $\begingroup$ 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 $\endgroup$ – asv Jul 3 at 13:56

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