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An algebraic extension $K/F$ is called a normal extension if any irreducible polynomial of $F[x]$, which has a root in $K$, can be completely decomposed in $K$.

I have know that a finite extension $K/F$ is a normal extension if and only if $K$ is a splitting field of a polynomial $f(x)\in F[x]$.

Is is right that if $K$ is the splitting field of a collection of polynomials in $F[x]$, this is, let $T$ denote a collection of polynomials and $S$ the set of roots of polynomials in $T$ ($S$ maybe be uncountable) then $K=F(S)$, $K/F$ is a normal extension? How to prove?

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  • $\begingroup$ Yes this is true. You can prove this by observing that the splitting of a polynomial can be tested over a finite sub-extension. $\endgroup$ – Victor Zhang Dec 17 '17 at 15:03
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Yes, this is true. To prove it, suppose $f\in F[x]$ is irreducible and has a root $\alpha$ in $K$. Then there is some finite subset $S_0\subseteq S$ such that $\alpha\in F(S_0)$. Let $S_1$ be the set of elements of $S$ which have the same minimal polynomial over $F$ as an element of $S_0$. Then $S_1$ is still finite, and $F(S_1)$ is now the splitting field of a finite set of polynomials and thus normal over $F$. But $\alpha\in F(S_0)\subseteq F(S_1)$, so all the roots of $f$ are in $F(S_1)$. Thus $f$ splits over $K$, and since $f$ was arbitrary, $K$ is normal over $F$.

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