## Existence and properties of normal closure

If $E$ is an algebraic extension field of $K$, then there exists an extension field $F$ of $E$ (called the normal closure of $E$ over $K$) such that
(i) $F$ is normal over $K$;
(ii) no proper subfield of $F$ containing $E$ is normal over $K$;
(iii) if $E$ is separable over $K$, then $F$ is Galois over $K$;
(iv) $[F:K]$ is finite if and only if $[E:K]$ is finite.

The field $F$ is uniquely determined up to an $E$-isomorphism.

Proof:
(i) Let $X=\{u_i\mid i\in I\}$ be a basis of $E$ over $K$ and let $f_i\in K[x]$ be the minimal polynomial of $u_i$. If $F$ is a splitting field of $S=\{f_i\mid i\in I\}$ over $E$, then $F=E(Y)$, where $Y\supseteq X$ is the set of roots of the $f_i$. Then $F=K(X)(Y)=K(Y)$ so $F$ is also a splitting field of $S$ over $K$, hence $F$ is normal over $K$ as it is the splitting field of a family of polynomials in $K[x]$.

(iii) If $E$ is separable over $K$, then each $f_i$ is separable. Therefore $F$ is Galois over $K$ as it is a splitting field over $K$ of a set of separable polynomials in $K[x]$.

(iv) If $[E:K]$ is finite, then so is $X$ and hence $S$. Say $S=\{f_1,\dots,f_n\}$. Then $F=E(Y)$, where $Y$ is the set of roots of the $f_i$. Then $F$ is finitely generated and algebraic, thus a finite extension. So $[F:K]$ is finite.

(ii) A subfield $F_0$ of $F$ that contains $E$ necessarily contains the root $u_i$ of $f_i\in S$ for every $i$. If $F_0$ is normal over $K$ (so that each $f_i$ splits in $F_0$ by definition), then $F\subset F_0$ (since $F$ is the splitting field) and hence $F=F_0$.

Finally let $F_1$ be another extension field of $E$ with properties (i) and (ii). Since $F_1$ is normal over $K$ and contains each $u_i$, $F_1$ must contain a splitting field $F_2$ of $S$ over $K$ with $E\subset F_2$. $F_2$ is normal over $K$ (splitting field over $K$ of family of polynomials in $K[x]$), hence $F_2=F_1$ by (ii).

Therefore both $F$ and $F_1$ are splitting fields of $S$ over $K$ and hence of $S$ over $E$: If $F=K(Y)$ (where $Y$ is set of roots of $f_i$) then $F\subseteq E(Y)$ since $E(Y)$ contains $K$ and $Y$. Since $Y\supseteq X$, so $K(Y)$ contains $E=K(X)$ and $Y$, hence $F=E(Y)$. Hence the identity map on $E$ extends to an $E$-isomorphism $F\cong F_1$.

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