If is an algebraic extension field of , then there exists an extension field of (called the **normal closure** of over ) such that

(i) is normal over ;

(ii) no proper subfield of containing is normal over ;

(iii) if is separable over , then is Galois over ;

(iv) is finite if and only if is finite.

The field is uniquely determined up to an -isomorphism.

**Proof:**

(i) Let be a basis of over and let be the minimal polynomial of . If is a splitting field of over , then , where is the set of roots of the . Then so is also a splitting field of over , hence is normal over as it is the splitting field of a family of polynomials in .

(iii) If is separable over , then each is separable. Therefore is Galois over as it is a splitting field over of a set of separable polynomials in .

(iv) If is finite, then so is and hence . Say . Then , where is the set of roots of the . Then is finitely generated and algebraic, thus a finite extension. So is finite.

(ii) A subfield of that contains necessarily contains the root of for every . If is normal over (so that each splits in by definition), then (since is the splitting field) and hence .

Finally let be another extension field of with properties (i) and (ii). Since is normal over and contains each , must contain a splitting field of over with . is normal over (splitting field over of family of polynomials in ), hence by (ii).

Therefore both and are splitting fields of over and hence of over : If (where is set of roots of ) then since contains and . Since , so contains and , hence . Hence the identity map on extends to an -isomorphism .