If is an algebraic extension, then any -algebra homomorphism
is surjective.
Proof
Let . We claim that .
Since is algebraic over , it satisfies some polynomial over .
Let
be its minimal polynomial.
Let be the subfield of containing all roots of that lie in .
We have
because if is a root of , then is also a root of the same polynomial.
Hence .
Since
we get
Therefore,
Thus is surjective.
Normal Extension
Let be an algebraic extension and let be an algebraic closure of containing .
If for every -algebra homomorphism
we have
then is called a normal extension.
Example
Let
where is a primitive cube root of unity.
Then is normal, since it contains all roots of
If instead
then is not normal, because a -embedding may send
and
Theorem
Let be an algebraic extension and let be an algebraic closure containing .
Then the following are equivalent:
is normal.
is the splitting field of some family of polynomials in .
If an irreducible polynomial has one root in , then it splits completely in .
Proof
(1) (2)
Since is algebraic, each is algebraic over .
Let
be its minimal polynomial.
Let
Take a -embedding
Since is algebraic, extends to
Because is normal,
Thus every conjugate of over lies in .
Hence splits in .
Therefore is the splitting field of .
(2) (1)
Let be a family of polynomials over such that is their splitting field.
Let
be any -algebra homomorphism.
If is a root of some , then is also a root of .
Since contains all roots of every , we get
Since is generated by these roots,
Because is algebraic, is surjective (by the previous proposition), hence
Thus is normal.
(2) (3)
If is a splitting field, then any irreducible polynomial having one root in must split completely in .
Conversely, if every irreducible polynomial over that has a root in splits in , then is generated by roots of such polynomials and hence is a splitting field.