Let be a polynomial of degree , and let be the splitting field of over .
Show that
Question 2
Let be a field and let be an algebraic closure of .
(i) Prove that is infinite.
(ii) If is countable, prove that is countable.
Question 3
Let be a field and let be the field of fractions of .
Prove that is not algebraically closed.
Question 4
Let be a field extension and let be an embedding into an algebraically closed field .
Give an example to show that if is not an algebraic extension, then there may not exist an extension of to .
Question 5
Find the splitting field over of:
(i)
(ii)
Question 6
Let be a real root of
(i) Prove that
is a normal extension.
(ii) Prove that
is normal.
(iii) Prove that
is not normal.
Question 7
(i) Let be a field and let
Show that is a subgroup of .
(ii) Let denote the group of all -algebra automorphisms of .
Show that there is a group homomorphism
where
Prove that there is a homomorphism
where
for
Question 8
Let be the polynomial ring over and let be its field of fractions.
Let be generated by
Find
Let be generated by
Show that is an infinite cyclic group and find
Question 9
Let and be two field extensions.
Let be algebraic over and let be its minimal polynomial.
Prove that there is a one-to-one correspondence between
the set of -algebra embeddings, , given by
the set of all roots of in , given by .
Solutions
Question 1
Let be a root of and let . Then .
Over , the polynomial has one root removed, so the remaining factor has degree at most . If is the splitting field of over , then is also the splitting field of a polynomial of degree at most over .
By induction,
Therefore
So
Question 2
(i) is infinite
If were finite, say , then every element would satisfy . So the polynomial
would have no root in , because for every .
This contradicts algebraic closedness. Hence is infinite.
(ii) If is countable, then is countable
There are only countably many monic polynomials in when is countable. Each such polynomial has only finitely many roots in . Since is the union of the root sets of all monic polynomials over , it is a countable union of finite sets.
Therefore is countable.
So
Question 3
Consider
If it had a root in , then would be a square in . But is irreducible over by Eisenstein’s criterion at the prime polynomial in .
Hence contains a polynomial with no root, so it is not algebraically closed.
Therefore
Question 4
Take
with transcendental over , and let be the inclusion.
If extended , then would have to be transcendental over , because is transcendental over . But contains only algebraic elements over .
So no extension exists.
Hence a valid example is
Question 5
(i)
Let . The roots are for , where is a primitive sixth root of unity. So the splitting field is
(ii)
Let . Then , and the roots are for .
Thus the splitting field is
Question 6
The handwritten statement in (i) appears to contain a typo. As written, is not normal for a root of ; the standard normal extension is , the splitting field of .
Using the corrected standard version:
Let . Then the roots of are
So the splitting field is , which is normal over .
For the tower
one gets a quadratic extension, hence it is normal.
But the total extension over is not normal, because does not contain all roots of the minimal polynomial of over ; its normal closure is the larger splitting field .
Question 7
(i) is a subgroup of
An element of has the form
Its inverse is
and the product of two such matrices is again of the same form:
So .
(ii) The homomorphism to
For
define and extend uniquely to a -algebra endomorphism of . Since , it has inverse , so it is an automorphism.
Moreover,
so is a group homomorphism.
(iii) The homomorphism to
For
define
Because , this is an invertible fractional linear transformation, with inverse given by . Hence it extends to a -automorphism of .
Again composition matches matrix multiplication, so
is a group homomorphism.
Question 8
(1) The group generated by and
These two Möbius transformations permute , and together they generate the full permutation group of that set, which is isomorphic to .
A standard invariant is
This is fixed by both generators, and in fact
So
(2) The group generated by
Let . Then , so only when . Therefore the group generated by is infinite cyclic:
If satisfies , then any pole of would give infinitely many translated poles. A rational function cannot have infinitely many poles, so must be constant.
Hence
Question 9
Define
If two -embeddings agree on , then they agree on all of , since every element of is a rational function in with coefficients in . So is injective.
Conversely, if is a root of , then the map respects the minimal polynomial relation of , so it extends uniquely to a -embedding . Thus every root arises from an embedding, so is surjective.
Therefore is a bijection, and we have the desired one-to-one correspondence: