Let be a field extension and let . Show that is algebraic over if and only if
Solution
We prove both directions.
If is algebraic over
Let be the minimal polynomial of over . Then every element of can be written as a rational function in with denominator not divisible by .
Because , any power for can be reduced to a -linear combination of , where .
Hence is spanned over by , so every element of lies in . The reverse inclusion is obvious, so
If
Suppose were transcendental over . Then would be isomorphic to the polynomial ring , which is not a field. But is a field.
So cannot be transcendental. Therefore is algebraic over .
Thus
Question 2
Let be an algebraic extension, and let be a subring containing . Show that is a field.
Solution
Take any nonzero element .
Since is algebraic, is algebraic over . Let its minimal polynomial over be
Because , the constant term is nonzero. Indeed, if , then divides , so would be a root and the minimal polynomial of would not be irreducible unless .
Now evaluate at :
Rearrange to solve for :
Since and the right-hand side is in , we get
So every nonzero element of is invertible in . Therefore is a field.
Hence
Question 3
If and are two distinct primes, show that
Solution
Let
Clearly , so
For the reverse inclusion, compute
Hence
Now use
Since , we have , and therefore
Adding and subtracting gives
Thus , so
Therefore
Question 4
Let . Is the polynomial irreducible over ?
Solution
We show that has no root in .
Suppose is a root, with . Then
Expanding gives
Equating imaginary parts,
So either or .
If , then , impossible in .
If , then , also impossible in .
Hence has no root in and therefore is irreducible over .
So the answer is
Question 5
Let be a field extension and let be algebraic over .
(i) If is odd, show that .
(ii) If , does this necessarily mean that is odd?
Solution
Let
Then and is algebraic over because it satisfies
So
(i) If is odd
By the tower law,
The left-hand side is odd, while . Therefore cannot be , so it must be .
Hence
i.e.
(ii) Does imply is odd?
No. The converse is false.
Take
Then is algebraic of degree over , so
which is even.
But
so and hence
Thus equality of the fields does not force the degree to be odd.
So the answer is:
Question 6
Let be a field extension, and let be two subfields of a common overfield containing . Assume that and are finite extensions.
(i) If
prove that
(ii) Show that the converse of (i) holds if or .
Solution
Because and are finite over , the standard degree formula applies:
(i) If
Substitute into the formula above:
Since the numerator is nonzero, we get
Therefore
(ii) Converse when one degree is
In fact, for finite extensions the converse is true without extra assumptions:
If , then the same degree formula gives
So certainly the converse holds when or .
Thus
Question 7
Let be a field and let be transcendental over .
(i) Prove that is algebraic over and that is transcendental over .
(ii) Find .
Solution
(i) is algebraic over , and is transcendental over
Since , the element satisfies
over .
So is algebraic over .
Now suppose were algebraic over . Then would be an algebraic extension of , and since is algebraic over , transitivity of algebraicity would imply that is algebraic over .
That contradicts the hypothesis that is transcendental over .
Therefore is transcendental over .
So
(ii) Compute
Let . Since is transcendental over , the field is a rational function field, and is obtained from it by adjoining an -th root of .
Consider the polynomial
This polynomial is irreducible over . One way to see this is to view it in and apply Eisenstein’s criterion at the prime element .
Since is a root of , the minimal polynomial of over has degree .