Any field automorphism fixes the prime subfield. Since the prime subfield of is itself, it is enough to show that fixes every rational number.
Because is a field homomorphism, it fixes :
Hence it fixes every integer by additivity:
For a nonzero rational number with and , we get
So fixes every element of .
Therefore
Question 2
Let be a prime integer and let be an automorphism. Prove that .
Solution
This is the same idea as in Question 1.
The field has prime subfield itself. Any field automorphism fixes the prime subfield, so must fix every element of .
Equivalently, every field automorphism of a finite prime field is the identity.
Hence
Question 3
Let be a finite field extension and let be an irreducible polynomial of degree . If and are co-prime, then prove that does not have any root in .
Solution
Assume, for contradiction, that has a root .
Since is irreducible over and is a root, the minimal polynomial of over is exactly . Therefore
Now is an intermediate field:
By the tower law,
So divides .
But , and by hypothesis and are coprime. The only positive integer dividing and equal to is impossible.
This contradiction shows that cannot have a root in .
Therefore
Question 4
Let be two prime integers.
Prove that
are isomorphic as -vector spaces but not isomorphic as fields.
Solution
Both fields are -dimensional vector spaces over :
Therefore they are isomorphic as -vector spaces, since any two -dimensional vector spaces over the same field are isomorphic.
Now suppose there were a field isomorphism
Because fixes , it must send the element to another root of the polynomial in .
If as fields, then the two quadratic extensions would be the same up to -isomorphism. That would force to satisfy the same square class relation as inside the other field, which in turn implies is a square in .
But for distinct primes and , this is impossible. So there is no field isomorphism.
Hence:
Question 5
Let be a finite field with elements, where is a prime integer and .
Prove that
Solution
The additive group of a finite field has order . In particular, the characteristic of must be a prime divisor of .
More concretely, the characteristic of any finite field is prime, say , and the field contains the prime subfield . Since has exactly elements, the size of its prime subfield must divide .
The only prime divisor of is itself. Therefore the characteristic must be .
So
Question 6
Let be a field and let be the field of fractions of the polynomial ring .
Prove that is an infinite extension.
Solution
We show that has infinite degree over .
The element is transcendental over , so the powers
are linearly independent over .
Indeed, if there were a nontrivial linear relation
with not all zero, then would satisfy a nonzero polynomial over , contradicting transcendence.
Thus contains infinitely many -linearly independent elements, so it cannot be a finite-dimensional vector space over .
Therefore
Question 7
Let be a field and .
Prove that there exists a field extension such that each has a root in .
Solution
For each polynomial , choose one of its roots in some algebraic closure of .
Let be a root of in , and define
Then is a field extension of , and by construction each lies in and satisfies