Let F be a field of prime characteristic p.?

Question

Prove that K={a E F:a^p=a} is a subfield of F.
Thanks in advance

Answer 1

Doug forgot to check that a^p+b^p and a^p x b^p are both in K. The second is easy, the first comes from the formula that (a+b)^p=a^p+C(p,p-1)a^{p-1}b+...+C(p,1)ab^{p-1}+b^p. Now in that expansion, except for the first and the last element, each term is divisible by p, hence are zero in a field of characteristic zero. That should do it.

Answer 2

You can completely determine the set K as a familiar object. Look at the elements 0,1,2,3,...,p-1. These elements exist in the field F because F has characteristic p. By Fermat's Little Theorem, a^p=a (mod p) for a=0,1,2,...,p-1. This shows that all elements of the field Z/pZ lie in K. Are there any more? Well, no. All elements in K satisfy x^p-x=0. K is a *field*, so there are at most p roots of a polynomial of degree p. Since we have already exhibited p roots, that must be all of them. Since Z/pZ is a field, K is a field (they are isomorphic).

Answer 3

What's to prove? If F is a field, then it's a commutative division ring with identity. That means that K isn't the null set since it will contain, as a minimum, the multiplicative identity of F. And if a,b ε F then a^p, b^p ε F and, by definition, a^p+b^p and a^p*b^p are in F. So for any a, b (if it exists) such that a^p=a, b^p=b ε K => (ab)^p ε K.


Doug