# Math Help - subfield isomorphism

1. ## subfield isomorphism

How do I show that if p > 0 is the characteristic of the finite field F (p is prime obviously) that F contains a subfield {k * 1F: k=0,1,...,p-1} that is isomorphic to Zp?

Note k*1F= 1+1+1...+1 (k times)

Also, how can I show that the set {k * 1F: k=0,1,...,p-1} actually is a subfield?

2. Originally Posted by PvtBillPilgrim
How do I show that if p > 0 is the characteristic of the finite field F (p is prime obviously) that F contains a subfield {k * 1F: k=0,1,...,p-1} that is isomorphic to Zp?
This is one of the elegant results about fields.

Theorem: If $R$ is a ring with unity and with charachteristic $n>0$, then $R$ contains $\mathbb{Z}_n$ as a subring up to isomorphism.

Thus, given a field with a charachetristic $n>1$ then $n$ must be a prime. Because otherwise it will have zero-divisors, which a field does not have. Thus it contains $\mathbb{Z}_p$ as a subfield up to isomorphism.

Originally Posted by PvtBillPilgrim
Also, how can I show that the set {k * 1F: k=0,1,...,p-1} actually is a subfield?
This is a known fact that $\mathbb{Z}_p$ is always a field. Because it is a finite integral domain, and a finite integral domain is always a field.

3. How do you prove that initial part though (the theorem)?

4. Originally Posted by PvtBillPilgrim
How do you prove that initial part though (the theorem)?
The map $\phi:\mathbb{Z}\to R$ which we define as $\phi(m)=m\cdot 1$ is a homomorphism (excerise, prove it). By the fundamental theorem the kernel is an ideal of $\mathbb{Z}$. Now all ideals of $\mathbb{Z}$ are of the form $k\mathbb{Z}$ because a subgroup of a cyclic group is cyclic. Thus, if $R$ has charachteristic $n>0$ then, $\ker \phi =n\mathbb{Z}$. Again by the fundamental theorem, $\phi [\mathbb{Z} ] \simeq \mathbb{Z}/n\mathbb{Z}\simeq \mathbb{Z}_n$. Thus, the subring $\phi[\mathbb{Z}]$ confirms the theorem. Q.E.D.

What is happening to you in this class? I seriously doubt it they will ask you to prove this theorem, is it not in your textbook?

5. Yeah, we have no textbook.

Thanks for the help.

6. Originally Posted by PvtBillPilgrim
Yeah, we have no textbook.
Why not? Is the professor insane?