# subfield isomorphism

• Jan 27th 2007, 06:13 PM
PvtBillPilgrim
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?
• Jan 27th 2007, 06:46 PM
ThePerfectHacker
Quote:

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

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

Quote:

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 $\displaystyle \mathbb{Z}_p$ is always a field. Because it is a finite integral domain, and a finite integral domain is always a field.
• Jan 27th 2007, 06:53 PM
PvtBillPilgrim
How do you prove that initial part though (the theorem)?
• Jan 27th 2007, 07:25 PM
ThePerfectHacker
Quote:

Originally Posted by PvtBillPilgrim
How do you prove that initial part though (the theorem)?

The map $\displaystyle \phi:\mathbb{Z}\to R$ which we define as $\displaystyle \phi(m)=m\cdot 1$ is a homomorphism (excerise, prove it). By the fundamental theorem the kernel is an ideal of $\displaystyle \mathbb{Z}$. Now all ideals of $\displaystyle \mathbb{Z}$ are of the form $\displaystyle k\mathbb{Z}$ because a subgroup of a cyclic group is cyclic. Thus, if $\displaystyle R$ has charachteristic $\displaystyle n>0$ then, $\displaystyle \ker \phi =n\mathbb{Z}$. Again by the fundamental theorem, $\displaystyle \phi [\mathbb{Z} ] \simeq \mathbb{Z}/n\mathbb{Z}\simeq \mathbb{Z}_n$. Thus, the subring $\displaystyle \phi[\mathbb{Z}]$ 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?
• Jan 27th 2007, 08:11 PM
PvtBillPilgrim
Yeah, we have no textbook.

Thanks for the help.
• Jan 28th 2007, 05:06 AM
ThePerfectHacker
Quote:

Originally Posted by PvtBillPilgrim
Yeah, we have no textbook.

Why not? Is the professor insane?