0

Let $K$ a finite field. I want to show that $|K|=p^d$. I consider an homomorphism $\Phi:\mathbb Z\to K$ which is clearly not injective, therefore $\ker\Phi\neq\{0\}$.

What I want is to prove that $\ker \Phi= p\mathbb Z$ for a certain prime, and thus $\mathbb Z/p\mathbb Z\cong \text{Im}(\Phi)$. In particular, I would like to have $\text{Im}(\Phi)=K$ and thus, since $\Phi(\mathbb Z/p\mathbb Z)=k$, we need to have $p\mid |K|$ and thus $|K|=p^d$.

Well, every thing is confuse in my mind, so any help si welcome.

PS: My hint is : consider $\Phi: \mathbb Z\to K$.

idm
  • 11,824
  • Ok, thanks for your answer. I'm sorry, but I don't understand every thing. How do you get that $\Phi$ is unique and that it's isomorphic to a subring of $K$ ? – idm Oct 14 '15 at 12:22
  • The first thing you should notice is that every finite field has characteristic $p$ for some prime $p$, so we have char $K = p$. This prime $p$ will be the one you then use to prove $|K| = p$. – layman Oct 14 '15 at 12:23

1 Answers1

2

You don't want to consider a homomorphism, but the unique ring homomorphism $\Phi\colon\mathbb{Z}\to K$ (sending $1$ to $1$, of course). It's unique, because it's an additive group homomorphism and must map $1$ to $1$. It's a standard theorem that the additive group homomorphism sending $1$ to $1$ is also a ring homomorphism.

By the homomorphism theorem, $\mathbb{Z}/\ker\Phi$ is isomorphic to a subring of $K$, so it's a domain. Then $\ker\Phi=p\mathbb{Z}$, for a unique prime $p$ (it's impossible that $\ker\Phi=\{0\}$, as you remarked).

Now $K$ becomes a vector space over the image $F$ of $\Phi$, which is a field with $p$ elements, being a finite domain with $p$ elements. Hence $|K|=p^d$, where $d$ is the dimension of $K$ as vector space over $F$.

egreg
  • 238,574
  • I follow you util $\ker\Phi =p\mathbb Z$. By homomorphsim theorem, $\mathbb Z/\ker\Phi$ is isomorphic to a subgroup of $K$. I think is not difficult to prove that $\mathbb Z/\ker\Phi$ is a ring. But why $\ker\Phi =p\mathbb Z$ ? – idm Oct 14 '15 at 13:28
  • @idm Since $\Phi$ is a ring homomorphism (prove it), $\mathbb{Z}/\ker\Phi$ is isomorphic to a subring of $K$. Any subring of a field is a domain. When, for $n>0$, $\mathbb{Z}/n\mathbb{Z}$ is a domain? And what are the ideals of $\mathbb{Z}$, which $\ker\Phi$ is one of? – egreg Oct 14 '15 at 14:33
  • Actually, what is $\Phi$ ? I mean, what is $\Phi(h)$ ? – idm Oct 14 '15 at 14:55
  • @idm Let $\mathbf{1}$ denote the unity in $K$; then $\Phi(n)=n\mathbf{1}$. That's the only possibility. – egreg Oct 14 '15 at 14:56