Why if $K$ is a finite field, $|K|=p^d$ for a prime $p$?

Question

Why if $K$ is a finite field, $|K|=p^d$ for a prime $p$ ?

The solution goes like :

Consider $\varphi:\mathbb Z\longrightarrow K$. Since $K$ is finite $\ker \varphi=p\mathbb Z$ for a prime $p$.

Q1) Why $\ker\varphi=p\mathbb Z$ ? And why $p$ has to be prime ? I tryied to understand for couple times, but impossible to understand how. Any idea ?

Therefore $$\tilde \varphi: \mathbb Z/p\mathbb Z\longrightarrow K$$ is one to one. In particular $K/(\mathbb Z/p\mathbb Z)$ is a field extension, and thus $K$ is a vector space over $\mathbb Z/p\mathbb Z$. We conclude that $|K|$ is a power of $p$ (i.e. of the form $p^d$).

Q2) I agree with every thing except in what we can conclude that $|K|$ is a power of $p$. How can it be ?

Answer 1

Q1: You did not say how $\phi$ is defined. It follows from the definition that $\ker(\phi)$ is a principal ideal $n\mathbb{Z}$, because $\mathbb{Z}$ is a PID. Since the image is contained in a field, it follows that $n$ is prime ($\mathbb{Z}/n\mathbb{Z}$ is a domain iff $n=p$ is prime).

Q2: Let $(e_1,\ldots ,e_n)$ a basis over $\mathbb{F}_p$. Hence every element $x$ can be written in $K$ as $x=a_1e_1+\cdots a_ne_n$ with $a_i\in \mathbb{F}$. This yields $p^n$ possibilities. Hence $|K|=p^n$.

Answer 2

The image of $\varphi: \mathbb{Z} \to K$ is a subset of a field, and hence is a domain. Then $\ker \varphi$ is prime and thus has the form $(p)$ for some prime $p \in \mathbb{Z}$. This is because the ideal is of the form $(d)$ and divisors $m,n \in \mathbb{Z}$ of $d$ are zero divisors in the quotient.

To see that $|K|$ is a power of $p$, try to count the number of K-linear combinations with $n$ basis vectors. (It's $p^n$)

Answer 3

1. For any ring $R$, we can define $$\phi: \Bbb Z \to R$$ to be the homomorphism $$n \mapsto n \cdot 1 = \underbrace{1 + \cdots + 1}_n .$$ (Presumably this is the unspecified map $\varphi$ in the question.) Since $\Bbb Z$ is a principal ideal domain, $\ker \phi = n \Bbb Z$ for some $n$ (in fact, there are two choices for $n$, which differ only in sign; the positive choice is the characteristic of the ring). If $R$ finite, since $\Bbb Z$ is not, $\phi$ must have nontrivial kernel, and hence $n \neq 0$. If $n$ is composite, say, $n = a b$ for $a, b \neq 1$, then by construction $\phi(a), \phi(b)$ are nonzero but $\phi(a) \phi(b) = \phi(ab) = \phi(n) = 0$; so $\phi(a)$ and $\phi(b)$ are zero divisors, and in particular, $R$ is not a field. Thus, if $R$ is a field, $n$ must be some prime $p$.

2. Like you say, any finite field $K$ is a vector space over its prime field $\phi(\Bbb Z)$, which (since $\phi$ is a homomorphism) is isomorphic to $\Bbb F_p := \Bbb Z / p \Bbb Z$. Any basis $(E_1, \ldots, E_m)$ of $K$ over $\Bbb F_p$ defines a vector space isomorphism $\Bbb F_p^m \to K$ by $(a_1, \ldots, a_m) \mapsto \sum_i a_i E_i$, and so $|K| = |\Bbb F_p^m| = p^m$.