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

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*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 $p$.

*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 $|K| = |\Bbb F_p^m| = p^m$.
