Why does it stand that #$\mathbb{Z}_p(a)=p^n$?

Question

If $\mathbb{Z}_p \leq K$ an algebraic extension, then $K$ has the identity $$\forall a \in K, \exists b \in K \text{ with } a=b^p$$

The proof is the following:

Let $a \in K$.

We take $\mathbb{Z}_p \leq \mathbb{Z}_p(a)$, $a$ algebraic over $\mathbb{Z}_p$.

So, $[\mathbb{Z}_p(a) : \mathbb{Z}_p ]=n<\infty$, so $dim_{\mathbb{Z}_p}\mathbb{Z}_p(a)=n<\infty$.

So, #$\mathbb{Z}_p(a)=p^n$, that means that it is a finite field of characteristic $p$.

From that it follows that $$\exists b \in \mathbb{Z}_p(a) \subseteq K \text{ with } a=b^p$$

Do we have that $[\mathbb{Z}_p(a) : \mathbb{Z}_p ]=n<\infty$ because of the fact that $a$ algebraic over $\mathbb{Z}_p$ ??

Also could you explain me how we conclude that #$\mathbb{Z}_p(a)=p^n$ ??

Answer 1

An $n$-dimensional space $V$ over a field $K$ of cardinality $p$ has $p^n$ elements, because you can chose a basis $\mathcal B = \{v_i\}_{i=1}^n$ such that every element of $V$ can be written as $$v = \sum_{i=1}^n \alpha_i v_i$$ Where $\alpha_i \in K$ are unique for each $v$. Thus there are $n$ choices of $p$ elements each yielding $p^n$ combinations for $v$.

Here $V = \mathbb Z_p (a)$ and $K = \mathbb Z_p$

Answer 2

Let $a\in K$; then $\mathbb{Z}_p(a)$ is a finite dimensional vector space over $\mathbb{Z}_p$; hence it has $p^n$ elements, where $n$ is the dimension. Why is it finite dimensional? Because, if $f$ is the minimum polynomial of $a$ and it has degree $n$, then $\{1,a,\dots,a^{n-1}\}$ is a basis of $\mathbb{Z}_p(a)$ over $\mathbb{Z}_p$.

Now, the map $\varphi\colon \mathbb{Z}_p(a)\to\mathbb{Z}_p(a)$ defined by $\varphi(x)=x^p$ is a field homomorphism (prove it), so it is injective. But, since the domain is equal to the codomain and it is finite, $\varphi$ is also surjective.