Finite field with algebraic element over $\mathbb{Z}_p$ Let $F$ be a finite field of characteristic $p$. Show that every element of $F$ is algebraic over $\mathbb{Z}_p$.
Since char$(F)$ = $p$, $\forall a \in F $ not zero, $ap$ = 0. We also have the same characteristic for $\mathbb{Z}_p$.
How can I use the prime characteristic to show there is a polynomial $P(x) \in \mathbb{Z}_p[x]$ such that $P(a) = 0$?
 A: As $F$ is finite, every element of $F^*$ has finite (multiplicative) order. So, for $\alpha \in F^*$, there is some $n \in {\mathbb N}$ such that $\alpha^n = 1$, making $\alpha$ a root of $X^n - 1 \in {\mathbb Z}_p[X]$.
(In fact, $F^*$ is cyclic and you can take the same $n$ for all elements of $F^*$.)
Alternatively, the dimension of $F$ over ${\mathbb Z}_p$ must be finite, as $F$ itself is finite. This makes it a finite and hence algebraic extension.
A: For every polynomial $f(x)$ over $\mathbf{F}_p$, the value $f(a)$ is also in $\mathbf{F}_p$.
There are infinitely many such polynomials, but only finitely many possible values for $f(a)$. In particular, there must be two polynomial $f(x)$ and $g(x)$ such that $f(a) = g(a)$. Can you use this information to find a polynomial that has $a$ as a root?
This is a standard method for reasoning about finite structures: you take whatever 'constructions' you're considering, and use the fact there are more constructions than there are actual elements (usually because you can make infinitely many constructions), so two of those constructions must yield the same value. Then you infer what you can from that fact.
