I have to prove the following theorem :
Let $p$ be a prime number and let $n \ge 1$,be any integer, then there exists a field of order $p^n$.
My attempt
I started off by considering the polynomial $f(x)$=$x^{p^n}-x \in \Bbb Z_p[x]$.
I took $F$ to be the splitting field of $f(x)$ over $\Bbb Z_p$.
Since $F$ is the splitting field of $f(x)$ over $\Bbb Z_p$ therefore $f(x)$ has exactly $p^n$ zeros in $F$ counting multiplicity.
Also since $f'(x)$ = $p^nx^{p^n-1}-1$ , therefore $f(x)$ will not have any multiple zeros and hence it will have $p^n$ distinct zeros in $F$.
Then I took $K =\{k \in F | k^{p^n} = k\}$.
I was trying to prove that $K$ is a sub-field of $F$.
$1 \in K$ thus $K$ is non-empty.
For $a,b \in K$ $a^{p^n}=a$ and $b^{p^n}=b$
So $(a+b)^{p^n}$ (should be) = $a^{p^n}$+$b^{p^n}$, this is where I am currently stuck!
I know that this is true when $p$ is the characteristic of a field, but why would this expansion be true here?
Best Answer
In $\mathbb Z_p$ and hence its extension $F$ we have $$ p a = \underbrace{a + \cdots + a}_{p \text{ times}} = (\underbrace{1 + \cdots + 1}_{p \text{ times}})a = 0 \cdot a = 0. $$
By the binomial theorem $$ (a + b)^{p^n} = \sum_{k=0}^{p^n}\binom{p^n}{k} a^k b^{p^n - k}. $$
All terms in the latter expansion are multiples of $p$ except for $a^{p^n}$ and $b^{p^n}$. It follows that in $\mathbb Z_p$ and $F$ we have $$(a + b)^{p^n} = a^{p^n} + b^{p^n}.$$
Now show that the roots of $f(x)$ are closed under addition, multiplication and taking additive and multiplicative inverses. This shows that $F$ consists entirely of the roots of $f(x)$. We conclude that the order of $F$ is $p^n$ as desired.