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My problem:

Let $n$ be a natural number and $p$ be a prime. Prove that there exists an irreducible polynomial of degree $n$ in $mathbb{Z}_p[x]$ (note: $Z_p$ is the quotient field $mathbb{Z}/pmathbb{Z}$)

Basically I don't have a clear idea to even think about. But somehow I think this relates to the splitting field of $f(x) = x^{p^n} - x$ in $Z_p[x]$, which serves as the extension of $Z_p$ to $p^n$ elements. Is it possible that there exists a factor of this polynomial which has degree $n$?

Please give me a hint. Anything is greatly appreciated.

I have basic knowledge about field extensions, spliting fiels and finite fields. Thank you.

This is all standard theory. The main steps are:

1. Show that the splitting field $K$ of $x^{p^n}-x$ has order $p^n$,


2. Show that $K$'s multiplicative group is cyclic, so has a generator $alpha$,


3. Observe that $K=Bbb Z_p(alpha)$,


4. Show the minimal polynomial of $alpha$ over $Bbb Z_p$ has degree $n$.

OK, now that the standard theory's up, I'll add my favorite argument. We prove that irreducible polynomials of every degree exist by counting them.

Let $F$ be a field with $q$ elements, for some finite $q$. We count the number of monic polynomials of degree $n$ over $F$ in two ways.

First, a monic polynomial $x^n + a_{n-1}x^{n-1}+cdots + a_1 x + a_0$ has $n$ coefficients that can freely vary among $q$ values each, for a total of $q^n$ polynomials.

Second, a monic polynomial can be uniquely factored as a product of monic irreducible polynomials. We choose the power each polynomial is raised to, make sure the sum of the degrees is $n$ - it's an intimidating counting problem, but we can make it easier by thinking in terms of generating functions. Let $P_k$ be the number of monic irreducible polynomials of degree $k$, and $M_n$ the total number of monic polynomials of degree $n$. This model gives us that
$$sum_{n=0}^{infty} M_n t^n = prod_{k=1}^{infty}frac1{(1-t^k)^{P_k}}$$
(Each $frac1{1-t^k}=1+t^k+t^{2k}+cdots$ factor represents the choice of how many copies of a particular irreducible polynomial of degree $k$ to include)

So, what now? First, we use that earlier count to simplify the left side:
$$sum_{n=0}^{infty} M_n t^n = sum_{n=0}^{infty} q^n t^n = frac1{1-qt}$$
Next, take logarithms* and differentiate with respect to $t$:
begin{align*}frac1{1-qt} &= prod_{k=1}^{infty}frac1{(1-t^k)^{P_k}}\
-log(1-qt) &= sum_{k=1}^{infty} -P_klog(1-t^k)\
frac{q}{1-qt} &= sum_{k=1}^{infty} frac{kP_kt^{k-1}}{1-t^k}\
sum_{n=1}^{infty} q^nt^n = frac{qt}{1-qt} &= sum_{k=1}^{infty} frac{kP_kt^k}{1-t^k} = sum_{n=1}^{infty}t^nsum_{k|n}kP_k\
q^n &= sum_{d|n}dP_dend{align*}
Now, we can extract the values of $P_n$ by a standard technique; Mobius inversion. We get
$$nP_n = sum_{d|n} mu(d)q^{frac{n}{d}}$$
where $mu$ is the Mobius function of number theory.

Now, what we really wanted to know is that $P_n$ is positive for $nge 1$. For this, one key fact: $mu(d)in {-1,0,1}$ for all natural $d$ and $mu(1)=1$, so
begin{align*}nP_n &= q^n +sum_{k|n, 0<k<n}mu(frac{n}{k})q^k\
nP_n &ge q^n - sum_{k|n, 0<k<n} q^k ge q^n - sum_{0<k<n}q^k = q^n - frac{q^n-q}{q-1} > 0end{align*}
We use that $qge 2$ here; we do need at least two elements to have a field, after all. And there it is: any finite field with $q$ elements has irreducible polynomials for every degree. This includes both the prime fields $mathbb{Z}/p$ and fields with prime power order. That such fields exist to have irreducible polynomials over - well, we'll just have to apply this result over the prime fields first.

*Taking a logarithm? Technically, it's all still formal power series manipulation, that doesn't rely on any form of convergence - but if you feel uncomfortable, we're just taking a derivative and then dividing by the original.