*This is our first contributed gemstone! Submitted by user Anon1.*

In the following, $p$ denotes a prime. We wish to prove that, for all positive integers $n$, there is a finite field of order $p^{n}$.

**Step 1. Restating the problem. **

**Claim: **It suffices to show that, for some power of $p$ (call it $q$), there exists a finite field of order $q^{n}$.

**Proof. **Suppose there is a field $F$ such that $|F| = q^{n}$. The claim is that the solution set to $x^{p^{n}} = x$ in $F$ is a subfield of order $p^{n}$.

Since $q$ is a power of $p$, we have $$p^{n}-1 | q^{n}-1.$$ Since $q^{n}-1$ is the order of the cyclic group $F^{ \times }$, we know that $F^{ \times }$ has a unique cyclic subgroup of order $p^{n}-1$. This contributes $p^{n}-1$ solutions to $x^{p^{n}} = x$ in $F$. But $0$ is another solution, since it is not an element of $F^{ \times }$. This gives $p^{n}$ solutions, so this gives all solutions.

It remains to show that these solutions form a subfield of $F$. Since they form a finite set, it suffices to show that they are closed under addition and, since they include 0, it suffices to show they are closed under multiplication.

Closure under multiplication is immediate: $x^{p^{n}} = x$ and $y^{p^{n}} = y$ implies $$(xy)^{p^{n}} = x^{p^{n}}y^{p^{n}} = xy.$$

Closure under addition is similarly easy: Since $|F| = q$ and $q$ is a power of $p$, the field $F$ has characteristic $p$. This means that, for all $x,y \in F$, $$(x+y)^{p} = x^{p} + y^{p}.$$ Applying this $n$ times yields $(x+y)^{p^{n}} = x^{p^{n}} + y^{p^{n}}$. If $x^{p^{n}} = x$ and $y^{p^{n}} = y$, then it follows that $(x+y)^{p^{n}} = x + y$. We are done with Step 1. $\square$

**Step 2. Reducing to an asymptotic problem.**

**Claim:** There are arbitrarily large $k$ such that there is a finite field of order $q = p^{k}$.

**Proof.** If $f \in \mathbb{F}_{p}[x]$ is irreducible and of degree $k$, then $\mathbb{F}_{p}[x]/(f)$ is a field of order $p^{k}$. So we need to prove that the degrees of irreducible polynomials in $\mathbb{F}_{p}[x]$ get arbitrarily high. Since $\mathbb{F}_{p}[x]$ has only $(p-1)p^{k}$ polynomials of degree $k$, it suffices to prove that $\mathbb{F}_{p}[x]$ has infinitely many irreducible polynomials.

At this point we use Euclid’s argument: if $f_{1}, \ldots , f_{N}$ are all the irreducibles in $\mathbb{F}_{p}[x]$, then $1 + f_{1} \cdot \ldots \cdot f_{N}$ cannot be factored into irreducibles (and isn’t constant), contradicting the fact that $\mathbb{F}_{p}[x]$ is a unique factorization domain. Therefore $\mathbb{F}_{p}[x]$ has infinitely many irreducibles and we are done. $\square$

**Step 3. Counting the irreducible polynomials of degree $d$ over various $\mathbb{F}_q$. **

**Claim:** If $q$ is a power of $p$ for which there is a finite field of order $q$ (here denoted $\mathbb{F}_{q}$), then the number of irreducible monic polynomials of degree $d$ over $\mathbb{F}_{q}$ is a polynomial (call it $I_{d}(q)$) depending only on $d$, evaluated at $q$.

**Proof.** This proceeds by strong induction on $d$. The base case is $d = 1$: any linear polynomial is irreducible, and there are $q$ monic linear polynomials over $\mathbb{F}_{q}$.

Now suppose $d > 1$. There are $q^{d}$ monic degree $d$ polynomials over $\mathbb{F}_{q}$. To count how many are irreducible, we remove the ones which are reducible: If a monic polynomial is reducible, it can be written uniquely as a product of monic irreducibles of lower degree. These degrees add up to $d$, so they give a partition $\lambda$ of $d$ which can be any partition except $(d)$ itself.

Specifically, for each such partition $\lambda$ suppose $a_{i}$ is the number of parts of size $i$. Then

$$I_{d}(q) = q^{d} – \sum_{ \lambda \vdash d, \lambda \neq (d)}{ \prod_{i=1}^{d} \binom{I_{i}(q) + a_{i} – 1}{a_{i}} },$$ which is a polynomial in $q$ by the strong induction hypothesis. $\square$

Now, it suffices to show that for any given $n$, $I_n(q)>0$ for sufficiently large $q$. For, by step 2, this will give us a power of $p$ called $q$ for which $\mathbb{F}_q$ exists and such that there is a monic irreducible polynomial over $\mathbb{F}_q$ of degree $n$. By step 1 we are then done.

Since $I_{n}$ is a polynomial, this amounts to showing that $I_{n}$ has a positive leading coefficient for all $n$. This we do now:

**Step 4. Showing that the counting polynomial $I_n$ has positive leading coefficient.**

**Claim: **For all $n$, $I_{n}$ has degree $n$ and leading coefficient $1/n$.

**Proof.** Again we proceed by strong induction on $n$. When $n = 1$, we have $I_{n}(q) = q$ and we are done. So suppose $n>1$. Then as in step 3, we have

$$I_{n}(q) = q^{n} – \sum_{ \lambda \vdash n, \lambda \neq (n) }{ \Pi_{i=1}^{n} \binom{I_{i}(q) + a_{i} – 1}{a_{i}} }.$$

Since $a_{n} = 0$, we can restrict the products to only $i \lt n$. Then the degree of each of these products is $\sum_{i=1}^{n} i\cdot a_i=n$ by the induction hypothesis, and so $I_n(q)$ has degree at most $n$. Moreover, assuming by the induction hypothesis that the coefficient of $q^i$ in $I_i(q)$ is $1/i$, we have that the coefficient of $q^{i\cdot a_i}$ in $\binom{I_i(q)+a_i-1}{a_i}$ is $$\frac{1}{a_i!\cdot i^{a_i}}.$$

Finally, it follows that the coefficient of $q^n$ in $I_n(q)$ is $$1-\sum_{ \lambda \vdash n, \lambda \neq (n) }{ \Pi_{i=1}^{n}{ \frac{1}{a_{i}!i^{a_{i}}}}}.$$ Now, notice that for a given $\lambda$, the product $$z_\lambda=a_1!\cdot 1^{a_1}\cdot a_2!\cdot 2^{a_2}\cdot\cdots\cdot a_n!\cdot n^{a_n}$$ is the well-known demoninator of the formula $$n!/z_\lambda$$ for the size of the conjugacy class of elements of shape $\lambda$ in the symmetric group $S_n$. Thus we have $$\sum_{\lambda\vdash n} \frac{n!}{z_\lambda}=n!,$$ and so $$\sum_{\lambda\vdash n}\frac{1}{z_\lambda}=1.$$ It follows from this and the formula for our coefficient above that the coefficient is simply $$\sum_{\lambda=(n)}\frac{1}{z_\lambda}=1/n,$$ and the claim is proven. $\square$