# The Arithmetic of Polynomial Rings over Finite Fields

Today we’re going to talk about something quite exciting and unexpected: the arithmetic of polynomial rings over finite fields and its similarity to the arithmetic of the integers. We’ll first run though some preliminary observations about the arithmetic of the polynomial rings which will suggest that it behaves similar to the arithmetic of the integers. Then we’ll look at analogues of the $\phi$-function as well as the classic theorems of Euler and Fermat, and conclude with some discussion of a zeta-function over the polynomial rings.

As usual, let $\mathbb{F}_{q}[T]$ be the polynomial ring in $T$ over a finite field $\mathbb{F}_{q}$. We note here that the role of positive integers in $\mathbb{F}_{q}[T]$ are played by monic polynomials, and prime integers are played by irreducible monic polynomials. In fact, we can interchange “irreducible” with “prime” because like $\mathbb{Z}$, $\mathbb{F}_{q}[T]$ is also a PID. The proof for both cases are standard facts in an introductory abstract algebra course, and are even analogous; they both deal with looking at a minimal element of an ideal, with respect to a norm, and then using the division algorithm. Both rings also happen to have infinitely many primes. The proofs here are analogous as well, if you use Euclid’s argument for infinitely many primes. Moreover, the residue class rings modulo a non-zero ideal of both rings are finite. This follows quickly by using the division algorithm to find a complete set of representatives for the residue class ring. If we restrict the ideal to be generated by a prime element, then the residue class rings become residue class fields in both cases. There is also the discussion of units. Both $\mathbb{Z}$ and $\mathbb{F}_{q}[T]$ have a finite system of units, the former being $\{\pm1\}$ and the latter being $\mathbb{F}_{q}^{*}$ by degree considerations. It is here that we see the role of $2$ being played by $q-1$ in $\mathbb{F}_{q}[T]$. These basic properties suggest that the arithmetic of $\mathbb{F}_{q}^{*}$ should be similar to that of $\mathbb{Z}$. As said above, this is the case. However, it is surprisingly easier to understand and requires the use of much less machinery than $\mathbb{Z}$. For example, there is an analogous zeta-function for $\mathbb{F}_{q}[T]$, and therefore an analogous Riemann Hypothesis. Here is where $\mathbb{F}_{q}[T]$ differs from $\mathbb{Z}$, because we have proven that the hypothesis is true for $\mathbb{F}_{q}[T]$. Before digging myself into that hole too quickly, let’s talk about some simpler analogous number theoretic statements in $\mathbb{F}_{q}[T]$.

#### The $\phi$-function, Euler, and Fermat

Before looking at analogues to the $\phi$-function and theorems of Euler and Fermat there is a small technical note to be made aware of. If $f \in \mathbb{F}_{q}[T]$, then we define $|f| = q^{deg(f)}$. By finding a full set of representatives for $A/fA$ using the division algorithm it’s easy to check that the order of $A/fA$ is $|f|$.

Let’s start with Euler’s $\phi$-function. Euler proved a product formula for $\phi(n)$, rightly named Euler’s product formula, which says

$\displaystyle{\phi(n) = n\prod_{p|n}(1-\frac{1}{p})}$,

where $p$ is a prime. The proof rests on the fact that the $\phi$-function is multiplicative and that for a prime $p$ and positive integer $k$ we have $\phi(p^{k}) = p^{k}(1-\frac{1}{p})$. The analogous function for $\mathbb{F}_{q}[T]$ is $\Phi$ where $\Phi(f)$, for $f$ non-zero, returns the order of the group $(A/fA)^{*}$ or equivalently the number of polynomials of degree less than $f$ and relatively prime to $f$. The analogous product formula is

$\displaystyle{\Phi(f) = |f|\prod_{P|f}(1-\frac{1}{|P|})}$,

where $P$ is an irreducible monic polynomial. This looks strikingly similar to the Euler product formula above, and keeping in a similar fashion as before, the proof is somewhat analogous. The Chinese remainder theorem in $\mathbb{F}_{q}[T]$ is used to show $\Phi$ is multiplicative. Yet, we need a different argument to compute $\Phi(P^{k})$. It is essentially a counting argument; the ring $A/P^{k}A$ is shown to be local, and so by counting the complement of a maximal ideal in $A/P^{k}A$ we have also counted $(A/P^{k}A)^{*}$. Like in the case of the Euler product formula, the proof follows easily from these two facts.

We also have an analogues to the famous number theory statements of Euler and Fermat. In $\mathbb{F}_{q}[T]$ Euler’s theorem takes the form

$g^{\Phi(f)} \equiv 1 \pmod{fA}$,

where $f \in \mathbb{F}_{q}[T]$, $f \ne 0$, and $g$ is relatively prime to $f$. As usual, the proof is completely analogous since the order of $(A/fA)^{*}$ is by definition $latex \Phi(f)$, and since $g$ is relatively prime to $f$ it’s an element of $(A/fA)^{*}$ (this last statement follows from the Chinese remainder theorem in $\mathbb{F}_{q}[T]$). If we take $f$ to be some irreducible monic polynomial $P$ then we obtain the analogue to Fermat’s little theorem:

$g^{|P|-1} \equiv 1 \pmod{PA}$.

Indeed, $\Phi(P)$ is the order of $(A/PA)^{*}$ which is \$latex $|P|-1$ because $A/PA$ is a field as we have discussed above.

#### The Zeta-Function $\zeta_{\mathbb{F}_{q}[T]}$

As mentioned, there is an analogue to the zeta-function in $\mathbb{Z}$. In $\mathbb{F}_{q}[T]$, the zeta-function $\zeta_{\mathbb{F}_{q}[T]}$ on complex numbers $s$ with $\mathcal{R}(s) > 1$ is defined as

$\displaystyle{\zeta_{\mathbb{F}_{q}[T]}(s) := \sum_{\substack{f \in \mathbb{F}_{q}[T] \\ f monic}}\frac{1}{|f|^{s}} = \prod_{P}(1-\frac{1}{|P|^{s}})^{-1}} = \frac{1}{1-q^{1-s}}$.

The product form of $\zeta_{\mathbb{F}_{q}[T]}(s)$ can be obtained in the exact same style of how Riemann did for the usual zeta-function by sieving out multiples of $\frac{1}{|P|^{s}}$ for all monic irreducible polynomials $P$. The closed form expression can be deduced by noticing that the summation form is an infinite geometric series by using the definition of $|f|$ and collecting terms by degree. Similar to the usual zeta-function $\zeta$, $\zeta_{\mathbb{F}_{q}[T]}$ can be extended to a meromorphic function on $\mathbb{C}$ with a simple pole at $1$. However, $\zeta_{\mathbb{F}_{q}[T]}$ is significantly more simple to work with.

That’s all for today. I hope to continue this topic in the future with a heavier discussion about $\zeta_{\mathbb{F}_{q}[T]}$, so keep an eye out. Thanks for reading!

References

Number Theory in Function Fields – Michael Rosen