If $K$ is a finite field of size $q$ and $f$ is a degree $n$ polynomial in $K[x]$, then we can form the quotient field by modding out this polynomial. Elements of this quotient field are of the form $\phi + (f)$ where $f$ is the ideal generated by $f$. I've been trying to figure out what the order of $x^2 + (f)$ within the multiplicative group of this field, but I am struggling with how to incorporate the coefficients of this arbitrary polynomial.
I would like to understand the case where $f$ is any polynomial, but I'm struggling to even do the simpler case of when $f$ is monic and irreducible.
Best Answer
First of all, being monic is a trivial restriction; because your polynomial $f = a_n x^n + \cdots + a_1 x + a_0$ has coefficients in your field $K$, adjoining a root of $f$ to $K$ is the same as adjoining a root of the monic polynomial $a_0^{-1} f$. In fancier language, $a_0^{-1} f$ and $f$ are associates in the ring $K[x]$ and so generate the same principal ideal.
The quotient $K[x]/(f)$ is a field only in the simpler case where the polynomial is irreducible, so it does make sense to focus on that case. Otherwise the set of nonzero elements doesn't form a group (consider the ring $\mathbb{F}_2[x]/(x^2)$).
Assuming $f$ is irreducible, we know that the order of $x^2 + (f)$ will divide $q^{\deg(f)} - 1$, but what it is varies depending what $f$ is.
For example, both the polynomials $f = x^2 + 1$ and $g = x^2 + x + 2$ are irreducible over $K = \mathbb{F}_3$, so $K[x]/(f) \cong K[x]/(g) \cong \mathbb{F}_9$, but the multiplicative order of $x + (f)$ is 4 in the former and 8 in the latter, so $x^2 + (f)$ will have respective orders 2 and 4 in the former and the latter.