# Counting couples of square-free polynomials over finite fields

I have a curve defined by the following equations over the finite field $\mathbb{F}_q$ with $q=p^r$ with $p \geq 3$: $$C_{h_1,h_2}:\begin{cases} y_1^2=h_1(t) & \\ y_2^2=h_2(t) & \\ \end{cases}$$ with $h_1,h_2$ monic, square-free polynomials in $\mathbb{F}_q[t]$, and $h_1\neq h_2$. This is a so-called biquadratic curve, because its function field is a biquadratic extension of $\mathbb{F}_q(t)$, that we can write as $\mathbb{F}_q(t)(\sqrt{h_1(t)},\sqrt{h_2(t)})$.

If I want to count these curves, I use the formula for counting square-free monic polynomials over finite fields, that is $$\# \lbrace \mbox{square-free monic polynomials of degree n in } \mathbb{F}_q \rbrace= (q-1)q^{n-1}$$ (this is Exercise 3 at page 20 of Rosen's "Number Theory in Function Fields", that uses the zeta function associated to $\mathbb{F}_q[t]$)

So it's easy to see that $$\# \{ C_{h_1,h_2} | \deg h_1=n_1, \deg h_2 = n_2 \}=\frac{(q-1)^2q^{n_1+n_2-2}}{1+\delta_{n_1n_2}} - \delta_{n_1n_2}\frac{(q-1)q^{n_1-1}}{2}$$ where $\delta_{n_1n_2}$ is the Kronecker delta.

My question is:

Is there a formula to compute the following cardinality? $$\# \{ C_{h_1,h_2} | \deg h_1=n_1, \deg h_2 = n_2, \deg f = n \}$$ where $f=(h_1,h_2)$, that is the greatest common divisor of $h_1$ and $h_2$.

• Can you do this by: first fixing $f$ using the formula from Rosen, then picking a pair $g_1 = f/h_1$, $g_2 = f/h_2$ squarefree, of degree $n_1 - n,\, n_2 - n$ respectively? I don't know if there will be trouble counting square-free polynomials relatively prime to a certain given polynomial. – John Binder Feb 20 '15 at 15:05
• You mean $g_1=h_1/f$, $g_2=h_2/f$, I guess. Thank you for your comment. I'll try and I'll let you know. – Miles Feb 20 '15 at 19:16

## 1 Answer

Let $\beta(k) = \frac 1k\sum_{d|k}\mu(d)q^{k/d}$, the number of irreducible monic polynomials of degree $k$ over $\mathbb{F}_q$. Let $f(n_1,n_2,n)$ be the number of pairs $(f,g)$ of squarefree monic polynomials over $\mathbb{F}_q$ such that $\deg f=n_1$, $\deg g=n_2$, and $\deg \mathrm{gcd}(f,g)=n$. Then $f(n_1,n_2,n)$ is the coefficient of $x^ny^{n_1-n}z^{n_2-n}$ in $$\prod_{k\geq 1}( 1+x^k+y^k+z^k)^{\beta(k)}.$$ This gives a way to compute $f(n_1,n_2,n)$ (as a polynomial in $q$) for small $n_1,n_2,n$, but I don't see how to obtain a nice'' formula.

Incidentally, if you want to drop the squarefree condition, then replace the above product by $$\prod_{k\geq 1}\left( 1+\frac{x^k}{1-x^k}+ \frac{y^k}{1-y^k}+\frac{z^k}{1-z^k}\right)^{\beta(k)}.$$