Upper bound on greatest prime of bad reduction for a plane curve Background
We are given a curve with integer coefficients. I want to make a suggestion in another question (Computationally bounding a curve's genus from below?) into a deterministic algorithm for finding the genus of a plane curve.
The suggestion is: reduce modulo a random prime and find all singular points there. If the prime was of good reduction, then these are the reductions of all of the algebraic singular points, and you can compute the genus easily from here.
Question
What is an effective bound on the largest prime of bad reduction?
What I imagine
Say $C$ is given by $\sum a_{ij} x^i y^j$, then I imagine a bound similar to: $\displaystyle\sum_{\sigma} \prod a_{i\sigma^{-1} (i)}$
 A: @Dror: I will address your last comment in the question about bounding the genus from below.
Here I wanted to add a remark: the explicit bound you found is such that any prime larger than this bound will be of "good" reduction.  Unless the bound you compute is at most two, you are guaranteed the existence of smaller primes not dividing the resultant you computed.  Since I assume from your question that you simply want to find a prime of "good" reduction, it is probably much more efficient to compute explicitly the resultant (over Z) and then look for primes not dividing it.  In particular, it seems like you might necessarily have one such prime of the order not bigger than $d^3M$ (and possibly much smaller than this).
Finally, why did you include the computation of the gcd of the resultant in $\mathbb{F}_p$?
A: The primes that are "bad" in your sense will divide the number $Res_x(Res_y(f,\frac{\partial f}{\partial x}), Res_y(f,\frac{\partial f}{\partial y}))$. (If I interpreted damiano's comment correctly).
All that is left is to bound this number. So:
Let $M := max (|a_{ij}|)$.
$\parallel Res_y(f,\frac{\partial f}{\partial x})\parallel and \parallel Res_y(f,\frac{\partial f}{\partial y})\parallel are < (2d)!M^{2d}$
$\Rightarrow \parallel Res_x(Res_y(f,\frac{\partial f}{\partial x}), Res_y(f,\frac{\partial f}{\partial y}))\parallel < (2d^2)^{2d^2}((2d)!M^{2d})^{2d^2} \ll (dM)^{4d^3+O(d^2)}$ 
So pick a random prime larger than this and then compute $gcd(Res_y(f,\frac{\partial f}{\partial x}), Res_y(f,\frac{\partial f}{\partial y}))$ in $\mathbb{F}_p$. The complexity is $O(poly(d)\times poly(\log(M))$. Is this better than Groebner computations in $\mathbb{Q}$? I have no idea...
