Singularities of Curves in Positive Characteristic Given a collection of polynomials $\mathscr{F}\subset\mathbb{Z}[x_1,\ldots,x_n]$, we can associate to each prime ideal of $\mathbb{Z}$ an affine variety as follows:
$$
(p)\longmapsto Z(\mathscr{F})\subset\mathbb{A}^n_{k}
$$
where $k=\overline{\mathbb{F}_p}$ (in the case of $p=0$, define $\mathbb{F}_0:=\mathbb{Q}$).  Under this association, which prime ideals $(p)$ correspond to singular varieties?
This question is motivated by exercise I$.5.1$ in Hartshorne.  Part $(c)$ of the exercise asks us to find the singular points of the curve $X$ defined by $x^3=y^2+x^4+y^4$ in $\mathbb{A}^2$.  Having done this exercise I've found that if char $k=0$, then $(0,0)$ is the only singular point of $X$, whereas for characteristics $2$,$7$, and $13$, the curve has other singular points (you can check my result if you'd like; the exercise excludes characteristic $2$, but I've included this case anyway).
Finding the extra singular points and their characteristics involved a fair deal of computing Jacobians and solving simultaneous equations.  Is there another way to find the characteristics which give extra singular points?
I guess this is really a question about the existence of solutions to a system of polynomial equations mod $p$.  In the case of a plane curve, defined by $f(x,y)=0$, we want to know in which characteristics the system of equations
$$
f(x,y)=0
$$
$$
\partial_xf(x,y)=0
$$
$$
\partial_yf(x,y)=0
$$
has a solution.  Given $p$, is there any quick way to tell that this system is consistent?
 A: As you say, the question is equivalent to finding those primes for which the polynomials $f$, $\partial f/\partial x$ and $\partial f/\partial y$ have a common root. More generally, consider three polynomials $f(x,y)$, $g(x,y)$ and $h(x,y)$ with integer coefficients. We can ask for which primes $p$ they have a common root.
It is more common to ask this question for homogenous polynomials $f(x,y,z)$, $g(x,y,z)$ and $h(x,y,z)$, asking when the corresponding equations have a common root in $\mathbb{P}^2$. This is because of the issue I point out in my comment: $x(x-py-1)=0$ is singular for every prime other than $p$, but not for $p$. If you work in projective space, then the set of primes for which there exists a common root will always either be all primes or a finite set. (In this example, the homogenous equation $x(x-py-z)=0$ is singular for all $p$.)
The concept you want here is the multivariate resultant: A polynomial in the coefficients of $f(x,y,z)$, $g(x,y,z)$ and $h(x,y,z)$ which vanishes if and only if $f$, $g$ and $h$ have a common root in $\mathbb{P}^2$.  
If you compute the multivariate resultant of $f(x,y,z)$, $f_x(x,y,z)$ and $f_y(x,y,z)$, you'll get an integer $R$. The curves $f$, $f_x$ and $f_y$ in $\mathbb{P}^2$ will have a common root for the primes which divide $R$. (Because $R=0$ in characteristic $p$ if $p$ divides $R$.) This isn't quite the same as asking when the original polynomial $f(x,y)=0$ is singular, but it is pretty close.
When I started writing this answer, I knew that what you wanted was the multivariate resultant, but I didn't know how to compute it. After digging around online, I think that is because the problem is hard! Here are a few links I fond that might help: Chapter 4 in Sturmfels's book Solving Systems of Polynomial Equations, MO question, D'Andrea and Dickenstein.
Dealing with varieties that aren't hypersurfaces are going to be harder.
