This is a follow-up to this post.
Nilpotent infinitesimals allow you to define objects like the “double point”, which is the solution set of x2 = 0 on the line. Intuitively, the double point is the point x = 0, plus another point infinitesimally close to it. We can mimic this in nonstandard analysis by considering the solution set to x(x-h) = 0, where h is an infinitesimal. This has the property that if you evaluate any differentiable f at h, you get
f(h) = f(0) + f'(h) + o(h),
which also holds for nilpotent infinitesimals if you drop the o(h) term. (Here, a nonstandard number k is o(h) if k/h is infinitesimal.)
We can follow the same route to define the intersection of two double lines in the plane as the solution set to x(x-h) = 0 and y(y-k) = 0 where h and k are infinitesimals. In this case, we get a subtle difference from the nilpotent infinitesimal definition as the solution set for x2 = 0 and y2 = 0. In nonstandard analysis, xy is necessarily o(h) and o(k). If we neglect lower-order terms, we get the additional equation xy = 0 which is not satisfied by the intersection of two double lines defined using nilpotent infinitesimals. So seemingly we can’t simulate the intersection of two double lines using nonstandard analysis.
(Reasonably, you might think that maybe the nilpotent infinitesimal definition is just wrong, and that the intersection of two double lines really does morally satisfy the equation xy = 0. Just counting points, the intersection of two double lines should be some sort of quadruple point. For an ordinary set of n points, the ring of all differentiable functions from a set of four points to the real line is a vector space of dimension n. Nilpotent infinitesimals preserve this property: the ring of differentiable functions on the intersection of two double lines is a vector space of dimension 4, spanned by 1, x, y, xy. The nonstandard simulation by neglecting lower-order terms gives you a vector space of only dimension 3, spanned by 1, x, y.)