I am learning number theory, specifically ramification indices, and I was looking at the example of what primes in $\Bbb{Z}$ ramify in $\Bbb{Z}[i]$. Of course, the only one to do so is $2$, because it is the only prime $p$ such that $x^2+1$ can be written as $f(x)^2$ modulo $p$.
From a scheme-theoretic point of view, we have a map $Spec(\Bbb{Z}[i])\to Spec(\Bbb{Z})$, and we can look at the fiber of each prime $p\in\Bbb{Z}$. From this point of view, the fiber over $2$ turns out to be $Spec(\Bbb{Z}/2[x]/(x+1)^2)$, and in this sense it is the only prime whose fiber is a single point where the ring of global functions has nilpotents (i.e. the point has "fuzz" around it as Vakil likes to say).
I am wondering what the connection is here? What is the intuition for $2$ being the only prime which both ramifies and has "fuzz"?
What follows is more intuitive than precise.
If you have a map $f: C \to D$ of curves (say) (in your example $C = \text{Spec}\,\mathbb{Z}[i]$, $D = \text{Spec}\,\mathbb{Z}$), and $f(P) = Q$ (these are points on those curves, for example $P = (1 + i)$, $Q = (2)$; in your case all points are nonsingular), then $f$ induces a linear map of the tangent space at $P$ to $C$ to the tangent space to $Q$ at $D$. Both these tangent spaces are $1$-dimensional (because the points are nonsingular) so the linear map is either zero or an isomorphism.
(What I am saying is not very precise, as the residue fields can also change$\ldots$ but geometrically it gives the right intuition.)
In "decent" cases, the former behavior (the linear map is zero) is exceptional, and then $P$ is a ramification point of the map. When one defines all these things algebraically, then nilpotents are not the only culprit; inseparable residue field extensions also give ramification.