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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"?

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    I'm a bit confused about what your question is. Since $(2) = (1+i)^2$ in $\mathbb{Z}[i]$, then this ramification can be detected in the quotient: $\mathbb{Z}[i]/(2) = \mathbb{Z}[i]/(1+i)^2 \cong \frac{(\mathbb{Z}/2\mathbb{Z})[x]}{(1+x)^2}$. This holds more generally: if $(p) = \mathfrak{p}_1^{e_1} \cdots \mathfrak{p}_g^{e_g}$ and $p$ is ramified, i.e., $e_i \geq 2$ for some $i$, then we again get ramification in the quotient. Is this what you are looking for, or something else? – 2017-01-30
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    @SpamIAm I guess what I am looking for is some intuitive description of what ramification means geometrically, using this particular example. That is, I'd like to have a better intuitive picture of what the map of schemes I described "looks like", and if there's any geometric reason that $2$ is the only ramified prime in this case. – 2017-01-30
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    From my viewpoint this just says that the discriminant of the binary quadratic form $f(x,y) = x^2 + y^2$ has discriminant $-4.$ For quadratic fields, the primes that ramify must divide the discriminant. In case you can stand looking at (part of) the relationship of quadratic fields and (binary) quadratic forms, http://math.blogoverflow.com/2014/08/23/binary-quadratic-forms-over-the-rational-integers-and-class-numbers-of-quadratic-%EF%AC%81elds/ – 2017-01-30
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    Well, to see something by a person who knows both worlds well, Franz Lemmermeyer has at least one book. It was in process last I checked, http://www.rzuser.uni-heidelberg.de/~hb3/publ/bf.pdf also https://arxiv.org/abs/1108.5687 – 2017-01-30
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    @AlexMathers Crap, that was supposed to be "*nilpotents* in the quotient." I suspect there's an analogy to be made between completions of the local rings: power series and $p$-adic expansions both encode ramification data. But maybe this is not what you're looking for either. Also, [obligatory pictures](http://pbelmans.ncag.info/blog/atlas/) from Mumford's Red Book. – 2017-01-30
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    @SpamIAm No worries, it was clear what you meant. Thanks for that link. – 2017-01-31
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    @WillJagy I'm definitely going to spend some time with that book, it looks nice. Thanks for the response – 2017-01-31

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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.