$\def\A{\mathbb{A}} \def\spec{\mbox{Spec}} \def\R{\mathbb{R}} \def\Q{\mathbb{Q}} \def\Z{\mathbb{Z}}$
Here is an example of the sort of thing that I'm hoping to learn about from this math.SE question. This is copied directly from an email that I sent a while back, hence the tone.
subject:
TIL (today I learned)...
body:
...an incredibly neat way of computing the algebraic de Rham cohomology of certain affine spaces $\A^n_R=\spec(R[x_1,...,x_n])$ (and their formal analogs, if you know what that means -- the story is almost exactly the same). This beautifully illustrates the interplay of differential geometry and algebraic geometry: on the one hand, algebraic geometry imports ideas from differential geometry all the time (although occasionally the reverse is true!), but then the much more general algebraic setting affords new and exciting techniques and results. Of course, the analogies extend much deeper: in the algebro-geometric setting one can define vector fields, connections, curvature, integrability, etc.; you name it, Grothendieck's probably already done it.
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Definition
Recall that the sheaf $\Omega^1$ of Kähler differentials of $\A^n_R$ (relative to $R$ -- this essentially means that we're treating $R$ as the "constants") is supposed to be the "sheaf of 1-forms on $\A^n_R$". As a quasicoherent sheaf, this is determined by an $R[x_1,...,x_n]$-module, namely the free module generated by the symbols $\{dx_i\}$. More generally, the sheaf of $n$-forms on $\A^n_R$ is just defined to be the $n^{th}$ alternating power $\Lambda^n(\Omega^1)$ (as an $R[x_1,...,x_n]$-module), as it should be; in particular, $\Omega^0=\Lambda^0(\Omega^1)=R[x_1,...,x_n]$, the functions on $\A^n_R$. Of course. These all fit together into a chain complex
$$ 0\rightarrow \Omega^0\rightarrow \Omega^1\rightarrow \Omega^2\rightarrow \cdots, $$
where the differential is given by the usual formula
$$d(f \cdot dx_{i_1} \wedge ... \wedge dx_{i_k}) = \sum_j \frac{\partial f}{\partial x_j} \cdot dx_j \wedge dx_{i_1} \wedge ... \wedge dx_{i_k},$$
and the cohomology of this complex is by definition the de Rham cohomology of our scheme. Excellent.
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Technique
Now, suppose that $R$ is torsion-free, and let $K=R\otimes \Q$. Recall the Poincare lemma, a standard differential-geometric fact, which says that $\R^n$ has the de Rham cohomology of a point. (This is what makes all of differential de Rham cohomology tick.) You will note from the wikipedia link that all that's needed to define the necessary chain-homotopy is a notion of integral. But actually this is entirely formal, and in our setting where all our functions are polynomials, we merely need an appropriate notion of integration of polynomials: this is precisely what tensoring with $\Q$ buys us. That is to say,
$$ \begin{array}{ll}
H^0(\A^n_K)=K & \mbox{ (given by the constant functions),} \\
H^i(A^n_K)=0 & \mbox{ for }i>0. \end{array} $$
Now, here we go. I'll stick to 1-dimensional cohomology for simplicity. Suppose that $\omega \in \Omega^1_{\A^n_R}$ is a closed 1-form. This means that $d\omega = 0 \in \Omega^2_{\A^n_R}$. However, we can also consider $\omega \in \Omega^1_{\A^n_K}$; it's still closed of course, but $\A^n_K$ has no 1-dimensional cohomology, so every closed form is exact, so $\omega=df$ for some $f \in \Omega^0_{\A^n_K}=K[x_1,...,x_n]$, and f is unique up to adding a constant. Thus we have a canonical isomorphism
$$H^1(\A^n_R) = \frac{\{ f \in K[x_1,...,x_n] : df \in \Omega^1_{\A^n_R}\mbox{ and }f(0)=0\}}{\{f \in R[x_1,...,x_n] : f(0)=0\}}.$$
This is fantastic!
To illustrate, let's start with the simplest case $R=\Z$, $n=1$. We're looking for polynomials $f(x) \in \Q[x]$ such that $f(0)=0$ and $df$ is an integral 1-form, and we're hoping that $f(x)$ itself isn't integral. So, write $f(x)=\sum a_k x^k$. Then $df = \sum k\cdot a_k x^{k-1} dx$. So we're demanding that $k\cdot a_k \in \Z$, but allowing that $a_k \notin \Z$. Of course, we can choose these $a_k$ distinctly since everything in sight is linear. Moreover, any choice $a_k \in \Z$ is considered trivial. Thus, the first de Rham cohomology group of the affine line over $\Z$ is
$$ H^1(\A^1_\Z) = \Z/1\Z \oplus \Z/2\Z \oplus \Z/3\Z \oplus \Z/4\Z \oplus \Z/5\Z \oplus \ldots !$$
Now, if you extend to multivariable functions and write $f(x) = \sum_{J} a_J x^J$ (using multiindex notation), then you'll get that each $a_J \in \Q$ must have denominator a divisor of all the $j_1,\ldots,j_n$, so $a_J$ can be chosen precisely from $\Z[1/\mbox{gcd}(j_k)]$, and hence
$$H^1(\A^n_\Z) = \bigoplus_J \Z/\mbox{gcd}(j_k)\Z,$$
where $J=(j_1,...,j_n)$ runs over all length-$n$ multiindices (and every number divides $0$, of course).
...all of which is the cutest thing I've seen all week. Maybe I need to get out more.