Sheafyness and relative chinese remainder theorem The relative chinese remainder theorem says that for any ring $R$ with two ideals $I,J$ we have an iso $R/(I\cap J)\cong R/I\times_{R/(I+J)}R/J$.
Let's take $R=\Bbbk [x_1,\dots ,x_n]$ for $\Bbbk $ algebraically closed. If $I,J$ are radical, the standard dictionary tells us $R(I\cap J)$ is the coordinate ring of the variety $\mathbf V(I)\cup \mathbf V(J)$. Furthermore, if $I+J$ is radical then $R/(I+J)$ is the coordinate ring of the intersection $\mathbf V(I)\cap\mathbf V(J)$. Now the elements in the pullback are just pairs of functions which are consistent on the intersection, and the isomorphism tells us we can glue them to get a function defined on the union.
This has a very sheafy feel to it, yet I find the need for $I,J$ to be radical somewhat disconcerting. I don't know any scheme theory, and I'm not sure exactly how to phrase my question except:

What's the underlying sheaf here and in what context is it most
  natural?

I guess what I'm hoping for is a setting in which every ideal of a ring has some geometric analog, not just radicals.
 A: I don't know if this is what you want, but here are some related facts :
Let $F$ be any sheaf (of sets) on a topological space. If $Z$ is a closed subset, denote by $F_Z$ the sheaf $F_Z=i_*i^{-1}F$ where $i:Z\rightarrow X$ is the inclusion. Note that if $Y\subset Z$, there is an restriction mapping $F_Z\rightarrow F_Y$. Now let $Y,Z$ be two subset of $X$, then there is a cartesian square of sheaves on $X$ :
$$\require{AMScd}\begin{CD}F_{Y\cup Z}@>>>F_Y\\
@VVV@VVV\\
F_Z@>>> F_{Y\cap Z}
\end{CD}$$
(To prove this is indeed cartesian, just check stalks). In an abelian context, this yields a short exact sequence (giving rise to a useful long exact sequence in cohomology) :
$$0\longrightarrow F_{Y\cup Z}\longrightarrow F_Y\oplus F_Z\longrightarrow F_{Y\cap Z}\longrightarrow 0.$$
The Chinese Remainder Theorem says that the same is true for schemes. More precisely, let $X$ be any scheme, $Y,Z$ be two closed subschemes defined by ideals $\mathcal{I,J}$. Let $Y\cup Z$ be the closed subscheme defined by $\mathcal{I\cap J}$ and $Y\cap Z$ be the closed subscheme defined by $\mathcal{I+J}$. Then the square
$$\require{AMScd}\begin{CD}\mathcal{O}_{Y\cup Z}@>>>\mathcal{O}_Y\\
@VVV@VVV\\
\mathcal{O}_Z@>>> \mathcal{O}_{Y\cap Z}
\end{CD}$$
is cartesian.
A: I haven't read all of your questions (and answers) on this topic, so I hope I'm not being redundant, but you might take a look at this article by Ernst Kleinert.
