sheaves of rings and maps to classifying topos Let $R$ be the category of finitely presented commutative rings (but I don't know how necessary the hypothesis of finite presentation is for my question). Let $Set^R=Fun(R, Set)$ be the category of functors from $R$ to the category $Set$ of sets, so $Set^R$ is the category of pre sheaves on $R^{op}$. Let $X$ be a topological space, and $Sh(X)$ the category of sheaves on $X$. 
My question is, why is the data of a geometric morphism of topoi 
$$f: Sh(X) \to Set^R$$
(meaning a pair of adjoint maps $f^*:Set^R \to Sh(X)$ and $f_*:Sh(X) \to Set^R$ such that the left adjoint $f^*$ preserves finite limits) equivalent to giving a sheaf $F$ of commutative rings on $X$? For example, given such a map $f$, and an open set $U \subseteq X$ what ring is $F(U)$? What is the universal model (of a ring theory) on $Set^R$? Also, for the sheaf $\mathcal{O}_X$ of continuous real valued functions on $X$, what is the corresponding map $f$? 
(The fact that $R$ is freely generated (under colimits) by $\mathbb{Z}[x]$ might be relevant, I don't know).
 A: The classifying topos of rings is discussed in gory detail in Section VIII.5 of Mac Lane and Moerdijk, Sheaves in Geometry and Logic. 
But for a start, Diaconescu's Theorem (which is discussed in the same book, Corollary VII.9.4) identifies left exact left adjoints $\mathsf{Set}^{\mathcal{C}^\mathrm{op}} \to \mathcal{E}$ (where $\mathcal{E}$ is a Grothendieck topos) with left exact functors $\mathcal{C} \to \mathcal{E}$ which respect covers in a certain way. The correspondence is simple to state: it comes by precomposing with the Yoneda embedding $\mathcal{C} \to \mathrm{Set}^{\mathcal{C}^\mathrm{op}}$ in one direction, and taking the left Kan extension along the Yoneda embedding in the other. Personally I haven't studied how the details with coverings work out, but in the case $\mathcal{E} = \mathrm{Set}$, where you can ignore coverings, this fact is central to the theory of locally presentable categories.
Now take $\mathcal{C} = R^\mathrm{op}$ and use the fact that $R$ is generated under finite colimits by the free ring on one generator, so that a left exact functor out of $R^\mathrm{op}$ amounts to a choice of where to send $\mathbb{Z}[x]$. This isn't a complete explanation because we haven't analyzed how the covering condition restricts where we can send $\mathbb{Z}[x]$, but it's a start.
