Let $A$ be a commutative ring with unity. If $\varphi_1 : A \rightarrow K_1$ and $\varphi_2 : A \rightarrow K_2$ are ring morphisms from $A$ to fields, $\varphi_1$ and $\varphi_2$ are said to $1$-equivalent (resp. $2$-equivalent) if (and only if, but it is a definition) there exists a ring morphism $\varphi : A \rightarrow K$ to a field $K$ and field morphisms $i_1 : K\rightarrow K_1$ and $i_2 : K\rightarrow K_2$ (resp. field morphisms $i_1 : K_1 \rightarrow K$ and $i_2 : K_2\rightarrow K$ ) such that $i_{\alpha} \circ \varphi = \varphi_{\alpha}$ (resp. such that $i_{\alpha} \circ \varphi_{\alpha} = \varphi$) for $\alpha = 1,2$. Obviously $\varphi_1$ and $\varphi_2$ are $1$-equivalent (resp. $2$-equivalent) if and only if they have the same kernel $\mathfrak{p}$ (a prime ideal of $A$), and then "everything will go through the field of fractions $\kappa(\mathfrak{p})$ of $A / \mathfrak{p}$, also residue field of $A$ at $\mathfrak{p}$. One gets $X = \textrm{Spec}(A)$ as a set, and one can easily after that get its Zariski site.

This is classic indeed. The only question remaining is : on what is the $1$ (or $2$)-equivalence defined ? On something that we could write $\coprod_{K} \text{Hom}_{{\text{Rings}}}(A,K)$. All Hom's are sets indeed, but $K$ runs through the object "class" of the category of fields, which is not a set. Which leads to my question : the rationale of the previous paragraph must be right, but how to give a proper meaning to the previous disjoint sum ?

mot-valisefor something which is not an set. What do you mean exactly with a theory of classes ? I'd rather go with the third theory, as Grothendieck's EGA's and SGA's are firmly based on it, aren't they ? (Won't ask more set-or-not-theoretic question here, as I would lead myself off-topic.) $\endgroup$2more comments