# (Affine) Schemes and the point of view of morphisms with values in a field

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 ?

• It depends on the set-theoretic framework you place yourself. Within ZFC, you have no other option than fixing a set of fields; within a theory with classes, the morphisms from $A$ to a field will constitute a class; within Bourbaki's theory, there is a notion of an equivalence relation without mentioning an underlying set, but you will need to prove (this is easy) that this relation is "collectivisante" so that its equivalence classes form a set.
– ACL
Dec 17 '14 at 10:52
• I was using the word "class" as a mot-valise for 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.) Dec 17 '14 at 12:27
• I think that the best way to describe this point of view is to describe the set of points as the colimit of the functor $Hom(A,-)$ over the category of fields. In this way you can talk about the colimit even if the source category is not small (but you need to prove it exists!) Dec 17 '14 at 13:55
• @ACL : and - I forgot - thx for your comment. Dec 17 '14 at 14:05
• @DenisNardin Yes, I think that this is the best way also, thx. Dec 17 '14 at 14:06