# Looking for a pair of adjoint functors

I'm struggling with finding the left adjoint to a functor, and the right adjoint to another one. Here's some context.

Given any ring $$R$$, we can associate two categories to it. The first one is a POset, which I will denote by Hom(R): its elements are ordered pairs, $$(a,M)$$, where $$a=\ker\phi$$ and $$M=\phi^{-1}(U(S))$$, for some given ring morphism $$\phi$$ of $$R$$ into an arbitrary ring $$S$$ ($$U(S)$$ denotes the group of invertible elements of the ring $$S$$); the partial order on it is defined as follows: $$(a,M)\leq(a',M')\quad\iff\quad a\subseteq a', M\subseteq M'.$$

The second one, which I will denote by R-Rings, is the one whose objects are the ring morphisms from $$R$$ to another ring $$S$$, $$\phi\colon R \to S$$, and such that, given any pair of objects, $$\phi\colon R\to S$$ and $$\phi'\colon R\to S'$$, the morphism between them are exactly the ring morphism $$f\colon S\to S'$$ that form commutative triangles with the given morphisms, i.e. that make the following diagram commute

Given any pair $$(a,M)$$ defined as above, it is possible to prove that there is a canonical ring morphism that "realizes" that pair, which I will denote by $$\psi\colon R\to S_{(R/a, M/a)}$$; besides, this morphism is universal in this sense: given any ring morphism $$\phi'\colon R\to S'$$ such that $$(a,M)\leq(ker(\phi'),\phi'^{-1}(U(S))$$, there exists one and only one morphism $$g\colon S_{(R/a, M/a)}\to S'$$ such that $$g\psi=\phi'$$. Given these facts, we can observe that the functor $$A\colon Hom(R)\to R-Rings$$ which maps an ordered pair $$(a,M)$$ to the canonical morphism that realizes it, and the functor $$B\colon R-Rings\to Hom(R)$$ which maps a morphism $$\phi\colon R\to S$$ to the pair $$(\ker(\phi),\phi^{-1}(U(S)))$$, form a pair of adjoint functors (A is left-adjoint to B and B is right adjoint to A), since for every $$(a,M)\in Hom(R)$$ and every $$\phi\colon R\to S \in R-Rings$$, we have $$Hom_{R-Rings}(A(a,M),\phi\colon R\to S)\cong Hom_{Hom(R)}((a,M),B(\phi\colon R\to S)).$$ I've been trying to find a left adjoint to $$A$$ and a right adjoint to $$B$$, but I couldn't figure them out. I'll be thankful to anybody who'd like to give me some advice.

• Why the functor $A$ would have a left adjoint? – Fabio Lucchini Oct 14 '18 at 20:45
• I was asked to prove that A is left-adjoint to B and B is right adjoint to A (and this was quite easy, as shown above); moreover, I was asked to find a left adjoint to A (and a right adjoint to B), but I have no guarantee they exist. – Vladimir Oct 15 '18 at 8:17