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

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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.

  • $\begingroup$ Why the functor $A $ would have a left adjoint? $\endgroup$ – Fabio Lucchini Oct 14 '18 at 20:45
  • $\begingroup$ 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. $\endgroup$ – Vladimir Oct 15 '18 at 8:17

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