# Adjoints for radical and socle functors

Let $$R$$ be a ring and $$M$$ be a $$R$$-module. Let $$rad(M)$$ be the radical of $$M$$, that is, the intersection of all maximal submodules of $$M$$. Moreover, let $$soc(M)$$ be the socle of $$M$$, that is, the sum of all simple submodules of $$M$$.

We know that both rad and soc define covariant subfunctors of $$Id:Mod_R\rightarrow Mod_R$$. Do radical and socle functors admit left or right adjoints? Thanks in advance for answers.

• Have you tried to check whether they preserve limits and/or colimits? Nov 20, 2019 at 19:16
• It seems I can use the Adjoint Functor Theorem to check existence. But, beyond that, I want to know who the adjoints are, if they exist. Nov 20, 2019 at 19:21
• I'd say that the importance of socle and radical construction lies not in the fact that they are well-behaved under co/limits, but on the fact that they are monads; but I may remember this incorrectly, are they? Nov 20, 2019 at 20:11

While it does not work for general rings, for Artin algebras one has that the left adjoint of the socle functor is the functor $$M \rightarrow M/rad(M)$$. I would think that for general rings that is the only choice in case a left adjoint exists.

• Similarly, in the case of Artin algebras, would a possible candidate for (left or right?) adjoint of radical functor be $M\rightarrow M/soc(M)$? Nov 20, 2019 at 20:34
• @ClaytonCristiano No, since the radical functor takes simple modules to zero, an adjoint would take every module to a module with no maps to (or from, depending on which adjoint) any simple module. So for nonsemisimple Artin algebras, the radical functor can't have either adjoint. Nov 21, 2019 at 10:02
• @JeremyRickard Do you know any other rings for which the socle functor has this left adjoint (or another)? Maybe one can use this to give a characterisation of certain rings (semiperfect, Artin algebras...?)
– Mare
Nov 21, 2019 at 11:30
• @Mare Semilocal rings, maybe? Nov 21, 2019 at 12:57
• The functor $soc$ is left-exact, it has a left adjoint iff it commutes with products. This condition is equivalent to both following conditions: - each simple module is endofinite (is it free over a commutative ring, for example); - there is only a finite number of iso-classes of simple modules. Nov 22, 2019 at 6:55

In abelian groups: $$\text{soc}\left(\prod_{p\text{ prime}} \mathbb{Z}/p\mathbb{Z}\right) = \bigoplus_{p\text{ prime}} \mathbb{Z}/p\mathbb{Z}\not\cong \prod_{p\text{ prime}}\mathbb{Z}/p\mathbb{Z} = \prod_{p\text{ prime}} \text{soc}(\mathbb{Z}/p\mathbb{Z})$$ so the socle functor does not preserve limits and thus does not have a left adjoint. I also doubt that the radical functor preserves infinite products, but I don't have an example off the top of my head.

Also, we have a coequalizer diagram $$\mathbb{Z}\rightrightarrows \mathbb{Z} \to \mathbb{Z}/4\mathbb{Z}$$ where the arrows on the left are the identity and the multiplication by $$4$$ map. We have $$\text{rad}(\mathbb{Z}) = \text{soc}(\mathbb{Z}) = \{0\}$$ and $$\text{rad}(\mathbb{Z}/4\mathbb{Z}) = \text{soc}(\mathbb{Z}/4\mathbb{Z}) = \{0,2\}$$. So taking radicals or socles gives $$\{0\}\rightrightarrows \{0\} \to \{0,2\}$$ which is not a coequalizer diagram. So neither functor preserves colimits, and neither has a right adjoint.