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.

  • 3
    $\begingroup$ Have you tried to check whether they preserve limits and/or colimits? $\endgroup$ Nov 20, 2019 at 19:16
  • $\begingroup$ 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. $\endgroup$ Nov 20, 2019 at 19:21
  • $\begingroup$ 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? $\endgroup$
    – fosco
    Nov 20, 2019 at 20:11

2 Answers 2


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.

  • $\begingroup$ Similarly, in the case of Artin algebras, would a possible candidate for (left or right?) adjoint of radical functor be $M\rightarrow M/soc(M)$? $\endgroup$ Nov 20, 2019 at 20:34
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    $\begingroup$ @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. $\endgroup$ Nov 21, 2019 at 10:02
  • $\begingroup$ @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...?) $\endgroup$
    – Mare
    Nov 21, 2019 at 11:30
  • $\begingroup$ @Mare Semilocal rings, maybe? $\endgroup$ Nov 21, 2019 at 12:57
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    $\begingroup$ 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. $\endgroup$ 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.


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