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? – Alex Kruckman Nov 20 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. – cl4y70n____ Nov 20 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? – Fosco Nov 20 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 $M\rightarrow M/soc(M)$? – cl4y70n____ Nov 20 at 20:34
• 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. – Aurélien Djament Nov 22 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.