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.
 A: 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. 
A: 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.
