# Whether hom functor and socle (radical) are commutative?

Let $A$ be an algebra. We know that $Soc M =\oplus _A Soc M_{\alpha}$ and $Rad M =\oplus _A Rad M_{\alpha}$ if $M= \oplus _A M_{\alpha}$ as $A$-modules.

Now let $M,N$ be $A$-modules, $\Lambda:=End_A(N)$, whether $Soc_{\Lambda}(Hom_A(N,M)) =Hom_A(N,Soc_A(M))$ and $Rad_{\Lambda}(Hom_A(N,M)) =Hom_A(N, Rad_A(M))$ hold(here $Soc_A$ means the socle functor acts on $A$-modules, so does $Rad_A$ )?

## 1 Answer

Take a symmetric Nakayama algebra with Kupisch series [3,3]. (see for example the book by assem simson skowronski in the chapter about nakayama algebras for the meaning) Then $Hom(e_0 A, soc(e_1A))=Hom(e_0 A, S_1)=0$, but $Hom(e_0A , e_1A)$ is nonzero and thus has nonzero socle.

Next try: Let $A=K[x]/(x^2)$ with simple modules $S$ and $N=A \oplus S$ and $M=S$. Then $Hom(N,rad(M))=0$. But $Hom(N,M)$ is a projective indecomposable $End(N)$-module which is not simple and thus has nonzero radical.