3

I managed to show $\forall$ R-Module M, $\exists$ unique semi-simple submodule $sM \subset M$ containing every semi-simple submodule of M, by showing that the direct sum of semi-simple submodules are also semi-simple modules and using zorns lemma: $\exists$ a maximal semi-simple submodule, and then to prove uniqueness I allowed two maximal semi-simple submodules and then arrived to a contradiction.

Now I am having difficulties showing the following:

$f:M \longrightarrow N$ where $M$ and $N$ are modules and $f$ is a homomorphism, I need to show that $sM$ is sent into $sN$ by $f$ and denoting $s(f):sM \longrightarrow sN$ & letting $g:N \longrightarrow P$ be an $R$-module then how would i show $s(g \circ f) = s(g) \circ s(f)$

Also I've been told that: $N \subset M$ then $sM = sN \cap M$ , but why is this actually true? Is there some sort of short proof showing this ?

Thanks.

user26857
  • 53,190
Piri
  • 470
  • 3
  • 10
  • See my answer here: http://math.stackexchange.com/questions/1031258/socle-of-submodule-relative-to-the-module/1031288#1031288 for the question in your final sentence. – Jim Nov 20 '14 at 19:31

1 Answers1

1

To see that $f(sM) \subseteq sN$ note that you already know that $sM$ is the sum of the simple submodules of $M$. If $S \subseteq M$ is simple show that $f(S)$ is either simple or zero. Either way you get $f(S) \subseteq sN$. As $sM$ is generated by the simple $S$ you get that $f(sM)$ is generated by the $f(S)$. Hence $f(sM) \subseteq sN$.

Now $s(f)\colon sM \to sN$ is just restriction of $f$ and it doesn't matter if you do restriction before or after composition, hence $s(g\circ f) = s(g)\circ s(f)$.

Jim
  • 31,547