in these notes i am reading i am told that the socle of $K$ (where $K \subset M$ , and $M$ is a module) is = $K \cap$ Soc $ M$
But why is this? i see the intuition but cannot formalize a proof
any help would be great thanks
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1What's the theorem that this is a corollary of? – Avi Steiner Nov 20 '14 at 19:11
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if f is the R homomorphism $f:M \longrightarrow N$ then f(soc) $\leq$ Soc N @AviSteiner – Peter A Nov 20 '14 at 19:15
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There are two ways to see this.
First, one characterization of the socle of a module $N$ is that it is the sum of the simple submodules of $N$. The simple submodules of $K$ are exactly the simple submodules of $M$ that lie in $K$.
Second, using the theorem you stated in the comments let $f\colon K \to M$ be the inclusion map. Then that theorem gives $\mathrm{soc} \ K \subseteq \mathrm{soc} \ M$ and clearly it's in $K$ so $\mathrm{soc} \ K \subseteq K \cap \mathrm{soc} \ M$. But $K \cap \mathrm{soc} \ M$ is semisimple and a submodule of $K$ so $K \cap \mathrm{soc} \ M \subseteq \mathrm{soc} \ K$. Hence they are equal.
Jim
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Also, just out of interest is there any way i could use this fact to show that given $ 0 \longrightarrow M \longrightarrow N \longrightarrow P$ is exact, then $0 \longrightarrow soc(M) \longrightarrow soc(N) \longrightarrow soc(P)$ is exact ? – Peter A Nov 20 '14 at 19:50
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Yes, this is how you show exactness at $\mathrm{soc} \ N$. Since the maps are restrictions the kernel of $\mathrm{soc} \ N \to \mathrm{soc} \ P$ is $\ker(N \to P) \cap \mathrm{soc} \ N$. – Jim Nov 20 '14 at 23:00