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Let $M$ be an $A$-module. How do I show that Soc$(M)$ is the intersection $Q$ of all essential submodules of $M$? One direction is easy enough (Soc$(M)\subset Q$), but I can't seem to show the other direction. Only if I could show that every submodule of $Q$ is a direct summand of $Q$... Does anyone have any idea?

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Is your ring arbitrary? – Mariano Suárez-Alvarez Sep 5 '11 at 22:01
I think it is true for non-commutative rings with unit, but I would be happy to prove it for commutative rings. – ashpool Sep 5 '11 at 22:04
up vote 2 down vote accepted

This is e.g. proven in Proposition 7.19 of the representation theory lecture notes of Ringel and Schröer (here essential submodules are called large submodules). Here is a link

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A hint version might be to prove a module is semi-simple iff it has no proper essential submodules. (Again the direction you've already done is the easy one since direct summands are not usually essential, and the other direction looks at maximal complements as in the notes). – Jack Schmidt Sep 5 '11 at 22:41
Thanks! I also found a nice proof in Rings and Categories of modules, Frank Anderson & Kent Fuller. – ashpool Sep 5 '11 at 22:53

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