Let $A$ be an associative algebra over a field $k$ and let $M$ be some $A$-module. If $\mathfrak{a}$ is an ideal of $A$ which annihilates $M$, i.e. if $\mathfrak{a}M=0$, then why must every composition factor of $M$ be a composition factor of $A/\mathfrak{a}$?
Composition factors and annihilation
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abstract-algebra
modules
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0I think most of the content will be contained in the fact that every $A$-submodule of $M$ is "actually" an $A/\mathfrak{a}$-submodule. – 2011-05-18
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0ANSWER: Let $E$ be a composition factor of $M$. Then $E$ is an irreducible $A$-submodule of $M$, and since $\mathfrak{a}M=0$ we can consider $E$ as an irreducible $A/\mathfrak{a}$-module. Since every irreducible $A/\mathfrak{a}$-module must occur as a composition factor in its regular representation, we are done. – 2011-05-18
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0Sounds good to me. – 2011-05-18
1 Answers
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Let E be a composition factor of M. Then E is an irreducible A-submodule of M, and since aM=0 we can consider E as an irreducible A/a-module. Since every irreducible A/a-module must occur as a composition factor in its regular representation, we are done.