Why is $A/\mathrm{ann}(M)$ necessarily Artinian?

Asked Sep 22 '15 at 03:54

Active Sep 23 '15 at 02:44

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Let $M$ be a finitely generated Artinian module over a commutative ring $A$. Why is $A/\mathrm{ann}(M)$ Artinian?

edited Sep 22 '15 at 20:59

user26857

asked Sep 22 '15 at 03:54

D_S

This is probably another duplicate: http://math.stackexchange.com/q/378563/29335 – rschwieb Sep 22 '15 at 04:32
2

That link explains that if M is generated by n elements, the R embeds into $M^n$, which is Artinian. Since a submodule of an Artinian module is Artinian, R is Artinian. – rschwieb Sep 22 '15 at 04:34

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