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May I refer you to page $32$ of:

http://www.staff.city.ac.uk/a.g.cox/LTCC/Week3.pdf

Part (b) of proposition $3.2.4$. Why the fact that $A/\operatorname{rad}(A)$ is isomorphic to $k^m$ for some $m$ implies that the dimension of any simple $A$-module is $1$?

Julian Kuelshammer
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user10
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1 Answers1

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There are two claims:

  1. Every simple $A$-module is a simple $A/\mathrm{rad}A$-module.
  2. Every simple $k^m$-module is $1$-dimensional.

First note that if $I \subseteq A$ is an ideal of any ring $A$ and $M$ is an $A$-module such that $IM = 0$, then $M$ is an $A/I$-module. Moreover, we haven't changed the submodules of $M$, i.e., any $N \subseteq M$ is a submodule for the $A$-action on $M$ if and only if it is a submodule for the $A/I$-action on $M$. (If you don't already know this you should stop and figure out why this is true, it's an important point)

Now, for (1) you use the Nakayama lemma which says that for any finitely generated module $M$ the submodule $(\mathrm{rad}A)M$ is proper. If $M$ is simple this forces $(\mathrm{rad}A)M = 0$ so that $M$ is an $A/\mathrm{rad}A$-module. It is still simple because we haven't changed the submodules of $M$.

For (2) let $e_1, \ldots, e_m$ be the standard basis for $k^m$. Then given any nonzero $k^m$-module $M$ we can form the submodule $e_iM$. As the unit $1 \in k^m$ is $1 = e_1 + e_2 + \cdots + e_m$ it cannot be the case that $e_iM = 0$ for all $i$ (otherwise we'd have $1M = 0$). So fix $i$ such that $e_iM \neq 0$.

If $M$ is simple we must have $e_iM = M$ (there are no proper non-zero submodules). Note $e_ie_j = 0$ so $e_jM = e_je_iM = 0$ if $i \neq j$. Thus the ideal generated by $e_j$ with $i \neq j$ annihilates $M$. When we factor the action again we get that $M$ is a simple $k$-module. Now module = vector space and submodule = subspace. That $M$ is a simple module means it is a nonzero vector space with no proper non-zero subspace. Hence $M$ must be a $1$-dimensional vector space.

Jim
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