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Proceedings of the 44th Symposium on Ring Theory and ...

Proceedings of the 44th Symposium on Ring Theory and ...

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principal ideal Rr ⊂ ker ϖ. Thus, ker f = 0 if <strong>and</strong> <strong>on</strong>ly if ker ϖ c<strong>on</strong>tains no n<strong>on</strong>zero left<br />

ideals. The pro<str<strong>on</strong>g>of</str<strong>on</strong>g> for right generating characters is similar.<br />

□<br />

Propositi<strong>on</strong> 8 ([31, Theorem 4.3]). A character ϱ <str<strong>on</strong>g>of</str<strong>on</strong>g> a finite ring R is a left generating<br />

character if <strong>and</strong> <strong>on</strong>ly if it is a right generating character.<br />

Pro<str<strong>on</strong>g>of</str<strong>on</strong>g>. Suppose ϱ is a left generating character, <strong>and</strong> suppose that I ⊂ ker ϱ is a right<br />

ideal. Then for every r ∈ R, Ir ⊂ ker ϱ, so that I ⊂ ker(rϱ), for all r ∈ R. But<br />

every character <str<strong>on</strong>g>of</str<strong>on</strong>g> R is <str<strong>on</strong>g>of</str<strong>on</strong>g> <str<strong>on</strong>g>the</str<strong>on</strong>g> form rϱ, because ϱ is a left generating character. Thus <str<strong>on</strong>g>the</str<strong>on</strong>g><br />

annihilator ( ̂R : I) = ̂R, <strong>and</strong> it follows from (2.5) that I = 0. By Propositi<strong>on</strong> 7, ϱ is a<br />

right generating character.<br />

□<br />

Propositi<strong>on</strong> 9 ([33, Propositi<strong>on</strong> 3.3]). Let A be a finite left R-module. Then soc(Â) ∼ =<br />

(A/ rad(R)A)̂.<br />

Pro<str<strong>on</strong>g>of</str<strong>on</strong>g>. There is a short exact sequence <str<strong>on</strong>g>of</str<strong>on</strong>g> left R-modules<br />

0 → rad(R)A → A → A/ rad(R)A → 0.<br />

Taking character modules, as in (2.4), yields<br />

0 → (A/ rad(R)A)̂ → Â → (rad(R)A)̂ → 0.<br />

Because A/ rad(R)A is a sum <str<strong>on</strong>g>of</str<strong>on</strong>g> simple modules, <str<strong>on</strong>g>the</str<strong>on</strong>g> same is true for (A/ rad(R)A)̂ ∼ =<br />

(Â : rad(R)A). Thus (Â : rad(R)A) ⊂ soc(Â).<br />

C<strong>on</strong>versely, soc(Â) rad(R) = 0, because <str<strong>on</strong>g>the</str<strong>on</strong>g> radical annihilates simple modules [7, Exercise<br />

25.4]. Thus soc(Â) ⊂ (Â : rad(R)A), <strong>and</strong> we have <str<strong>on</strong>g>the</str<strong>on</strong>g> equality soc(Â) = (Â :<br />

rad(R)A). Now remember that (Â : rad(R)A) ∼ = (A/ rad(R)A)̂.<br />

□<br />

Using Propositi<strong>on</strong> 7 as a model, we extend <str<strong>on</strong>g>the</str<strong>on</strong>g> definiti<strong>on</strong> <str<strong>on</strong>g>of</str<strong>on</strong>g> a generating character to<br />

modules. Let A be a finite left (resp., right) R-module. A character ϖ <str<strong>on</strong>g>of</str<strong>on</strong>g> A is a generating<br />

character <str<strong>on</strong>g>of</str<strong>on</strong>g> A if ker ϖ c<strong>on</strong>tains no n<strong>on</strong>zero left (resp., right) R-submodules <str<strong>on</strong>g>of</str<strong>on</strong>g> A.<br />

Lemma 10 (†). Let A be a finite left R-module, <strong>and</strong> let B ⊂ A be a submodule. If A<br />

admits a left generating character, <str<strong>on</strong>g>the</str<strong>on</strong>g>n B admits a left generating character.<br />

Pro<str<strong>on</strong>g>of</str<strong>on</strong>g>. Simply restrict a generating character <str<strong>on</strong>g>of</str<strong>on</strong>g> A to B. Any submodule <str<strong>on</strong>g>of</str<strong>on</strong>g> B inside <str<strong>on</strong>g>the</str<strong>on</strong>g><br />

kernel <str<strong>on</strong>g>of</str<strong>on</strong>g> <str<strong>on</strong>g>the</str<strong>on</strong>g> restricti<strong>on</strong> will also be a submodule <str<strong>on</strong>g>of</str<strong>on</strong>g> A inside <str<strong>on</strong>g>the</str<strong>on</strong>g> kernel <str<strong>on</strong>g>of</str<strong>on</strong>g> <str<strong>on</strong>g>the</str<strong>on</strong>g> original<br />

generating character.<br />

□<br />

Lemma 11 (†). Let R be any finite ring. Define ϱ : ̂R → Q/Z by ϱ(ϖ) = ϖ(1), evaluati<strong>on</strong><br />

at 1 ∈ R, for ϖ ∈ ̂R. Then ϱ is a left <strong>and</strong> right generating character <str<strong>on</strong>g>of</str<strong>on</strong>g> ̂R.<br />

Pro<str<strong>on</strong>g>of</str<strong>on</strong>g>. Suppose ϖ 0 ≠ 0 has <str<strong>on</strong>g>the</str<strong>on</strong>g> property that Rϖ 0 ⊂ ker ϱ. This means that for every<br />

r ∈ R, 0 = ϱ(rϖ 0 ) = (rϖ 0 )(1) = ϖ 0 (r), so that ϖ 0 = 0. Thus ϱ is a left generating<br />

character by definiti<strong>on</strong>. Similarly for ϱ being a right generating character.<br />

□<br />

Propositi<strong>on</strong> 12 (†). Let A be a finite left R-module. Then A admits a left generating<br />

character if <strong>and</strong> <strong>on</strong>ly if A can be embedded in ̂R.<br />

–228–

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