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In any ring $R$ define the socle as the sum of all minimal right ideals of $R$.

Say we have two minimal ideals $A,B$. If $a\in A,b\in B$, then $a+b$ is in the socle. If $x\in R$, then $(a+b)x=ax+bx$. As $ax\in A$ and $bx\in B$, this element is in $A+B$ and hence in the socle. So the socle is a right ideal.

Is it also a left ideal, and if so, how can I prove it?

Stefan4024
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man_in_green_shirt
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1 Answers1

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If $r\in R$ and $A$ is a minimal right ideal, then $rA$ is either $0$ or a minimal right ideal, because the mapping $R\to R$ defined by $x\mapsto rx$ is a homomorphism of right modules. In any case $rA$ is contained in the sum of the minimal right ideals.

egreg
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  • I get that that mapping is a homomorphism (it can easily be checked to preserve addition, multiplication and the 'zero' and 'one'), but why does that mapping being a homomorphism show that $rA$ is either $0$ or a minimal right ideal? – man_in_green_shirt Jul 17 '15 at 22:02
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    @man_in_green_shirt It induces a (surjective) homomorphism $A\to rA$, whose kernel is either $0$ or $A$, as $A$ is simple. In the latter case $rA=0$, in the former case $rA$ is isomorphic to $A$. – egreg Jul 17 '15 at 22:13
  • Ah ok, and then $A\cong rA$ by the first isomorphism theorem. Thanks! – man_in_green_shirt Jul 17 '15 at 22:15