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Let $R$ be a commutative ring, $I$ be a minimal ideal of $R$. Prove that for all $y$ belong to $Rad(R)$, $yI=0$. ($Rad(R)$ denotes the Jacobson radical of $R$)

$Rad(R)$ equals the intersection of all maximal right ideals of the ring. It is also true that $Rad(R)$ equals the intersection of all maximal left ideals within the ring. In that case, $Rad(R)$ equals the intersection of all maximal ideals of the ring. So take $y$ belongs to $Rad(R)$, $y$ must belong to some maximal ideal $M$ of $R$. Clearly, $yI$ belongs to $MI$. I want $MI=0$. Is that a trivial fact?

Thank you very much.

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    $\begingroup$ MI is not going to be zero in general. M here can be any maximal ideal whatsoever, so if this were true all the non-unit elements of your ring would be zero-divisors. $\endgroup$
    – Zavosh
    Nov 30, 2014 at 3:42
  • $\begingroup$ I see what you means. Actually we only need $yI=0$, not $MI=0$. Thus I 'll use another definition of Jacobson radical which especially for the case commutative. $\endgroup$
    – Truong
    Nov 30, 2014 at 3:52
  • $\begingroup$ Since you are assuming the ring is commutative, there is utterly no point going into the left-right ideal distinction... $\endgroup$
    – rschwieb
    Nov 30, 2014 at 4:02
  • $\begingroup$ Okay I 'll remove the tags. $\endgroup$
    – Truong
    Nov 30, 2014 at 4:04

1 Answer 1

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Since $I$ is simple, $R/M \cong I$ For some maximal ideal M of R.

Clearly the annihilator of $R/M$ is M, therefore... (Can you take it from here?)

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  • $\begingroup$ @chuyenvien94 it was a typo (of course...) $\endgroup$
    – rschwieb
    Nov 30, 2014 at 4:04
  • $\begingroup$ Thank you very much @rschwieb. I've got it. $\endgroup$
    – Truong
    Nov 30, 2014 at 4:05
  • $\begingroup$ @chuyenvien94 great! No problem... $\endgroup$
    – rschwieb
    Nov 30, 2014 at 4:06

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