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I have the factor ring $\mathbb Z[i]/\langle3-i\rangle$ and am asked to find elements zero in this ,they are $0,3-i,i(3-i),(3-i)+i(3-i)$.

But I can't understand how do we guarantee these are the only zero elements and there are no more.....

Please help I'm STRUCK.......

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  • $\begingroup$ You want "zero elements" or zero divisors , @cool ? $\endgroup$ – Timbuc Oct 18 '14 at 15:22
  • $\begingroup$ @Timbuc zero elements ..... $\endgroup$ – coool Oct 18 '14 at 15:23
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    $\begingroup$ Well, that's odd @cool :as user's answer mentions, in any ring there is one, unique zero element, which in this quotient ring case is represented by the ideal $\;I:=\langle 3-1\rangle\;$, and the elements of this ideal are of the form $\;(3-i)x\;,\;\;x\in \Bbb Z[i]\;$, as written there. $\endgroup$ – Timbuc Oct 18 '14 at 15:24
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    $\begingroup$ @coool, please do read the answer below! An element belongs to a principal ideal iff it is a multiple (from the ring) of the element generating that ideal, thus in this case there are infinite elements that represent the zero element in this quotient ring: $$0=0\cdot(3-i)\;,\;\;3-i=1\cdot (3-i)\;,\;\;(3-i)+i(3-i)=(3-i)(1+i)\;,\;\;etc.$$ $\endgroup$ – Timbuc Oct 18 '14 at 15:33
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    $\begingroup$ @Timbuc thanks for explanation.... $\endgroup$ – coool Oct 18 '14 at 15:36
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The zero element (which is always unique) of the quotient ring $R/I$ is the ideal $I$. In present question, $I$ consists of all elements $(3-i)x$ with $x\in\mathbb{Z}[i]$.

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  • $\begingroup$ @Timbuc I don't see any mention of zero divisors in the question. $\endgroup$ – user2097 Oct 18 '14 at 15:13
  • $\begingroup$ As odd as the question is, this answer addresses and explains what the OP asked. +1 $\endgroup$ – Timbuc Oct 18 '14 at 15:27

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