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Let $R$ denote the factor ring $\mathbb{Z}[i]/(1+3i)$. Show that $i-3 \in (1+3i)$ and that $i+(1+3i) = 3 + (1+ 3i)$ in R. I am unsure how to find the elements of this factor ring?

I know how to find the elements of, for example $\mathbb{F}_{3}[x]/(x^{2}+1)$, but unsure about this example with Gaussian integers.

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    $\begingroup$ So in the factor ring, $1 + 3i = 0$, or $1 = -3i$, or $i = 3$. $\endgroup$ – M. Vinay Apr 18 '16 at 16:44
  • $\begingroup$ Note that $i-3=i(1+3i)$ and hence, by definition, your first statement. This implies that $[i-3]=[0]$ in $R$ and hence that $[i]=[3]$, which is your second statement. $\endgroup$ – Cla Apr 18 '16 at 16:45
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You obtain a ring isomorphism $\;\begin{aligned}[t]\mathbf Z[i]/(i-3)&\longrightarrow\mathbf Z,\\a+ib&\longmapsto a+3b.\end{aligned}$

Note the ideals $(1+3i)$ and $(i-3)$ are the same since their generators are associate (one is a unit times the other).

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  • $\begingroup$ $\mathbf{Z}/10\mathbf{Z}$, no? $\endgroup$ – André 3000 Apr 19 '16 at 1:51

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