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Let $R=\mathbb{Z}[i]=\{a+bi\mid a,b\in\mathbb{Z}\}$, where $i^2 =−1$, and let $I=(3+i)$. Find the exact number of elements in $R/I$.

I know that all element can be written as a real number reminder but does that just mean there is only $0$, $1$, and $2$. But if we take $1+i$ this is equivalent to $-2$ but surely this would go on forever.

user26857
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Maths22
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    Hint: If $I \subset J$ then $R/J \simeq (R/I) /(J/I)$. Try with $I=(10)$ and $J=(3+i)$ – Sabino Di Trani Apr 02 '17 at 10:50
  • I'm not sure how that helps – Maths22 Apr 02 '17 at 11:58
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    Since $|R/(10)| = 10^2 = 100$, it reduces it to a finite problem, although I agree that it is still not particularly easy. It might help to factorize $3+i$ into irreducibles in $R$, $3+i=(1-i)(1+2i)$, so $R/I \cong R/(1-i) \oplus R/(1+2i)$. Then, since $1-i$ and $1+2i$ are irreducible, the ideals they generate are maximal, and the two direct summands are finite fields. – Derek Holt Apr 02 '17 at 12:58
  • Related posts: 1, 2 – Viktor Vaughn Apr 02 '17 at 17:25

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