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Studying for an exam in Algebra.

Let $R=\mathbb{Z}[i]$ with the usual normfuction $N, z = 5+3i$ and $I = \, <z>$

Show that z isn't a prime element in $R$ and that $R/I$ isn't a field.

I solved the first part by factorizing $(5+3i) = (4-i)(1+i)$ where $5+3i$ divides neither of the two factors in $R$.

I got stuck on the second part. I know that $R/I \iff I$ is an maximal ideal in $R$, and further on that I is maximal if $ I \subsetneq J$ implies that $J=R$. I thought about using the fact that $ y \;| \; x \iff \langle x \rangle \subseteq \langle y \rangle$ for $x,y \in R$, but was unable to solve the problem.

Thanks in advance.

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  • $\begingroup$ $\Bbb Z[i]$ is an Euclidean domain, so an ideal $(x)$ is maximal iff $x$ is irreducible. $\endgroup$
    – Pedro
    Jan 4, 2015 at 18:39

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You have almost solved it. $R/I$ is not even an integral domain. To see this use the factorization you got for $z = 5 + 3i.$

EDIT: We have the following factorization $5 + 3i = (4 - i)(1+i)$ in $R$ with $5 + 3i$ doesn't divide $4 - i$ and $1 + i.$ So in the quotient ring $R/I,$ the image $\overline{4 - i}$ and $\overline{1+i}$ are non-zero. But $\overline{4 - i} \cdot \overline{1+i} = \overline{5 + 3i} = \overline 0$ in $R/I.$

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  • $\begingroup$ I see that if $R/I$ is a field it implies that $R/I$ is a domain, thanks! But I'm unable to find a zero-divisor in $R/I$. It seems like I miss something obvious. @krish $\endgroup$
    – Harpunius
    Jan 4, 2015 at 19:03

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