I know $\mathbb{Z}[i]$, the Gaussian integers, is a PID. So $I$ is generated by a single element. At first I thought $I=(521)$, but $521$ can be reduced to $11^2 + 20^2$. Would $I=(11 + 20i)$ or $I=(20 + 11i)$ then be the maximal ideal needed to achieve this isomorphism? I need a little help.
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1I like your question! Note that in $\mathbb{Z}/521\mathbb{Z}$, the congruence $x^2 \equiv -1 \pmod{521}$ has solutions $x \equiv \pm 235$. So define $\varphi : \mathbb{Z}[i] \to \mathbb{Z}/521\mathbb{Z}$ by $1 \mapsto 1 + 521 \mathbb{Z}$ and $i \mapsto 235 + 521\mathbb{Z}$. I think the kernel of this map (which I hope we can compute) will be the ideal you're looking for. – Viktor Vaughn Feb 18 '14 at 04:57
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Note that $(11+20\mathbf i)$ and $(20+11\mathbf i)$ are different ideals of $\Bbb Z[\mathbf i]$. Both are valid answers to the question. – Marc van Leeuwen Feb 20 '14 at 10:06
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Given that $211$ is prime, it suffices to find an element $\def\i{\mathbf i}z=a+b\i$ of norm $a^2+b^2=211$, and then $\def\Z{\Bbb Z}\Z[\i]/(z)$ will be isomorphic to the field $\Z/211\Z$. As explained in this answer, this does not need the fact that $\Z[\i]$ is a principal ideal domain. Onr erasons simply that
- $(z)$ contains $z\overline z=a^2+b^2=211$, but not $1$, so $\Z[\i]/(z)$ contains $\Z/211\Z$ as subring;
- using Bezout coefficients (in $\Z$) for $(a,b)$ one finds an element of the form $c+\i\in(z)$;
- so every element of $\Z[\i]$ is congruent modulo$~(z)$ to some $k\in\Z$, showing that the subring is in fact all of $\Z[\i]/(z)$.
And your question already indicates two non-associated elements of norm $211$ (and in fact that is all one can find), so you have two different valid candidates for the ideal $I$. One can show that there are only two such ideals (as opposed to elements of norm$~211$) by showing that the ring $\Z[\i]/(211)\cong(\Z/211\Z)[X]/(X^2+1)$ is a product of two fields, so that it has exactly two maximal ideals.
Marc van Leeuwen
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