Find a maximal ideal $I$ in the ring $\mathbb{Z}[i]$ such that $\mathbb{Z}[i]/I$ is isomorphic to $\mathbb{Z}/521\mathbb{Z}$. 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.
 A: 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.
