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Theorem $\ \ w=a+bi,\ (a,b)\!=\!1\,\Rightarrow\,\Bbb Z[i]/(w)\,\cong\,\Bbb Z/(w\bar w)$$\ \ w=a+bi,\ \color{#90f}{(a,b)\!=\!1}\ \Rightarrow\ \Bbb Z[i]/(w)\,\cong\,\Bbb Z/(w\bar w)$

Proof $\,\ \overbrace{(a,b)\!=\!1\,\Rightarrow\, ak\!+\!bj=1}^{\rm Bezout}.\:$$\,\ \overbrace{(a,b)\!=\!1\,\Rightarrow\, ak\!+\!bj\!=\!1}^{\rm Bezout}.\:$ In $\, ( w)\,$ is $\,(a\!+\!b\:\!{\it i}\,)(j\!+\!k\:\!{\it i}\,) =\, \overbrace{aj\!-\!bk+ \it i}^{\Large \color{darkorange}{ e\,\ +\,\ {\it i}}} $$\,(a\!+\!b\:\!{\it i}\,)(j\!+\!k\:\!{\it i}\,) =\, \overbrace{aj\!-\!bk+ \it i}^{\Large \color{darkorange}{\ \ \ \ \ e\ \ \,+\, \ \ {\it i}}} $

henceso $\,\smash[t]{\Bbb Z\overset{h}{\to}}\, \Bbb Z[{\it i}\,]/( w)\ $$\ \smash[t]{\Bbb Z\overset{h}{\to}}\, \Bbb Z[{\it i}\,]/( w)\ $ is $\,\color{#0a0}{{\textit onto},\ }$$\,\color{#0a0}{{\rm onto},\,}$ by $\bmod w\!:\,\ \color{darkorange}{{\it i}\,\equiv -e}\phantom{\phantom{\dfrac{.}{.}}}\!\!\Rightarrow\:c\!+\!d\:\!{\it i}\:\equiv\, c\!-\!de\in \Bbb Z$$\!\bmod w\!:\,\ \color{darkorange}{{\it i}\,\equiv -e}\phantom{\phantom{\dfrac{.}{.}}}\!\!\Rightarrow\:c\!+\!d\:\!{\it\color{darkorange}i}\:\equiv\, c\!\color{darkorange}-\!d\color{darkorange}e\color{#0a0}{\in \Bbb Z}$. $\!\begin{eqnarray}{\rm Note}\ \ \color{#c00}{m\in {\rm ker}\ h} &\iff& w\mid m\!\iff \phantom{\dfrac{|}{|_|}}\!\!\!\!\!\! \dfrac{m}{w} = \dfrac{m\,\bar w}{\color{#0af}{w\,\bar w}}\!=\dfrac{ma\!-\!mb\,{\it i}}{\color{#0af}n}\in \Bbb Z[{\it i}\,]\\ &\iff& n\mid ma,mb\!\iff\! n\mid(ma,mb)\!=\!m(a,b)\!=\!m\!\iff\! \color{#c00}{n\mid m}\end{eqnarray} $$\!\begin{eqnarray}{\rm } \color{#c00}{m\in {\rm ker}\ h} &\iff& w\mid m\!\iff \phantom{\dfrac{|}{|_|}}\!\!\!\!\!\! \dfrac{m^{\phantom{|^|}}\!\!\!}{w} = \dfrac{m\,\bar w}{\color{#0af}{w\,\bar w}}\!=\dfrac{ma\!-\!mb\,{\it i}}{\color{#0af}n}\in \Bbb Z[{\it i}\,]\\ &\iff& n\mid ma,mb^{\phantom{|^|}}\!\!\!\!\iff\! \color{#c00}n\mid m\color{#90f}{(a,b)}\!=\!\color{#c00}m\end{eqnarray} $

${\rm Thus} \ \ \Bbb Z[{\it i}\,]/( w)\! \color{#0a0}{= {\rm Im}\:h}\,\cong\: \Bbb Z/\color{#c00}{{\rm ker}\:h} = \Bbb Z/\color{#c00}{n\Bbb Z}\,\ $${\rm So} \ \ \Bbb Z[{\it i}\,]/( w)\! \color{#0a0}{= {\rm Im}\:h}\,\cong\: \Bbb Z/\color{#c00}{{\rm ker}\:h} = \Bbb Z/\color{#c00}{n\Bbb Z}\, $ by the First Isomorphism Theorem.

Remark $ $ WeFor $\,\color{#c00}{{\rm ker}\, h}\,$ we tested divisibility by $\color{#0af}{\textit{rationalizing}}$$\color{#0af}{\text{rationalizing}}$ the denominator - which reduces division by an algebraic irrational $\, w\,$ to a simpler division by an integer (its norm $\, \color{#0af}{ww'=n}).\,$ This simplification is a special case of the method of simpler multiples.

Generally we can use this method to rewrite an ideal as a module in Hermite normal form.

Theorem $\ \ w=a+bi,\ (a,b)\!=\!1\,\Rightarrow\,\Bbb Z[i]/(w)\,\cong\,\Bbb Z/(w\bar w)$

Proof $\,\ \overbrace{(a,b)\!=\!1\,\Rightarrow\, ak\!+\!bj=1}^{\rm Bezout}.\:$ In $\, ( w)\,$ is $\,(a\!+\!b\:\!{\it i}\,)(j\!+\!k\:\!{\it i}\,) =\, \overbrace{aj\!-\!bk+ \it i}^{\Large \color{darkorange}{ e\,\ +\,\ {\it i}}} $

hence $\,\smash[t]{\Bbb Z\overset{h}{\to}}\, \Bbb Z[{\it i}\,]/( w)\ $ is $\,\color{#0a0}{{\textit onto},\ }$ by $\bmod w\!:\,\ \color{darkorange}{{\it i}\,\equiv -e}\phantom{\phantom{\dfrac{.}{.}}}\!\!\Rightarrow\:c\!+\!d\:\!{\it i}\:\equiv\, c\!-\!de\in \Bbb Z$. $\!\begin{eqnarray}{\rm Note}\ \ \color{#c00}{m\in {\rm ker}\ h} &\iff& w\mid m\!\iff \phantom{\dfrac{|}{|_|}}\!\!\!\!\!\! \dfrac{m}{w} = \dfrac{m\,\bar w}{\color{#0af}{w\,\bar w}}\!=\dfrac{ma\!-\!mb\,{\it i}}{\color{#0af}n}\in \Bbb Z[{\it i}\,]\\ &\iff& n\mid ma,mb\!\iff\! n\mid(ma,mb)\!=\!m(a,b)\!=\!m\!\iff\! \color{#c00}{n\mid m}\end{eqnarray} $

${\rm Thus} \ \ \Bbb Z[{\it i}\,]/( w)\! \color{#0a0}{= {\rm Im}\:h}\,\cong\: \Bbb Z/\color{#c00}{{\rm ker}\:h} = \Bbb Z/\color{#c00}{n\Bbb Z}\,\ $ by the First Isomorphism Theorem.

Remark $ $ We tested divisibility by $\color{#0af}{\textit{rationalizing}}$ the denominator - which reduces division by an algebraic irrational $\, w\,$ to a simpler division by an integer (its norm $\, \color{#0af}{ww'=n}).\,$ This simplification is a special case of the method of simpler multiples.

Generally we can use this method to rewrite an ideal as a module in Hermite normal form.

Theorem $\ \ w=a+bi,\ \color{#90f}{(a,b)\!=\!1}\ \Rightarrow\ \Bbb Z[i]/(w)\,\cong\,\Bbb Z/(w\bar w)$

Proof $\,\ \overbrace{(a,b)\!=\!1\,\Rightarrow\, ak\!+\!bj\!=\!1}^{\rm Bezout}.\:$ In $\, ( w)\,$ is $\,(a\!+\!b\:\!{\it i}\,)(j\!+\!k\:\!{\it i}\,) =\, \overbrace{aj\!-\!bk+ \it i}^{\Large \color{darkorange}{\ \ \ \ \ e\ \ \,+\, \ \ {\it i}}} $

so $\ \smash[t]{\Bbb Z\overset{h}{\to}}\, \Bbb Z[{\it i}\,]/( w)\ $ is $\,\color{#0a0}{{\rm onto},\,}$ by $\!\bmod w\!:\,\ \color{darkorange}{{\it i}\,\equiv -e}\phantom{\phantom{\dfrac{.}{.}}}\!\!\Rightarrow\:c\!+\!d\:\!{\it\color{darkorange}i}\:\equiv\, c\!\color{darkorange}-\!d\color{darkorange}e\color{#0a0}{\in \Bbb Z}$. $\!\begin{eqnarray}{\rm } \color{#c00}{m\in {\rm ker}\ h} &\iff& w\mid m\!\iff \phantom{\dfrac{|}{|_|}}\!\!\!\!\!\! \dfrac{m^{\phantom{|^|}}\!\!\!}{w} = \dfrac{m\,\bar w}{\color{#0af}{w\,\bar w}}\!=\dfrac{ma\!-\!mb\,{\it i}}{\color{#0af}n}\in \Bbb Z[{\it i}\,]\\ &\iff& n\mid ma,mb^{\phantom{|^|}}\!\!\!\!\iff\! \color{#c00}n\mid m\color{#90f}{(a,b)}\!=\!\color{#c00}m\end{eqnarray} $

${\rm So} \ \ \Bbb Z[{\it i}\,]/( w)\! \color{#0a0}{= {\rm Im}\:h}\,\cong\: \Bbb Z/\color{#c00}{{\rm ker}\:h} = \Bbb Z/\color{#c00}{n\Bbb Z}\, $ by First Isomorphism Theorem.

Remark $ $ For $\,\color{#c00}{{\rm ker}\, h}\,$ we tested divisibility by $\color{#0af}{\text{rationalizing}}$ the denominator - which reduces division by an algebraic irrational $\, w\,$ to a simpler division by an integer (its norm $\, \color{#0af}{ww'=n}).\,$ This simplification is a special case of the method of simpler multiples.

Generally we can use this method to rewrite an ideal as a module in Hermite normal form.

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Bill Dubuque
Bill Dubuque
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Theorem $\ \ (a,b)\!=\!1\,\Rightarrow\,\Bbb Z[i]/(\overbrace{a+bi}^{\large w})\,\cong\,\Bbb Z/(\overbrace{a^2+b^2}^{\large n\ =\ ww'})$$\ \ w=a+bi,\ (a,b)\!=\!1\,\Rightarrow\,\Bbb Z[i]/(w)\,\cong\,\Bbb Z/(w\bar w)$

Proof $\,\ \overbrace{(a,b)\!=\!1\,\Rightarrow\, ak\!+\!bj=1}^{\rm Bezout}.\:$ In $\, ( w)\,$ is $\,(a\!+\!b\:\!{\it i}\,)(j\!+\!k\:\!{\it i}\,) =\, \overbrace{aj\!-\!bk+ \it i}^{\Large \color{darkorange}{ e\,\ +\,\ {\it i}}} $

Notehence $\,\smash[t]{\Bbb Z\overset{h}{\to}}\, \Bbb Z[{\it i}\,]/( w)\ $ is $\,\color{#0a0}{{\textit onto},\ }$ by $\bmod w\!:\,\ \color{darkorange}{{\it i}\,\equiv -e}\phantom{\phantom{\dfrac{.}{.}}}\!\!\Rightarrow\:c\!+\!d\:\!{\it i}\:\equiv\, c\!-\!de\in \Bbb Z$.
  $\!\begin{eqnarray}{\rm Note}\ \ \color{#c00}{m\in {\rm ker}\ h} &\iff& w\mid m\!\iff \phantom{\dfrac{|}{|_|}}\!\!\!\!\!\! \dfrac{m}{w} = \dfrac{m\,w'}{\color{#0af}{ww'}}\!=\dfrac{ma\!-\!mb\,{\it i}}{\color{#0af}n}\in \Bbb Z[{\it i}\,]\\ &\iff& n\mid ma,mb\!\iff\! n\mid(ma,mb)\!=\!m(a,b)\!=\!m\!\iff\! \color{#c00}{n\mid m}\end{eqnarray} $$\!\begin{eqnarray}{\rm Note}\ \ \color{#c00}{m\in {\rm ker}\ h} &\iff& w\mid m\!\iff \phantom{\dfrac{|}{|_|}}\!\!\!\!\!\! \dfrac{m}{w} = \dfrac{m\,\bar w}{\color{#0af}{w\,\bar w}}\!=\dfrac{ma\!-\!mb\,{\it i}}{\color{#0af}n}\in \Bbb Z[{\it i}\,]\\ &\iff& n\mid ma,mb\!\iff\! n\mid(ma,mb)\!=\!m(a,b)\!=\!m\!\iff\! \color{#c00}{n\mid m}\end{eqnarray} $

${\rm Thus} \ \ \Bbb Z[{\it i}\,]/( w)\! \color{#0a0}{= {\rm Im}\:h}\,\cong\: \Bbb Z/\color{#c00}{{\rm ker}\:h} = \Bbb Z/\color{#c00}{n\Bbb Z}\,\ $ by the First Isomorphism Theorem.

Remark $ $ We tested divisibility by $\color{#0af}{\textit{rationalizing}}$ the denominator - which reduces division by an algebraic irrational $\, w\,$ to a simpler division by an integer (its norm $\, \color{#0af}{ww'=n})$.$\, \color{#0af}{ww'=n}).\,$ This simplification is a special case of the method of simpler multiples.

Generally we can use this method to rewrite thean ideal as a module in Hermite normal form.

Theorem $\ \ (a,b)\!=\!1\,\Rightarrow\,\Bbb Z[i]/(\overbrace{a+bi}^{\large w})\,\cong\,\Bbb Z/(\overbrace{a^2+b^2}^{\large n\ =\ ww'})$

Proof $\,\ \overbrace{(a,b)\!=\!1\,\Rightarrow\, ak\!+\!bj=1}^{\rm Bezout}.\:$ In $\, ( w)\,$ is $\,(a\!+\!b\:\!{\it i}\,)(j\!+\!k\:\!{\it i}\,) =\, \overbrace{aj\!-\!bk+ \it i}^{\Large \color{darkorange}{ e\,\ +\,\ {\it i}}} $

Note $\,\smash[t]{\Bbb Z\overset{h}{\to}}\, \Bbb Z[{\it i}\,]/( w)\ $ is $\,\color{#0a0}{{\textit onto},\ }$ by $\bmod w\!:\,\ \color{darkorange}{{\it i}\,\equiv -e}\phantom{\phantom{\dfrac{.}{.}}}\!\!\Rightarrow\:c\!+\!d\:\!{\it i}\:\equiv\, c\!-\!de\in \Bbb Z$.
  $\!\begin{eqnarray}{\rm Note}\ \ \color{#c00}{m\in {\rm ker}\ h} &\iff& w\mid m\!\iff \phantom{\dfrac{|}{|_|}}\!\!\!\!\!\! \dfrac{m}{w} = \dfrac{m\,w'}{\color{#0af}{ww'}}\!=\dfrac{ma\!-\!mb\,{\it i}}{\color{#0af}n}\in \Bbb Z[{\it i}\,]\\ &\iff& n\mid ma,mb\!\iff\! n\mid(ma,mb)\!=\!m(a,b)\!=\!m\!\iff\! \color{#c00}{n\mid m}\end{eqnarray} $

${\rm Thus} \ \ \Bbb Z[{\it i}\,]/( w)\! \color{#0a0}{= {\rm Im}\:h}\,\cong\: \Bbb Z/\color{#c00}{{\rm ker}\:h} = \Bbb Z/\color{#c00}{n\Bbb Z}\,\ $ by the First Isomorphism Theorem.

Remark $ $ We tested divisibility by $\color{#0af}{\textit{rationalizing}}$ the denominator - which reduces division by an algebraic irrational $\, w\,$ to a simpler division by an integer (its norm $\, \color{#0af}{ww'=n})$. This simplification is a special case of the method of simpler multiples.

Generally we can use this method to rewrite the ideal as a module in Hermite normal form.

Theorem $\ \ w=a+bi,\ (a,b)\!=\!1\,\Rightarrow\,\Bbb Z[i]/(w)\,\cong\,\Bbb Z/(w\bar w)$

Proof $\,\ \overbrace{(a,b)\!=\!1\,\Rightarrow\, ak\!+\!bj=1}^{\rm Bezout}.\:$ In $\, ( w)\,$ is $\,(a\!+\!b\:\!{\it i}\,)(j\!+\!k\:\!{\it i}\,) =\, \overbrace{aj\!-\!bk+ \it i}^{\Large \color{darkorange}{ e\,\ +\,\ {\it i}}} $

hence $\,\smash[t]{\Bbb Z\overset{h}{\to}}\, \Bbb Z[{\it i}\,]/( w)\ $ is $\,\color{#0a0}{{\textit onto},\ }$ by $\bmod w\!:\,\ \color{darkorange}{{\it i}\,\equiv -e}\phantom{\phantom{\dfrac{.}{.}}}\!\!\Rightarrow\:c\!+\!d\:\!{\it i}\:\equiv\, c\!-\!de\in \Bbb Z$. $\!\begin{eqnarray}{\rm Note}\ \ \color{#c00}{m\in {\rm ker}\ h} &\iff& w\mid m\!\iff \phantom{\dfrac{|}{|_|}}\!\!\!\!\!\! \dfrac{m}{w} = \dfrac{m\,\bar w}{\color{#0af}{w\,\bar w}}\!=\dfrac{ma\!-\!mb\,{\it i}}{\color{#0af}n}\in \Bbb Z[{\it i}\,]\\ &\iff& n\mid ma,mb\!\iff\! n\mid(ma,mb)\!=\!m(a,b)\!=\!m\!\iff\! \color{#c00}{n\mid m}\end{eqnarray} $

${\rm Thus} \ \ \Bbb Z[{\it i}\,]/( w)\! \color{#0a0}{= {\rm Im}\:h}\,\cong\: \Bbb Z/\color{#c00}{{\rm ker}\:h} = \Bbb Z/\color{#c00}{n\Bbb Z}\,\ $ by the First Isomorphism Theorem.

Remark $ $ We tested divisibility by $\color{#0af}{\textit{rationalizing}}$ the denominator - which reduces division by an algebraic irrational $\, w\,$ to a simpler division by an integer (its norm $\, \color{#0af}{ww'=n}).\,$ This simplification is a special case of the method of simpler multiples.

Generally we can use this method to rewrite an ideal as a module in Hermite normal form.

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Bill Dubuque
Bill Dubuque
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Theorem $\ \ (a,b)\!=\!1\,\Rightarrow\,\Bbb Z[i]/(\overbrace{a+bi}^{\large w})\,\cong\,\Bbb Z/(\overbrace{a^2+b^2}^{\large n\ =\ ww'})$

Proof $\,\ \overbrace{(a,b)\!=\!1\,\Rightarrow\, ak\!+\!bj=1}^{\rm Bezout}.\:$ In $\, ( w)\,$ is $\,(a\!+\!b\:\!{\it i}\,)(j\!+\!k\:\!{\it i}\,) =\, \overbrace{aj\!-\!bk+ \it i}^{\Large \color{darkorange}{ e\,\ +\,\ {\it i}}} $

Note $\,\smash[t]{\Bbb Z\overset{h}{\to}}\, \Bbb Z[{\it i}\,]/( w)\ $ is $\,\color{#0a0}{{\textit onto},\ }$ by $\bmod w\!:\,\ \color{darkorange}{{\it i}\,\equiv -e}\phantom{\phantom{\dfrac{.}{.}}}\!\!\Rightarrow\:c\!+\!d\:\!{\it i}\:\equiv\, c\!-\!de\in \Bbb Z$.
$\!\begin{eqnarray}{\rm Note}\ \ \color{#c00}{m\in {\rm ker}\ h} &\iff& w\mid m\!\iff \phantom{\dfrac{|}{|_|}}\!\!\!\!\!\! \dfrac{m}{w} = \dfrac{m\,w'}{\color{#0af}{ww'}}\!=\dfrac{ma\!-\!mb\,{\it i}}{\color{#0af}n}\in \Bbb Z[{\it i}\,]\\ &\iff& n\mid ma,mb\!\iff\! n\mid(ma,mb)\!=\!m(a,b)\!=\!m\!\iff\! \color{#c00}{n\mid m}\end{eqnarray} $

${\rm Thus} \ \ \Bbb Z[{\it i}\,]/( w)\! \color{#0a0}{= {\rm Im}\:h}\,\cong\: \Bbb Z/\color{#c00}{{\rm ker}\:h} = \Bbb Z/\color{#c00}{n\Bbb Z}\,\ $ by the First Isomorphism Theorem.

Remark $ $ We tested divisibility by $\color{#0af}{\textit{rationalizing}}$ the denominator - which reduces division by an algebraic irrational $\, w\,$ to a simpler division by an integer (its norm $\, \color{#0af}{ww'=n})$. This simplification is a special case of the method of simpler multiples.

Generally we can use this methodto to rewrite the ideal as a module in Hermite normal form.

Theorem $\ \ (a,b)\!=\!1\,\Rightarrow\,\Bbb Z[i]/(\overbrace{a+bi}^{\large w})\,\cong\,\Bbb Z/(\overbrace{a^2+b^2}^{\large n\ =\ ww'})$

Proof $\,\ \overbrace{(a,b)\!=\!1\,\Rightarrow\, ak\!+\!bj=1}^{\rm Bezout}.\:$ In $\, ( w)\,$ is $\,(a\!+\!b\:\!{\it i}\,)(j\!+\!k\:\!{\it i}\,) =\, \overbrace{aj\!-\!bk+ \it i}^{\Large \color{darkorange}{ e\,\ +\,\ {\it i}}} $

Note $\,\smash[t]{\Bbb Z\overset{h}{\to}}\, \Bbb Z[{\it i}\,]/( w)\ $ is $\,\color{#0a0}{{\textit onto},\ }$ by $\bmod w\!:\,\ \color{darkorange}{{\it i}\,\equiv -e}\phantom{\phantom{\dfrac{.}{.}}}\!\!\Rightarrow\:c\!+\!d\:\!{\it i}\:\equiv\, c\!-\!de\in \Bbb Z$.
$\!\begin{eqnarray}{\rm Note}\ \ \color{#c00}{m\in {\rm ker}\ h} &\iff& w\mid m\!\iff \phantom{\dfrac{|}{|_|}}\!\!\!\!\!\! \dfrac{m}{w} = \dfrac{m\,w'}{\color{#0af}{ww'}}\!=\dfrac{ma\!-\!mb\,{\it i}}{\color{#0af}n}\in \Bbb Z[{\it i}\,]\\ &\iff& n\mid ma,mb\!\iff\! n\mid(ma,mb)\!=\!m(a,b)\!=\!m\!\iff\! \color{#c00}{n\mid m}\end{eqnarray} $

${\rm Thus} \ \ \Bbb Z[{\it i}\,]/( w)\! \color{#0a0}{= {\rm Im}\:h}\,\cong\: \Bbb Z/\color{#c00}{{\rm ker}\:h} = \Bbb Z/\color{#c00}{n\Bbb Z}\,\ $ by the First Isomorphism Theorem.

Remark $ $ We tested divisibility by $\color{#0af}{\textit{rationalizing}}$ the denominator - which reduces division by an algebraic irrational $\, w\,$ to a simpler division by an integer (its norm $\, \color{#0af}{ww'=n})$. This simplification is a special case of the method of simpler multiples.

Generally we can use this methodto rewrite the ideal as a module in Hermite normal form.

Theorem $\ \ (a,b)\!=\!1\,\Rightarrow\,\Bbb Z[i]/(\overbrace{a+bi}^{\large w})\,\cong\,\Bbb Z/(\overbrace{a^2+b^2}^{\large n\ =\ ww'})$

Proof $\,\ \overbrace{(a,b)\!=\!1\,\Rightarrow\, ak\!+\!bj=1}^{\rm Bezout}.\:$ In $\, ( w)\,$ is $\,(a\!+\!b\:\!{\it i}\,)(j\!+\!k\:\!{\it i}\,) =\, \overbrace{aj\!-\!bk+ \it i}^{\Large \color{darkorange}{ e\,\ +\,\ {\it i}}} $

Note $\,\smash[t]{\Bbb Z\overset{h}{\to}}\, \Bbb Z[{\it i}\,]/( w)\ $ is $\,\color{#0a0}{{\textit onto},\ }$ by $\bmod w\!:\,\ \color{darkorange}{{\it i}\,\equiv -e}\phantom{\phantom{\dfrac{.}{.}}}\!\!\Rightarrow\:c\!+\!d\:\!{\it i}\:\equiv\, c\!-\!de\in \Bbb Z$.
$\!\begin{eqnarray}{\rm Note}\ \ \color{#c00}{m\in {\rm ker}\ h} &\iff& w\mid m\!\iff \phantom{\dfrac{|}{|_|}}\!\!\!\!\!\! \dfrac{m}{w} = \dfrac{m\,w'}{\color{#0af}{ww'}}\!=\dfrac{ma\!-\!mb\,{\it i}}{\color{#0af}n}\in \Bbb Z[{\it i}\,]\\ &\iff& n\mid ma,mb\!\iff\! n\mid(ma,mb)\!=\!m(a,b)\!=\!m\!\iff\! \color{#c00}{n\mid m}\end{eqnarray} $

${\rm Thus} \ \ \Bbb Z[{\it i}\,]/( w)\! \color{#0a0}{= {\rm Im}\:h}\,\cong\: \Bbb Z/\color{#c00}{{\rm ker}\:h} = \Bbb Z/\color{#c00}{n\Bbb Z}\,\ $ by the First Isomorphism Theorem.

Remark $ $ We tested divisibility by $\color{#0af}{\textit{rationalizing}}$ the denominator - which reduces division by an algebraic irrational $\, w\,$ to a simpler division by an integer (its norm $\, \color{#0af}{ww'=n})$. This simplification is a special case of the method of simpler multiples.

Generally we can use this method to rewrite the ideal as a module in Hermite normal form.

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Bill Dubuque
Bill Dubuque
  • 285k
  • 42
  • 343
  • 1.1k