Exercise 15.4.2

Question

\def\imp{\Rightarrow}
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ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

ewcommand{e}{\,\mathrm{Re}\,}

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ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

Let $\alpha \in \Z[i]$ be nonzero. The goal of this exercise is to prove part (a) of Lemma 15.4.2, which asserts that $|\Z[i]/\alpha \Z[i]| = N(\alpha)$. The idea is to forget multiplication and think of $\Z[i]$ and $\Z[i]/\alpha \Z[i]$ as groups under addition. let $m$ be the greatest common divisor of the real and imaginary parts of $\alpha$, so that $\alpha = m(a+bi)$, where $\gcd(a,b) = 1$. Then pick $c,d \in \Z$ such that $ad-bc = 1$.
\be
\item[(a)] Show that the map $\Z[i] 	o \Z \oplus \Z$ defined by
$$\mu + 
u i \mapsto \mu(d,-b) + 
u (-c,a) = (\mu d - 
u c, - \mu b + 
u a)$$
is a group isomorphism under addition.
\item[(b)] Show that the map of part (a) takes $\alpha$ and $i\alpha$ to $(m,0)$ and $-(m(ac+bd), m(a^2+b^2))$, respectively. Then use this to show that the map takes $\alpha \Z[i] \subset \Z[i]$ to the subgroup
$$m\Z \oplus m(a^2+b^2)\Z \subset \Z \oplus \Z.$$
\item[(c)] Use part (b) to conclude that $|\Z[i]/\alpha \Z[i]| = N(\alpha)$.
\ee

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\be} {\begin{enumerate}}

ewcommand{\ee} {\end{enumerate}}

ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\im}{\,\mathrm{Im}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{\Gal}{\mathrm{Gal}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\begin{proof} 
\item[(a)] Consider
$$
\psi 
\left\{
\begin{array}{ccl}
\Z[i] &	o& \Z \oplus \Z\
\mu + 
u i & \mapsto & \mu(d,-b) + 
u (-c,a) = (\mu d - 
u c, - \mu b + 
u a).
\end{array}
ight.
$$
We verify that $\varphi$ is a group homomorphism: if $z = \mu + 
u i, z' = \mu' + 
u' i \in \Z[i]$, then
\begin{align*}
\psi(z+ z') &= \psi((\mu+\mu') + i (
u + 
u')\
&= (\mu + \mu')(d,-b) + (
u + 
u')(-c,a)\
&= [\mu(d,-b) + 
u (-c,a)] + [\mu'(d,-b) + 
u' (-c,a) ]\
&= \psi(z) + \psi(z').
\end{align*}
Let $(u,v)$ be any element of $\Z \oplus \Z$. For all $\mu + i 
u \in \Z[i]$, since $ad - bc = 1$,
\begin{align*}
(u,v) = \psi(\mu + i 
u) & \iff
\left\{
\begin{array}{rl}
\mu d - 
u c &= u,\
-\mu b + 
u a &= v,
\end{array}
ight.\
&\phantom{\Leftarrow}\Rightarrow 
\left\{
\begin{array}{rl}
\mu(ad-bc) = au + cv,\

u(ad-bc) = bu + dv,
\end{array}
ight.\
&\phantom{\Leftarrow}\Rightarrow 
\left\{
\begin{array}{rl}
\mu = au + cv,\

u = bu + dv.
\end{array}
ight.\
\end{align*}
Conversely, if $\mu = au + cv, 
u = bu + dv$, then
$$
\left\{
\begin{array}{rl}
\mu d - 
u c &= \phantom{-}(au + cv) d - (bu + dv) c= u(ad-bc) = u, \
-\mu b + 
u a &= -(au + cv)b + (bu+dv)a = v(ad -bc) = v.
\end{array}
ight.\
$$
We have proved, for all $z \in \Z[i]$, for all $(u,v) \in \Z 	imes \Z$, that
$$(u,v) = \psi(z) \iff z = (au +cv) + i(bu+dv)$$
This shows that $\psi$ is bijective, and for all $(u,v) \in \Z \oplus \Z$,$$\psi^{-1}(u,v) = (au+cv) + i (bu +dv)= (a+ib) u + (c + id) v.$$
To conclude, $\psi$ is a group isomorphism.

\item[(b)] We compute the images of $\alpha$ and $i\alpha$ by the homomorphism $\psi$:
\begin{align*}
\psi(\alpha) &=\psi(ma + mbi)\
&=ma(d,-b) + mb(-c,a)\
&= (m(ad-bc), 0)\
&=(m,0),\
\psi(i\alpha) &= \psi(-mb, ma)\
&=-mb(d,-b)+ ma(-c,a)\
&=(-m(ac+bd), m(a^2+b^2)).
\end{align*}
$(\alpha, i\alpha)$ is a $\Z$-basis of $\alpha \Z[i]$, i.e. every element of $\alpha\Z[i]$ writes uniquely as a linear combination of $\alpha, i \alpha$ with integer coefficients. Moreover $\psi(\alpha) = (m,0) \in m\Z 	imes m(a^2+b^2)\Z$, and $\psi(i\alpha) = (-m(ac+bd), m(a^2+b^2)) \in m\Z 	imes (a^2+b^2)\Z = m\Z \oplus (a^2+b^2)\Z$, therefore 
$$\psi(\alpha\Z[i]) \subset m\Z \oplus (a^2+b^2)\Z.$$

Conversely, let $(u,v)$ be any element of $m\Z 	imes (a^2+b^2)\Z$. There are some $\lambda, \mu \in \Z$ such that
\begin{align*}
u &= \lambda m,\
v &= \mu(a^2+b^2).
\end{align*}
Using the formula which gives $\psi^{-1}(u,v)$ in part (a), we obtain
\begin{align*}
\psi^{-1}(u,v) &=(a+ib) \lambda m + (c+id)\mu(a^2+b^2)\
&=(a+ib) \left[ \lambda m + \mu(c+id)(a-ib)ight],\
\end{align*}
thus $\psi^{-1}(u,v) \in \alpha \Z[i]$, and $(u,v) \in \psi(\alpha \Z[i])$. This proves $m\Z \oplus (a^2+b^2)\Z  \subset \psi(\alpha\Z[i])$, thus
$$\psi(\alpha\Z[i]) = m\Z \oplus (a^2+b^2)\Z .$$

\item[(c)]
If $A,B$ are Abelian groups, and $I,J$ are subgroups of $A,B$ respectively, the surjective homomorphism
$$
\begin{array}{ccc}
A 	imes B & 	o A/I 	imes B/J\
(a,b) & \mapsto A/I 	imes B/J
\end{array}
$$
has kernel $I 	imes J$, so that 
$$(A	imes B) / (I 	imes J) \simeq A/I 	imes B/J.$$
This general property gives here
$$(\Z \oplus\Z)/(m\Z \oplus (a^2+b^2)\Z) \simeq (\Z/m\Z) 	imes (\Z/(a^2+b^2)\Z.$$
Since the isomorphism $\psi$ maps $\Z[i]$ on $\Z \oplus\Z$, and $\alpha \Z[i]$ on $m\Z \oplus (a^2+b^2)\Z$, this implies
$$\Z[i]/\alpha \Z[i] \simeq (\Z \oplus\Z)/(m\Z \oplus (a^2+b^2)\Z) \simeq \Z/m\Z 	imes \Z/(a^2+b^2)\Z.$$
Therefore
$$|\Z[i]/\alpha \Z[i] | = |\Z/m\Z|  \cdot |\Z/(a^2+b^2)\Z| = m(a^2+b^2) = N(\alpha).$$
\end{proof}

Exercise 15.4.2

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Exercise 15.4.2

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