Exercise 15.4.4

Question

\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}}

Prove (15.38).
\be
\item[(a)] $\alpha \beta$  is odd  $\iff \alpha$ and $\beta$ are odd.
\item[(b)] $\alpha + \beta$ is even $\iff$ $\alpha,\beta$ are both even or both odd.
\item[(c)] $\alpha$ is even $\iff$ $1+i$ divides $\alpha$.
\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} 
We say that a Gaussian integer $a+bi \in \Z[i]$ is {\it odd} if $a+b$ is odd ($b \equiv a+1 \pmod 2$ )and {\it even} if $a+b$ is even ($b\equiv a \pmod 2$).

\item[(a)] If $\alpha = a+bi$ and $\beta = c+di$ are odd, then $b \equiv a+1, d \equiv c+1 \pmod 2$, and $\alpha \beta = (a+bi)(c+di) = ac -bd + i (bc+ad) = A + Bi$, where
\begin{align*}
A + B &= ac - bd +bc +ad\
& \equiv ac - (a+1)(c+1) + (a+1)c +a(c+1)\ 
&\equiv  1 \pmod 2,\
\end{align*}
thus $\alpha \beta$ is odd.

If $\alpha$ is even, then $b\equiv a \pmod 2$, thus
\begin{align*}
A+B &= ac - bd +bc +ad\
&\equiv ac -a (c+1) + ac + a(c+1)\
&\equiv 0 \pmod 2,
\end{align*}
thus $\alpha \beta$ is even, and symmetrically the same is true if $\beta$ is even.

This proves
\begin{center}
$\alpha \beta$  is odd  $\iff \alpha$ and $\beta$ are odd.
\end{center}
\item[(b)] Now $\alpha + \beta = (a+c) + (b+d)i = C + Di$. 
If $\alpha, \beta$ are both even, then
\begin{align*}
C + D &= a + c + b + d\
&\equiv a + c + a + c \equiv 0 \pmod 2.
\end{align*}
If $\alpha + \beta$ are both odd, then
\begin{align*}
C + D &= a + c + b + d\
&\equiv a + c + a + 1  + c+ 1 \equiv 0 \pmod 2.
\end{align*}
If $\alpha$ is odd, and $\beta$ is even, or symmetrically, if $\alpha$ is even and $\beta$ odd,
\begin{align*}
C + D &= a + c + b + d\
&\equiv a + c + a + c + 1 \equiv 0 \pmod 2.
\end{align*}
This proves the equivalence
\begin{center}
$\alpha + \beta$ is even $\iff$ $\alpha,\beta$ are both even or both odd.
\end{center}

\item[(c)] If $\alpha$ is even, then $b \equiv a \pmod 2$, and $1 +i \mid 2 = (1+i)(1-i)$, therefore $b \equiv a \pmod {1+i}$, thus
$$\alpha = a + ib \equiv (1+i)a \equiv 0 \pmod {1+i},$$
thus $1+i \mid \alpha$.

Conversely, if $1+i \mid \alpha$, then $\alpha = \lambda(1+i)$ for some $\lambda \in \Z[i]$. Therefore $N(\alpha) = N(\lambda) N(1+i)$, so that $a^2 +b^2 = \alpha \overline{\alpha} = 2 N(\lambda)$. Thus $a^2+b^2 \equiv 0 \pmod 2$, which proves that $a,b$ have same parity, thus $\alpha = a+bi$ is even. This shows the equivalence
\begin{center}
$\alpha$ is even $\iff$ $1+i$ divides $\alpha$.
\end{center}
\end{proof} 
Note: The equivalence of part (c) gives a shorter proof of parts (a),(b).

Since $\Z[i]/(1+i)\Z[i] \simeq \F_2$ by Exercise 2, for every $\alpha \in \Z[i]$, 
$$\alpha \equiv 0 \pmod {1+i} \quad  	ext{ or }\quad  \alpha \equiv 1 \pmod {1+i}.$$
Therefore, 
\begin{center}
$\alpha$ is even $\iff \alpha \equiv 0 \pmod {1+i}$,\
$\alpha$ is odd \,$\iff \alpha \equiv 1 \pmod {1+i}$.
\end{center}
Then 
\begin{align*}
\alpha \beta 	ext{ is odd} &\iff \alpha \beta \equiv 1 \pmod {1+i}\
&\iff \alpha \equiv 1 	ext{ and } \beta \equiv 1 \pmod{1+i}\
&\iff \alpha 	ext { is odd and } \beta 	ext { is odd.}\
\end{align*}
and similarly,
\begin{align*}
\alpha + \beta 	ext{ is even } &\iff \alpha + \beta \equiv 0 \pmod {1+i}\
&\iff \alpha \equiv \beta \equiv 0 	ext{ or } \alpha \equiv \beta \equiv 1 \pmod {1+i}\
&\iff \alpha,\beta 	ext{ are both even or both odd}.
\end{align*}

Exercise 15.4.4

Answers

Comments

Exercise 15.4.4

Answers

Comments

Add answer