Exercise 5.23

Question

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

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

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

ewcommand{
}{\mathrm{N}}

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

Suppose that $p \equiv 1 \pmod 4$. Show that there exist integers $s$ and $t$ such that $pt = 1 + s^2$.  Conclude that $p$ is not a prime in $\Z[i]$. Remember that $\Z[i]$ has unique factorization.

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

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

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

ewcommand{
}{\mathrm{N}}

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

\begin{proof}
As  $p \equiv 1 \pmod 4$, then $\legendre{-1}{p} = (-1)^{\frac{p-1}{2}} = 1$, thus $-1$ is a square modulo $p$.
 
 So  $-1 \equiv s^2 \pmod p,  s \in \mathbb{Z}$. Therefore
  there exist $s \in \mathbb{Z}, t \in \mathbb{Z}$ such that $pt = 1+s^2$.
 
In $\mathbb{Z}[i]$, $p \vert (s+i)(s-i)$.

If $p$ was a prime in $\mathbb{Z}[i]$, then  $p \mid s+i$ ou $p \mid s-i$.
  
 This implies $s \pm i = (a+bi)p, (a,b) \in \mathbb{Z}^2$, thus $\pm1 = b p , p\mid 1$ : it's impossible.
 
Conclusion : if $p \equiv 1\pmod 4$, $p$ is not a prime in  $\mathbb{Z}[i]$.
\end{proof}

Exercise 5.23

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Exercise 5.23

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