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Suppose $p$ be an odd prime such that $p \equiv 1 \pmod 4$. In the ring of Gaussian integers $\mathbb{Z}[i]$, $p$ factors as $p = v \cdot \overline{v}$ for a prime $v \in \mathbb{Z}[i].$

Prove that $v$ and $\overline{v}$ are not associates. i.e. prove that the two ideals $(v)$ and $(\overline{v})$ are distinct.

If $v = a + bi,$ then direct computation yields: $ v \cdot 1 \ne \overline{v}, \hspace{1mm} v \cdot -1 \ne \overline{v},\hspace{1mm} v \cdot i \ne \overline{v}, \hspace{1mm}v \cdot -i \ne \overline{v}.$ Thus $v$ and $\overline{v}$ are not associates.

$($an explicit calculation would be $v \cdot -i = (a+bi) \cdot -i = b - ai \ne a - bi)$

Am I missing any details?

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  • $\begingroup$ You never use that $p$ is odd, and it is not true for $p=2$. $\endgroup$ – Thomas Andrews May 11 '15 at 23:23
  • $\begingroup$ Thank you for the feedback. Where would I use that $p$ is odd? I am not sure where to implement this assumption $\endgroup$ – St Vincent May 11 '15 at 23:34
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$ p = v\bar v\, \Rightarrow\ \color{#c00}{p} \in (v).\,\ v = a\!+\!bi,\ (v) = (\bar v)\, \Rightarrow\ v\pm \bar v=\color{#0a0}{2a},2bi \in (v)\,\Rightarrow\, \color{#0a0}{2b}\in (v).$

Therefore $\ \gcd(\color{#c00}p,\color{#0a0}{2a,2b}) \in (v)\, \Rightarrow\, p\mid a,b\,\Rightarrow\ p^2\mid a^2+b^2 = p\,\Rightarrow\!\Leftarrow$

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  • $\begingroup$ I understand from you work that $v \mid p, 2a, 2b \Longrightarrow v \mid \gcd(p,2a,2b) \Longrightarrow \gcd(a,2a,2b) \in (v).$ But why does $\gcd(p,2a,2b) \in (v) \Longrightarrow p \mid a,b?$ $\endgroup$ – St Vincent May 12 '15 at 4:20
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    $\begingroup$ @StVincent Else e.g odd $\,p\,$ is coprime to $\,2a\,$ so $\,(v)\supseteq (\color{#c00}p,\color{#0a0}{2a}) = 1\,$ by Bezout. In the same way, if an ideal $\ne 1$ contains a prime integer then every integer in the ideal must be divisible by that prime. $\endgroup$ – Bill Dubuque May 12 '15 at 13:05
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If $v=a+\mathrm i\mkern1mu b$ and $\bar v$ were associates, this would mean that: $$a-\mathrm i\mkern1mu b=\begin{cases}\pm(a+\mathrm i\mkern1mu b)\\\pm\mathrm i\mkern1mu (a+\mathrm i\mkern1mu b) \end{cases}$$ since the only units in $\mathbf Z[\,\mathrm i\,]$ are $\,,\pm1\,$ and $\,\pm\mathrm i$.

The first two cases imply $v=a$ or $v=\mathrm i\mkern1mu b$, which both imply $p$ is a square in $\mathbf Z$.

The last two cases imply $b=\pm a$, and then $$p=v\bar v\iff p=2a^2$$ which is also impossible.

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  • $\begingroup$ So these both contradict our hypothesis that $p$ can be written as $a^2+b^2$? $\endgroup$ – St Vincent May 12 '15 at 4:33
  • $\begingroup$ Not exactly: they contradict the fact that $p$ is prime. $\endgroup$ – Bernard May 12 '15 at 7:16

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