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Here, $N(\alpha)$ stands for the norm of $\alpha\in\mathbb{G}$, $\mathbb{G}$ is the set of Gaussian Integers, and ($\alpha, \bar\alpha$) is the ideal generated by $\alpha$ and $\bar\alpha$. In other words,

$(\alpha, \bar\alpha) = \{\alpha\lambda + \bar\alpha\mu\ | \ \lambda,\mu\in\mathbb{G}\}$

This problem was asked on a sample exam as a true/false question. I ran into the following problem.

When I attempt to prove ($\alpha, \bar\alpha$) = $\mathbb{G}$, I can show that ($\alpha,\bar\alpha$) $\subset$ $\mathbb{G}$. However, I run into a wall when attempting to show the reverse inclusion. This suggests that the statement is false, but I'm not sure.

Hints would be greatly appreciated!

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    @Qiaochu, I didn't even know of the multiplicative group schemes - $\mathbb{G}$ is just the notation my professor uses in class.2012-10-08

2 Answers 2

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Let $\alpha=a+bi$. Then $\alpha+\overline{\alpha}=2a \in I$. Clearly, $(2a,p)=1$. Therefore there exist integers $x,y$ such that $1=x2a+yp$. Since $p=\alpha \overline{\alpha} \in I$, this shows that $1 \in I$.

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Let $\alpha = a + bi$. It is easy to see that $\bar\alpha \neq \epsilon\alpha$, where $\epsilon = 1, -1, i, -i$. Hence $(\alpha) \neq (\bar\alpha)$. Since $(\alpha)$ and $(\bar\alpha)$ are prime ideals, they are coprime. Hence $(\alpha, \bar\alpha) = (1) = \mathbb{G}$.

Remark For example, suppose $\bar\alpha = -\alpha$. $a - bi = -a - bi$. Hence $a = 0$. Hence $N(\alpha) = b^2$. This is a contradiction. Hence $\bar\alpha \neq -\alpha$

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    @DJM Since $\alpha$ and $\bar\alpha$ are coprime, $(\alpha, \bar\alpha) = (1)$.2012-10-08