# If $N(\alpha)$ = $p$ where $p$ is an odd prime, then is ($\alpha, \bar\alpha$) = $\mathbb{G}$?

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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Hint: Note that $\alpha + \bar{\alpha}$ is an integer, but is not divisible by $p.$ –  Geoff Robinson Oct 7 '12 at 22:10
I think in your displayed equation you mean for $\beta$ to be $\overline\alpha$ (or perhaps vice versa). Could you edit, please? –  Gerry Myerson Oct 7 '12 at 22:48
$\mathbb{G}$ is not good notation for the Gaussian integers, which are usually denoted by $\mathbb{Z}[i]$; it looks like you're trying to refer to either the additive or the multiplicative group schemes (en.wikipedia.org/wiki/Group_scheme) $\mathbb{G}_a, \mathbb{G}_m$. Also, since $(\alpha, \bar{\alpha})$ is an ideal, when you want to say that it is all of the Gaussian integers it is usually easier just to say that it is equal to the unit ideal $(1)$. –  Qiaochu Yuan Oct 7 '12 at 22:52
@Qiaochu, there is precedent for using $\bf G$ for the Gaussians. Stewart and Tall use it in their Algebraic Number Theory text. Of course, that book doesn't get anywhere near group schemes. –  Gerry Myerson Oct 8 '12 at 6:29
@Qiaochu, I didn't even know of the multiplicative group schemes - $\mathbb{G}$ is just the notation my professor uses in class. –  Domonic Mei Oct 8 '12 at 16:29

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 $1 \in I$.
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 $\bar\alpha \neq -\alpha$
1 $\in I$ implies $(\alpha,\bar\alpha)$ = $\mathbb{G}$ is something shown in a previous assignment, so I went with that one as the accepted answer. I understand your response though, and thank you! –  Domonic Mei Oct 8 '12 at 16:27
@DJM Since $\alpha$ and $\bar\alpha$ are coprime, $(\alpha, \bar\alpha) = (1)$. –  Makoto Kato Oct 8 '12 at 20:40