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$R = \mathbb{Z}\left[\sqrt{-7}\right] \le \mathbb{C}$

In other words, $R$ is the set of all integers and integers multiplied by the square root of $-7$. I believe this might be called $\mathbb{Z}$ adjoin root $-7$ but I am not completely sure.

I am asked to find all units in $R$. In $\mathbb{Z}$, the only units are $1$ and $-1$ which are obviously units of $R$ as well. I know $\mathbb{Z}[i]$ has units $1, -1, i$, and $-i$ but the latter two are not in $R$.

Are there any non-trivial units in this ring? All I can find is $1$ and $-1$, but that seems too simple.

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    $\begingroup$ The set $ \mathbb Z[\sqrt{-7}]$ is the set of all complex numbers of the form $a+b\sqrt{-7}$, with $a,b\in \mathbb Z$. $\endgroup$ – lhf Apr 2 '15 at 1:24
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Hint:

  • Consider $N(a+b\sqrt{-7}) = a^2+7b^2$, which coincides with the norm or absolute value of the complex number $a+b\sqrt{-7}$.

  • Conclude that $N$ is multiplicative.

  • Prove that $N(\alpha)=\pm1$ iff $\alpha$ is a unit in $\mathbb Z[\sqrt{-7}]$.

  • Conclude that you need to solve $a^2+7b^2=\pm 1$ with $a,b\in \mathbb Z$.

  • Solve it.

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  • $\begingroup$ Thank you, but that suggests to me that my trivial answer was right after all. If I have an element of R written as a + b*sqrt(-7), then its norm is a^2 + 7b^2, which is clearly an integer as a and b are both integers. So say I have two elements of R, x and y, and they are both units. Then abs(xy) = abs(x)*abs(y) = 1. So I need two integers abs(x) and abs(y) such that their product is 1. The only solutions are abs(x) = abs(y) = 1. And if b cannot 0 this would force the norm to be greater than 1, b must be zero. Thank you for reminding me of the technique of using norms to find units. $\endgroup$ – ocyeung Apr 2 '15 at 1:44
  • $\begingroup$ Yes, imaginary quadratic rings have only trivial units, except for two cases ($-1$ and $-3$ instead of $-7$). $\endgroup$ – lhf Apr 2 '15 at 2:36

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