What Are the Units in the Ring $\mathbb{Z}[\sqrt{-7}]$?

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SUMMARY

The units in the ring $\mathbb{Z}[\sqrt{-7}]$ are definitively identified as $\pm 1$. This conclusion is reached by analyzing the product of two elements in the form $a + b\sqrt{-7}$ and $c + d\sqrt{-7}$, leading to the equation $(a^2 + 7b^2)(c^2 + 7d^2) = 1$. The only integer solutions for this equation occur when both $a$ and $c$ equal either 1 or -1, confirming that the only units are indeed $\pm 1$.

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I have to find the units in $\mathbb{Z}[\sqrt{-7}]$ so take $x,y\in\mathbb{Z}[\sqrt{-7}]$ then x has the form $a+b\sqrt{-7}$ and y has the form $c+d\sqrt{-7}$ for arbitrary x and y.

If $(a+b\sqrt{-7})(c+d\sqrt{-7})=1$ then $(a-b\sqrt{-7})(c-d\sqrt{-7})=1$ and so:

$1=(a+b\sqrt{-7})(c+d\sqrt{-7})(a-b\sqrt{-7})(c-d\sqrt{-7})=1$ which gives $1=(a^2+7b^2)(c^2+7d^2)$ so $b=d=0$ and $1=a^2c^2$ so either $a=c=1$ or $a=c=-1$ and these are the only units in $\mathbb{Z}[\sqrt{-7}]$

Is the above correct? (Sorry for so many of these type of questions)

Thanks very much for any help
 
Last edited:
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Re: Finding the units in $\mathbb{Z}[\sqrt{-7}]$

hmmm16 said:
Is the above correct?
Yes.
 

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