I have the ring $R = \mathbb{Z}[i\sqrt{7}]$. I need to find (if exists) $\gcd(-5+3i\sqrt{7},10+2i\sqrt{7})$ and all invertible elements of $R$.

I said that gcd dosen't exist because $R$ is not an euclidian domain, but I'm not sure how to prove it. And I don't know how to find its invertible elements. But in general I want to know how should I prove if gcd can be found or not.


Let $x = a + b\sqrt{-7} \in R$. You can define the norm of $x$ by $N(x) = x \cdot \bar x = a^2 + 7b^2$. You can prove that the norm is multiplicative.

This means that if $x$ is invertible in $R$, then $N(x)N(x^{-1}) = 1$, so $N(x) = 1$. What are the possible $x$'s?

For your gcd problem, if there is a nontrivial $x$ such that $x$ divides both $-5 + 3\sqrt{-7}$ and $10 + 2\sqrt{-7}$, then $N(x)$ divides both $N(-5 + 3\sqrt{-7}) = 88$ and $N(10 + 2\sqrt{-7}) = 128$. So $N(x)$ divides $8$. What are the possible $x$'s? There are not much $x$ in this list. You can test which one can be the gcd.

Remark: Under the "natural norm" $N$ defined above, we can say that $R$ is not an Euclidean domain, by constructing $a, b \in R$ such that the division $a = bq + r$ is not possible. On the complex plane $\mathbb C$, consider the rectangle with vertex $0, 1, 1+\sqrt{-7}$ and $\sqrt{-7}$. Since the diagonal is of length $\sqrt{8} \approx 2.8$, there is a point $z \in \mathbb Q + \sqrt{-7}\mathbb Q$ such that the distance of $z$ to any vertex is $> 1$. Let $b \in \mathbb N$ so that $a = bz \in R$. Then you will see that the Euclid division of $a$ by $b$ does not exist.

  • $\begingroup$ So If $R$ is not an Euclidian Domain can we discuss about gcd on $R$? $\endgroup$ – Raducu Mihai Jan 13 '18 at 21:40
  • $\begingroup$ Yes. We still can say $c \ | \ a$ if $a = qc$ for some $q$. And a gcd of $a$ and $b$ is an element $d$ such that (i) $d \ | \ a$ and $d \ | \ b$; and (ii) for any $d'$ such that $d' \ | \ a$ and $d' \ | \ b$, $d \ | \ d'$. So it need not be unique (can be up to a unit), but is still defined. $\endgroup$ – Hw Chu Jan 13 '18 at 21:45
  • $\begingroup$ Ok, and one more thing. At the gcd problem, shouldn't I find an x in $R$ such that its norm is 8. Or why should its norm divide 8 instead of being 8? $\endgroup$ – Raducu Mihai Jan 13 '18 at 21:56
  • $\begingroup$ Because the norm of the gcd can actually be smaller. For instance, the gcd of $1 + \sqrt{-7}$ and $1 - \sqrt{-7}$ is 1 since $1 + \sqrt{-7}$ is not divisible by $1 - \sqrt{-7}$ or 2 (just divide them in $\mathbb Q[\sqrt{-7}]$ and verify that the quotient is not in $R$). $\endgroup$ – Hw Chu Jan 13 '18 at 21:59
  • $\begingroup$ Ok so In my case the gcd is 2? $\endgroup$ – Raducu Mihai Jan 13 '18 at 22:01

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