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I need to show that there is no greatest common divisor for $4$ and $2 \cdot(1 + \sqrt{-3})$ in the ring $R := \{ x + y \cdot\sqrt{-3} \,\,|\,\, x, y \in \mathbb{Z}\}$.

I found that both $2$ and $(1 + \sqrt{-3})$ divide $4$ as well as $2 \cdot(1 + \sqrt{-3})$. Therefore if $x$ was a greatest common divisor of the two, it has to be divisible by both $2$ and $(1 + \sqrt{-3})$.

How can I proceed from here?

user7802048
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1 Answers1

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Let the common divisor of $4$ and $2(1+\sqrt{-3})$ be $x+y\sqrt{-3}$. Then we have \begin{align} (x+y\sqrt{-3})(a+b\sqrt{-3})&=4,\\ (x+y\sqrt{-3})(c+d\sqrt{-3})&=2(1+\sqrt{-3}), \end{align} where $x,y,a,b,c,d\in\mathbb{Z}$. Multiplying $x-y\sqrt{-3}$ on both sides of the equations, one obtains \begin{align} a&=\frac{4x}{x^2+3y^2},\quad b=-\frac{4y}{x^2+3y^2},\\ c&=\frac{2(x+3y)}{x^2+3y^2},\quad d=\frac{2(x-y)}{x^2+3y^2}. \end{align} The denominators have higher degrees than the numerators, so the number of integer pairs of $(x,y)$ that makes $|a|\geq1$ or $|b|\geq 1$ or $|c|\geq 1$ or $|d|\geq 1$ is finite. One can therefore enumerate all such cases. After ruling out them all, the only possibility left is $a=b=c=d=0$, which cannot hold either (not allowed to have $x=y=0$ as they appear in the denominators), thus completing the proof.

Zhuoran He
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  • How can you divide (by $x^2$, etc) in the first place? Rings don't necessarily provide multiplicative inverses. – user7802048 Sep 29 '17 at 19:31
  • I have $(x^2+3y^2)(a+b\sqrt{-3})=4(x-y\sqrt{-3})$. By comparing the coefficients of 1 and $\sqrt{-3}$, the solutions are given by $a=4x/(x^2+3y^2)$ and $b=-4y/(x^2+3y^2)$ if $a,b\in\mathbb{Z}$ exist. – Zhuoran He Sep 29 '17 at 19:35
  • In a ring no such thing as $(x^2 + 3y^2)^{-1}$ exists. How can this be part of a solution? – user7802048 Sep 30 '17 at 09:13
  • $\mathbb{Z}$ is a ring. If $(x^2+3y^2)a=4x$, we know in general $(x^2+3y^2)$ is not invertible because $1/(x^2+3y^2)\notin\mathbb{Z}$. But if $a\in\mathbb{Z}$ exists, it must equal $a=4x/(x^2+3y^2)$. When abstract algebraic objects confuse us, think of our familiar friends $\mathbb{Z}$ (a ring), $\mathbb{Q},\mathbb{R},\mathbb{C}$ (fields). – Zhuoran He Sep 30 '17 at 17:02
  • Hi! I see it is quite some time since this question was posted, but I was wondering if @ZhuoranHe is still active or if anyone else knows the answer and perhaps want to help... I am simply wondering why we choose to multiplying the equations by $x - y \sqrt{-3}$ on each side? – Rory Feb 07 '22 at 14:41