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gcd of $a=6(1-\sqrt{-5})$ and $b=3(1+\sqrt{-5})(1-\sqrt{-5})$doesn't exist in $\Bbb Z[\sqrt{-5}]$.

Here we can see that $N(a)=216$ and $N(b)=324$. On a contrary if the gcd exists and is $d$ say then $N(d)|\gcd(216,324)=108$.

So if we have to find out all the elements whose norm is less than 108 and then divisor of $108$ and then we have to find whether actually divide $a,b$ or not by rationalizing the denominator and then showing by calculation that for every one of the common divisors $x$ of $a$ and $b$ in our list, there is a common divisor $y$ in our list such that $y$ does not divide $x$.

Our list of common divisors will not be short, so this task will take longer. Is there any short proof?

Ri-Li
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1 Answers1

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If the gcd of $a$ and $b$ exists, say it is $c$. Suppose $d$ is some common divisor of $a$ and $b$. Then $d$ divides $c$ as well, so we can define $a=da',b=db',c=dc'$. Now $c'$ divides both $a'$ and $b'$ as we are in an integral domain. We show that $c'$ is actually their gcd.

For a common divisor $d'$ of $a'$ and $b'$, we have that $dd'$ is a common divisor of $a$ and $b$. So $dd'$ divides $c$. Therefore $d'$ divides $c'$. The proof is hence finished.

Now in our case, notice that as $6$ divides both $a$ and $b$. If there's a gcd $c$ of them, then $a'=a/6, b'=b/6$ also has a gcd $c'$ by the above analysis. Here easily calculate $a'=1-\sqrt{-5}$ and $b'=3$. Calculating norm gives that $N(a')=6$ and $N(b')=9$. Therefore $N(c')$ divides $3$. Is this possible? The only possible case is $c'=1$. So $c=6$.

However $3(1-\sqrt{-5})$ is a common divisor of $a$ and $b$, and one could easily find out (e.g. by norm) that it doesn't divide $6$.

August Liu
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