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$R=\Bbb Z[\sqrt{-1}], I=\langle 13\rangle \cap \langle 12-5\sqrt{-1}\rangle$

Problem : represent $\langle 13\rangle \cap \langle 12-5\sqrt{-1}\rangle $ to principal ideal.

The following is my progress.

$\langle 13\rangle\supset \langle a\rangle$ and $\langle 12-5\sqrt{-1}\rangle\supset \langle a\rangle$

So $13|a\land12-5\sqrt{-1}|a\rightarrow 13b=a, (12-5\sqrt{-1})c=a\rightarrow 169b\bar b=a\bar a, 169c\bar c=a\bar a.$

$\langle 169\rangle \supset\langle a\rangle$ so $b,c\neq0, a|169.$ $169|a\bar a$

I'm stuck here..How can I prove whether $I$ is principal ideal and, if so, $I=\langle ? \rangle$

Arbitrary
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1 Answers1

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Since $\mathbb Z[i]$ is an euclidean domain, you can compute greatest common divisors easily, and then $$ \langle a\rangle \cap \langle b\rangle = \left< {ab}/{\gcd(a,b)} \right >$$

and $\gcd(13, 12-5i)$ is $3+2i$ ...

hmakholm left over Monica
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