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Show that the ideal $ I = (2, 1 + \sqrt{-7} ) $ in $ \mathbb{Z} [\sqrt{-7} ] $ is not principal.

My thoughts so far:

Work by contradiction. Assume that $ I $ is principal, i.e. that it is generated by some element $ z = a + b\sqrt{-7} \in \mathbb{Z}[\sqrt{-7}] $. I'm really not sure what to consider though - I can't really 'see' what $ I $ looks like.

Any help would be greatly appreciated. Thanks

  • 3
    Maybe this [answer](http://math.stackexchange.com/questions/32577/show-that-a-specific-ideal-is-not-principal/32585#32585) can help you since it is a similar problem.2011-05-06
  • 2
    Let me point out that the class number of $K=\mathbb{Q}(\sqrt{-7})$ is 1, so the ring of integers $\mathcal{O}_K$ is a PID. However, $\mathbb{Z}[\sqrt{-7}]$ is only an order of conductor $2$ in $\mathcal{O}_K$.2011-05-12

4 Answers 4