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Suppose we are working in $\Bbb{Q}(\sqrt{-41})$. Given a ideal, for example $(2-\sqrt{-41})$ (we especially work in $\Bbb{Z}[\omega_{-41}]$). We know that this is a Dedekind ring, thus we have unique factorization of ideals. I started to take the norm of this ideal and I get $45$ and this is equal to $5\cdot3\cdot3$. Can I say something like this:

$(2-\sqrt{-41})=(5)(3)(3)=(5,2+\sqrt{-41})(5,2-\sqrt{-41})(3,1+\sqrt{-41})^2(3,1-\sqrt{-41})^2$

The last step is deduced from the fact that the ideals $(3)$ and $(5)$ are not prime. Is this true?! Suppose this is true. Then we can do this also for $(3+\sqrt{-41})$ and we get this result:

$$(3+\sqrt{-41})=(5)(5)(2)=(5,2+\sqrt{-41})^2(5,2-\sqrt{-41})^2(2,1+\sqrt{-41})^2$$

Here we get the last step from the fact that $-41\equiv3\ \mod\ 4$. Is this all true? But how to do this for $(2-\sqrt{-41},3+\sqrt{-41})$? I hope someone can help me?!

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    $(5)$ does not have norm $5$ in $\Bbb Z[\sqrt{-41}]$. However, the norm being $45$ does mean that some prime (and only one) lying over $5$ does appear in the factorization of $(2-\sqrt{-41})$.2012-12-09

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