So for the ideal $I = (20+\sqrt{-5})_{R}$ when $K=\mathbb{Q}(\sqrt{-5})$, how do I factorise this into a product of prime ideals. Do you start by taking the norm of $I$ and decomposing it into a product of primes? So $$N(I) = 405 = 3^{4} \times 5.$$ If so, where do I go next?
Factorise an ideal into a product of prime ideals
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$\begingroup$
ring-theory
ideals
prime-factorization
1 Answers
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Recall that the primes $ 3, 5 $ split as $ 3R = \mathfrak p \mathfrak p' $ and $ 5R = (\sqrt{-5})^2 $ in $ R = \mathcal O_K = \mathbf Z[\sqrt{-5}] $. The norm suggests that the ideal $ I = (20 + \sqrt{-5}) $ factors as $ \mathfrak p^i \mathfrak p'^j (\sqrt{-5}) $, where $ i + j = 4 $. It follows upon division that
$$ \mathfrak p^i \mathfrak p'^j = (1 - 4\sqrt{-5}) $$
This is not divisible by $ 3 $, therefore either $ i = 0 $ or $ j = 0 $. The primes lying over $ 3 $ are $ (3, 1 \pm \sqrt{-5}) $; by calculation we have
$$ (3, 1 - \sqrt{-5})^2 = (9, 3 - 3 \sqrt{-5}, -4 - 2\sqrt{-5}) = (9, 7 - \sqrt{-5}, -4 - 2\sqrt{-5}) = (9, 7 - \sqrt{-5}) = (2 + \sqrt{-5}) $$
$$ (2 + \sqrt{-5})^2 = (1 - 4 \sqrt{-5}) $$
It follows that
$$ (20 + \sqrt{-5}) = (3, 1 - \sqrt{-5})^4 (\sqrt{-5}) $$
is the desired factorization.