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We are in the ring $ \mathbb{Z}[\sqrt{-5}]$

$$ 6 = 2\times3=(1+\sqrt{-5}) (1-\sqrt{-5}) $$

My question is how to show $2,3,(1+\sqrt{-5}), (1-\sqrt{-5}) $ are irreducible elements of the ring.

Mark Bennet
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SamC
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  • I corrected $\mathbb Z[\sqrt{-6}]$ to $\mathbb Z[\sqrt{-5}]$ in the first statement in the question to make it consistent with the rest. If this is wrong, do edit again. – Mark Bennet Mar 30 '14 at 12:25

1 Answers1

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Look at the norm on the ring,

$$N(x+y\sqrt{-5}) = x^2 + 5y^2.$$

You have $N(a\cdot b) = N(a)\cdot N(b)$. That constrains the possible divisors, and allows you to conclude that all divisors of e.g. $1+\sqrt{-5}$ are units or associated to $1+\sqrt{-5}$.

Daniel Fischer
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  • Thanks :) Is this a general way to show something is irreducible ? – SamC Mar 30 '14 at 12:25
  • In rings where you have a ($\mathbb{Z}$-valued, or with values in another nice enough similar ring) norm (that you can reasonably compute), yes. It's of course only practical when the norm of the elements in question has few enough prime divisors. – Daniel Fischer Mar 30 '14 at 12:29
  • @Daniel Only in certain number rings does irreducibility depend only on the norm. But the monoid of norms does reflect many of the properties of the number ring - see this post for further discussion and references. – Bill Dubuque Mar 30 '14 at 14:29
  • @Bill Of course it does in general not depend only on the norm. But the prime factorisation of the norm considerably restricts the search space, that's the point. – Daniel Fischer Mar 30 '14 at 16:03
  • @SamC My prior comment was meant for you. Daniel: one can say much more - see the papers I reference there. – Bill Dubuque Mar 30 '14 at 16:25