Why is the polynomial ring $R[x]$ not a unique factorization domain, where $R$ is the quadratic integer ring $\mathbb{Z}[2\sqrt{2}]$?
I'm trying to find an irreducible non-prime element or something but I don't know where to start.
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Pedro
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user220694
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It is enough to show that $\mathbb{Z}[2\sqrt{2}]$ is not a unique factorisation domain (why?).
The elements $2$ and $2\sqrt{2}$ are irreducible and $$ 8 = (2\sqrt{2})^2 = 2^3, $$ so the factorisation is not unique.
MichalisN
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2How would one show that those two elements are irreducible in $\mathbb{Z}[2\sqrt{2}]$? – user220694 Mar 06 '15 at 13:23
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1Try to solve $2= ab$ for $a,b\in\mathbb{Z}[2\sqrt{2}]$. You should use that $\sqrt{2}$ and $1$ are linearly independent over $\mathbb{Q}$. Do the same for $2\sqrt{2}$. – MichalisN Mar 06 '15 at 14:14
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Hint $ $ Over $\, R = \Bbb Z[w],\ w =2\sqrt{2},\,$ the proper fraction $\,w/2\,$ is a root of monic $\,x^2-2,\,$ contra the Rational Root Test (which is true in any UFD; i.e. UFDs are integrally closed).
Bill Dubuque
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Nice answer! Do you know what we can tell about $R[x]$ (is it a UFD), as asked in the question? – Watson Aug 23 '16 at 11:50
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2@Watson $\ R,$ not integrally closed $,\Rightarrow R,$ not UFD $,\Rightarrow, R[x]$ not UFD $\ $ – Bill Dubuque Aug 23 '16 at 14:04