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I want to show that $\Bbb Z[\sqrt{6}]$ is not a UFD. We know that this is a Dedekind domain. (Hint: For the ideals $\mathfrak{p}=(2,4+\sqrt{6})$, $\mathfrak{q}=(5,4+\sqrt{6})$, compute $\mathfrak{p}^2$, $\mathfrak{q}\bar{\mathfrak{q}}$, $\mathfrak{p}\mathfrak{q}$ and $\mathfrak{p}\bar{\mathfrak{q}}$.)

Note that $\mathfrak{p}^2=(4,8+2\sqrt{6},22+8\sqrt{6})$, $\mathfrak{q}\bar{\mathfrak{q}}=(5, 20-5\sqrt{6},20+5\sqrt{6})$, $\mathfrak{p}\mathfrak{q}=(10,8+2\sqrt{6},20+5\sqrt{6},22+8\sqrt{6})$, and $\mathfrak{p}\bar{\mathfrak{q}}=(10,8-2\sqrt{6},20+5\sqrt{6})$. Also, we observe that $10=(4+\sqrt{6})(4-\sqrt{6})=(-1+\sqrt{6})(2+\sqrt{6})(1+\sqrt{6})(-2+\sqrt{6})$ and $10=2.5=(2+\sqrt{6})(-2+\sqrt{6})(-1+\sqrt{6})(1+\sqrt{6})$. But, this doesn't imply that $\Bbb Z[\sqrt{6}]$ is not a UFD.

I am not sure how to use the hint. Thanks!

user26857
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Ergin Süer
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  • Closely related. – Cameron Buie May 27 '19 at 23:23
  • Note that you can simplify $\mathfrak{p}^2$; for example, $22+8\sqrt{6} = 4(5+2\sqrt{6}) + 2$, so you can replace $22+8\sqrt{6}$ by $2$; which means you actually get $(2)$. Similarly, $\mathfrak{q}\overline{\mathfrak{q}}=(5)$. You need to simplify those products; remember that in a Dedekind domain, every ideal is either principal or $2$-generated. – Arturo Magidin May 27 '19 at 23:53
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    So... $\mathbb{Q}(\sqrt{6})$ has class number $1$. So what exactly is going on here? – Arturo Magidin May 28 '19 at 01:59
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    It’s just that $-2=(2-\sqrt6,)(2+\sqrt6,)$, and $\sqrt6=(2+\sqrt6)(3-\sqrt6)$, so that $\gcd(2,\sqrt6,)=2+\sqrt6$. – Lubin May 28 '19 at 02:23

1 Answers1

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$\mathbb Z[\sqrt 6]$ is indeed a unique factorization domain. Notice, for one thing, that $\langle 2 \rangle$ and $\langle 4 + \sqrt 6 \rangle$ are both contained in $\langle 2 + \sqrt 6 \rangle$. As for $\langle 5, 4 + \sqrt 6 \rangle$, that turns out to be the whole ring.

Maybe you meant $\mathbb Z[\sqrt{-6}]$ is not UFD? In which case the ideal $\langle 2, \sqrt{-6} \rangle$ would be far more pertinent than either $\langle 2, 4 + \sqrt{-6} \rangle$ or $\langle 5, 4 + \sqrt{-6} \rangle$.

Robert Soupe
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