0
$\begingroup$

Can you give an example of a nonunique factorization in $\mathbb{Z}[e^{2\pi i/23}]$?

Can you write two different decompositions into irreducibles in $\mathbb{Z}[e^{2\pi i/23}]$?

  • 0
    I think $(1 - \sqrt{-23})(1 + \sqrt{-23}) = 24 = 2^3 \cdot 3$ works. You still have to show the factors in the two factorizations are relatively prime, but they are. – 2017-01-16
  • 0
    @SpamIAm yes but how do you know that $(1+\sqrt{-23})$ is irreducible? – 2017-01-16
  • 0
    @Mathmo123 You're right, I haven't shown that it's irreducible. But I did check that it's relatively prime to $2$ and $3$, so if it is reducible, its irreducible factors will work. – 2017-01-16
  • 0
    Is $\sqrt{-23}$ in $\mathbb{Z}[e^{2\pi i/23}]$? – 2017-01-17
  • 0
    @kato87 Yes, for any odd prime $\ell$, $\sqrt{\ell^*} \in \mathbb{Z}[\zeta_\ell]$, where $\ell^* = (-1)^{(\ell-1)/2} \ell = \left(\frac{-1}{\ell}\right) \ell$. See p. 51 of Neukirch. – 2017-01-17

1 Answers 1

2

An explicit counterexample is given by $6533$, which has two different decompositions, namely $$ 6533=47\cdot 139= (1 - α + α^{21})(1 - α^2 + α^{21})\cdot \ldots \cdot(1 - α^{21} + α^{21}), $$ for a certain $\alpha\in \mathbb{Z}[\zeta_{23}]$. For details see the proof here, which is taken from Edwards book.

Another proof goes as follows: Since the ring of integers $\Bbb{Z}[e^{2\pi i/23}]$ in $\mathbb{Q}(\zeta_{23})$ is a UFD if and only if it is a PID, it is enough to see that $\Bbb{Z}[e^{2\pi i/23}]$ is not a PID. In fact, the ideal $Q = (2, \frac{1 + \sqrt{-23}}{2})$ is not principal in $\Bbb{Z}[e^{2\pi i/23}]$. This also leads to an explicit counterexample for unique factorization.

  • 0
    Quite nice, thank you – 2017-01-18