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Is $\mathbb Z[ \sqrt{-3}]$ UFD ? If so , why $$4=2 \times 2 = (1+\sqrt{-3}) \times (1-\sqrt{-3} )$$ and every terms are irreducible ?

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    $\begingroup$ Every PID is UFD, and in a PID every irreducible is prime. $\endgroup$
    – Hassan Muhammad
    Nov 18, 2011 at 16:25

2 Answers 2

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I think your suspicion is correct. On the other hand, the integral closure of $\mathbf{Z}$ in $\mathbf{Q}(\sqrt{-3})$ is the larger ring $\mathbf{Z}[\frac{1 + \sqrt{-3}}{2}]$, since $-3 \equiv 1 \bmod{4}$. This is norm-Euclidean and in particular a UFD. Note that all UFDs are integrally closed, so this is another way to see that $\mathbf{Z}[\sqrt{-3}]$ does not have unique factorization.

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  • $\begingroup$ Thank you ! It's rather what I want . $\endgroup$
    – Mahan
    Nov 18, 2011 at 17:31
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FYI, $\mathbb{Z}[\sqrt{-3}]$ is not only not a UFD, but it's the unique imaginary order of a quadratic ring of algebraic integers that has the half-factorial property (Theorem 2.3)--ie any two factorizations of a nonzero nonunit have the same number of irreducibles.

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  • $\begingroup$ interesting theorem ! thanks for your useful comment . $\endgroup$
    – Mahan
    Nov 18, 2011 at 17:29
  • $\begingroup$ You're most welcome. $\endgroup$
    – user5137
    Nov 18, 2011 at 17:31

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