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Let $P_3=${$z\in Z[\sqrt{-3}],|z|^2 \text { is a prime number}, >3$}

Let $\alpha_1,...\alpha_n$ be distinct elements in $P_3$, and $l_1,...l_n\in Z^+$. Set $\alpha_1^{l_1}...\alpha_n^{l_n}=a+b\sqrt{-3}$.

Is it correct that $|\alpha_i|^2,a,b$ are mutual prime for all $i$?

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    $\begingroup$ $7$ is in $P_3$, since $|7|=7$ is prime and greater than $3$. Let $n=1$, $\alpha_1=7$, $\ell_1=1$, then $|\alpha_1|^2=49$, $a=7$, $b=0$, counterexample. Maybe you had something else in mind? $\endgroup$
    – Gerry Myerson
    Commented Jun 23, 2013 at 11:06
  • $\begingroup$ Sorry, I forgot to add that P do not contain intergers. Thanks. $\endgroup$
    – Kim Jun-Lee
    Commented Jun 23, 2013 at 15:06
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    $\begingroup$ I suspect the poster means the norm inherited from the field extension, rather than the absolute value, i.e. $|a+b\sqrt{-3}| = a^2+3b^2|$, akin to the Gaussian integers. That way the set $P_3$ corresponds to "primes", and the question is about unique factorization in this "deformation". Unfortunately for the poster, this is not the ring of integers for this extension; instead one should consider the Eisenstein integers, where the statement holds after accounting for units. As stated, the question can be resolved by taking two conjugates as the distinct elements, e.g. $2\pm \sqrt{-3}$. $\endgroup$
    – Zack Wolske
    Commented Jun 23, 2013 at 15:30

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