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While studying Arithmetics, I came across an interesting theorem on the properties of primes in a quadratic extension $K$ of the ring $\mathbb{Z}$. The theorem writes as follows:

Let $x\in K$. If $N(x)$ is prime, then $x$ is prime.

The provided proof of the theorem states:

Suppose that $x=yz$, $y,z\in K$. (If $N(x)$ is prime) Then $N(x)=N(y)N(z)$, so at least one of $N(y)$, $N(z)$ equals $\pm1$, i.e. either $y$ or $z$ is a unit, while the other one is (by definition) adjoint to $x$.

However, this theorem doesn't appear to make sense to me due to the following points:

  • Firstly, for a nonunit, nonzero number $x\in K$ to be prime, its only divisors must be the units of $K$ and elements adjoint to itself.

  • Secondly, by definition, $N(x)=x\overline{x}$.

  • Lastly, given that $N(x)$ is always the product of two other numbers in $K$, how can it be that $N(x)$ is ever a prime? For every prime $x$, $N(x)$ cannot be prime, as $x$ is one of its divisors, right?

I'm very confused by this Theorem and would appreciate some insight on it. Thanks! :)

Bill Dubuque
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Braxxla_13
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    The condition is $N(x)$ is a prime in $\mathbb{Z}$, not the overring, so $N(x)=x\bar{x}$ does not necessarily say anything about primality of $N(x)$ in $\mathbb{Z}$. – user10354138 Sep 09 '23 at 16:29
  • So, for example, given that $N(1+i)=2$, I could infer from this theorem that, since 2 is prime in $\mathbb{Z}$, so is $1+i$ in $\mathbb{Z}[i]$? – Braxxla_13 Sep 09 '23 at 17:16
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    Yes, indeed, $1+i$ is prime, because its norm is a prime number in $\Bbb Z$. Have a look here, and at many similar posts. – Dietrich Burde Sep 09 '23 at 17:48
  • The close votes here are very bizarre. In what way does this question need more focus?? – Eric Wofsey Sep 12 '23 at 20:37
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    The argument is flawed, but for another reason: irreducible does not imply prime, even in quadratic integer rings. – Lukas Heger Sep 13 '23 at 17:56

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