MHB Can $\mathbb{Z}[\sqrt{-3}]$ Be Proven as a Principal Ideal Domain?

[javi410]

Hi,
Im trying to prove that a prime $p\neq 3$ is of the form $p=x^2 + 3y^2$ if $p \equiv 1 \pmod{3}$.

I have think in a prove as follows:
As we know that $-3$ is a quadratic residue mod p, we know that the ideal $(p)$ must divide $(x^2 + 3) = (x + \sqrt{-3})(x - \sqrt{-3})$ in the ring $\mathbb{Z}[\sqrt{-3}]$.
$p$ don't divide any of the factors so it can not be prime in $\mathbb{Z}[\sqrt{-3}]$, so there are ideals $I,J$ of $\mathbb{Z}[\sqrt{-3}]$ such that
\[
(p) = I\cdot J
\]
and, if we prove that $\mathbb{Z}[\sqrt{-3}]$ is a PID, we have the result as we can take elements $u,v$ of $I$ and $J$ such that $u\cdot v$ has norm $p^2$, therefore the norm of $u = a^2 + b\sqrt{-3}$ is $p =a^2 + 3b^2$.

The problem is that I don't know how to prove that $\mathbb{Z}[\sqrt{-3}]$ is a PID.

Thanks!

 

[chisigma]

Hi,
Im trying to prove that a prime $p\neq 3$ is of the form $p=x^2 + 3y^2$ if $p \equiv 1 \pmod{3}$.

I have think in a prove as follows:
As we know that $-3$ is a quadratic residue mod p, we know that the ideal $(p)$ must divide $(x^2 + 3) = (x + \sqrt{-3})(x - \sqrt{-3})$ in the ring $\mathbb{Z}[\sqrt{-3}]$.
$p$ don't divide any of the factors so it can not be prime in $\mathbb{Z}[\sqrt{-3}]$, so there are ideals $I,J$ of $\mathbb{Z}[\sqrt{-3}]$ such that
\[
(p) = I\cdot J
\]
and, if we prove that $\mathbb{Z}[\sqrt{-3}]$ is a PID, we have the result as we can take elements $u,v$ of $I$ and $J$ such that $u\cdot v$ has norm $p^2$, therefore the norm of $u = a^2 + b\sqrt{-3}$ is $p =a^2 + 3b^2$.

The problem is that I don't know how to prove that $\mathbb{Z}[\sqrt{-3}]$ is a PID.

Thanks!

Wellcome on MHB javi410!...

Setting $p= 3 n + a$, where it must be $a=1$ or $a=2$, You have...

$\displaystyle x^{2} + 3\ y^{2} = 3\ n + a \implies x^{2} \equiv a\ \text{mod}\ 3\ (1)$

Now the equation (1) has solution only if $a = 1$...

Kind regards

$\chi$ $\sigma$

 

[mathbalarka]

The problem is that I don't know how to prove that $\mathbb{Z}[\sqrt{-3}]$ is a PID.

You won't be, because $\Bbb Z[\sqrt{-3}]$ is not a principal ideal domain.

Consider $4 = (1 + \sqrt{-3})(1 - \sqrt{-3})$. $2$ divides $4$, but doesn't divide either of $1 \pm \sqrt{-3}$, so it's not a prime. Thus, it's not an unique factorization domain, so cannot possibly be a PID either.