Verification of a factorization of ideals, $\langle 2 \rangle$

Question

Still going over Alaca & Williams (I might die before I fully understand that book).

In $\mathbb{Z}[\sqrt{-21}]$, the factorization of $\langle 2 \rangle$ is $\langle 2, 1 + \sqrt{-21} \rangle^2$. I can easily see that this ideal contains both the number $1 + \sqrt{-21}$ and the number $1 - \sqrt{-21}$.

In $\mathbb{Z}[\sqrt{23}]$, the factorization of $\langle 2 \rangle$ is $\langle 2, 1 + \sqrt{23} \rangle^2$. But $\mathbb{Z}[\sqrt{23}]$ is a principal ideal domain, which means that $\langle 2, 1 + \sqrt{23} \rangle$ is generated by a single number in this domain. Since $(5 - \sqrt{23})(5 + \sqrt{23}) = 2$, I think that number is $5 + \sqrt{23}$.

But the problem that I'm having is that I don't see how $5 - \sqrt{23}$ is contained in $\langle 5 + \sqrt{23} \rangle$. I tried $$\frac{5 + \sqrt{23}}{5 - \sqrt{23}}$$ on my scientific calculator, but I don't recognize what this number (approximately $0.020842$) could be in this ring.

Answer 1

You've done 99% of the work! Note that $$\frac{5-\sqrt{23}}{5+\sqrt{23}} =\frac{(5-\sqrt{23})^2}{5^2-{23}}=\frac{48-10\sqrt{23}}{2} =24-5\sqrt{23},$$ so $$5-\sqrt{23} = (24-5\sqrt{23})(5+\sqrt{23})\in\langle5+\sqrt{23}\rangle.$$

Answer 2

Since you get $2$ as $(5-\sqrt{23})(5+\sqrt{23})$, you have: \begin{align*}5-\sqrt{23}&=5\cdot 2-(5+\sqrt{23})=5 (5-\sqrt{23})(5+\sqrt{23})-(5+\sqrt{23})\\&=(24-5\sqrt{23})(5+\sqrt{23}).\end{align*}

Answer 3

$$(a+b\sqrt{23})(5+\sqrt{23})=5-\sqrt{23}\implies a+b\sqrt{23}=\frac{48-10\sqrt{23}}{2}=24-5\sqrt{23}$$

and certainly:

$$(24-5\sqrt{23})(5+\sqrt{23})=120-115-\sqrt{23}=5-\sqrt{23}$$

Answer 4

Hint: Since you have $2$ is in the ideal generated by $5+\sqrt23$, $5+\sqrt23-(2\times 5)=-5+\sqrt23$ is also in this ideal.