Let $I=(2,1+\sqrt{-5}), J=(3,1-\sqrt{-5}) \trianglelefteq \mathbb{Z}[\sqrt{-5}]$, show $IJ=(1-\sqrt{-5})$

Question

Let $R=\mathbb{Z}[\sqrt{-5}]$

1. What is $R^\times$
2. Let $I=(2,1+\sqrt{-5})$, is $I$ a principal ideal in $R$?
3. Let $J=(3,1-\sqrt{-5})$, prove $I+J=R$
4. Prove $IJ=(1-\sqrt{-5})$.

I was able to prove 1. to 3. but I'm somewhat stuck at 4.

I know since $I+J=R$ that $I, J$ are coprime and $IJ=I\cap J$. Which would simplify the last statement as "prove $I\cap J=(1-\sqrt{-5})$"

It's easy enough to prove $(1-\sqrt{-5})\subseteq I\cap J$ but the other way around seems a lot harder.

What I've tried

I've tried taking $z\in I \cap J$ and showing how $z\in (1-\sqrt{-5})$.

Let $z\in I\cap J$ then $z=2r_1+(1+\sqrt{-5})s_1$ for certain $r_1,s_1\in \mathbb{Z}[\sqrt{-5}]$, then $z= 2a_1+c_1-\sqrt{5}d_1 + i(2b_1+d_1+\sqrt{5}c_1)$ for $r_1=a_1+b_1\sqrt{-5}, s_1=c_1+d_1\sqrt{-5}$. The same is valid for $z\in J$ which implies something like this: $$ \begin{align*} z&= 2a_1+c_1-\sqrt{5}d_1 + i(2b_1+d_1+\sqrt{5}c_1)\\ &= 3a_2+c_2+\sqrt{5}d_2 + i(3b_2+d_2-\sqrt{5}c_2) \end{align*} $$

I would like to prove how $z$ is of the form $(1-\sqrt{5}i)(A+\sqrt{5}Bi) = A+\sqrt{5}B + i(-\sqrt{5}A+B)$. In the equations above this would imply $\color{red}{2a_1+c_1= -c_1}$ and $\color{blue}{2b_1+d_1=-d_1}$

the condition that $z\in I\cap J$ should reveal this (I guess) From the equations above I deduce: $-d_1=d_2$ and $-c_1=c_2$ which leads to: $$ \begin{align*} \color{red}{2a_1+c_1} &= 3a_2\color{red}{-c_1}\\ \color{blue}{2b_1+d_1} &= 3b_2\color{blue}{-d_1} \end{align*} $$ This seems pretty close, but I'm not there yet. I also don't see how to get there... I need to use the condition $z\in I\cap J$ more, but how?

Answer 1

I'm not sure how to continue your method. I think it's actually easier here to use $IJ$ rather than $I \cap J$.

First, note that $I$ can also be expressed as $I = (2, 1 - \sqrt{-5})$. Then $IJ$ is generated by elements of the form $(2r_1 + (1 - \sqrt{-5})s_1)(3r_2 + (1 - \sqrt{-5})s_2)$. Keeping in mind that $(1 - \sqrt{-5})|(2\cdot3)$, we can show that these elements are all divisible by $1 - \sqrt{-5}$.

Answer 2

I think that considering $I\cap J$ is doing it the hard way.

You need to prove that $IJ=\left<1-\sqrt{-5}\right>$, that is both $IJ\subseteq\left<1-\sqrt{-5}\right>$ and $IJ\supseteq\left<1-\sqrt{-5}\right>$. Now $$IJ=\left<6,2(1-\sqrt{-5}),3(1+\sqrt{-5}),(1-\sqrt{-5})(1+\sqrt{-5})\right>.$$

To prove $IJ\subseteq\left<1-\sqrt{-5}\right>$, all you have to do is show that all these generators ($6$, $2(1-\sqrt{-5})$ etc.) are divisible by $1-\sqrt{-5}$ in $R$.

To prove $IJ\supseteq\left<1-\sqrt{-5}\right>$, you need to find $\alpha,\beta,\ldots\in R$ such that $$1-\sqrt{-5}=6\alpha+2(1-\sqrt{-5})\beta+\cdots.$$

Answer 3

we have $(2,1+\sqrt{-5})=(2,1-\sqrt{-5})$

$IJ=(2,1+\sqrt{-5})(3,1-\sqrt{-5})=(2,1-\sqrt{-5})(3,1-\sqrt{-5})$

$=(6,2(1-\sqrt{-5}),3(1-\sqrt{-5}),(1-\sqrt{-5})^2)$; $6=(1+\sqrt{-5})(1-\sqrt{-5})$

$=(1-\sqrt{-5})(1+\sqrt{-5},2,3,1-\sqrt{-5})=(1-\sqrt{-5})(1)=(1-\sqrt{-5})$.

For more details, you can see the book: Introductory algebraic number theory, Saban Alaca, Kenneth Williams, p202-203.