Let $R=\mathbb{Z}\left[\sqrt{−5}\right]$. Let $I=\left(2,1+\sqrt{−5}\right)$. Show $I^2 = (2)$ in $R$.

Question

I don't know how to solve this problem. I tried to write out the sum of products in the definition of products of ideals, but that doesn't seem to lead anywhere.

Answer 1

The fact that $(2) \subseteq I^2$ follows immediatelly, as $2 = (1+\sqrt{-5})(1-\sqrt{-5}) - 2\cdot 2 \in I^2$

On the other hand we have that an element of $I^2$ is of the form $\sum_{finite} ab$, s.t. $a,b \in I$. Now we have $a = 2A_1 + (1+\sqrt{-5})A_2$ and $b = 2B_1 + (1+\sqrt{-5})B_2$ as $2$ and $1+\sqrt{-5}$ are generators of $I$. Now we have:

$$ab = 2^2A_1B_1 + 2(1+\sqrt{-5})(A_1B_2 + A_2B_1) + (1+\sqrt{-5})A_2B_2$$ $$ = 2\left(2A_1B_1 + (1+\sqrt{-5})(A_1B_2 + A_2B_1) + 3A_2B_2\right) \in (2)$$

So as each term in the sum is in $(2)$, so is their sum. Hence $I^2 \subseteq (2)$ and so $I^2 = (2)$

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In general it holds that if $I = (a,b)$, then $I^2 = (a^2,ab,b^2)$. Using this lemma we have that:

$$I^2 = \left(4,2(1+\sqrt{-5}),2\sqrt{-5} - 4\right) = \left(4,2(1+\sqrt{-5}),2\sqrt{-5}\right) = \left(4,2,2\sqrt{-5}\right)= (2)$$

Answer 2

Observe that

$$2^2=4\;,\;\;(1+\sqrt{-5})^2=-4+2\sqrt{-5}=2(-2+\sqrt{-5})$$

so

$$I^2\subset\langle\,4,\,2(1+\sqrt{-5})\,\rangle=\langle2\rangle$$

since both generators on the left are contained in the right hand.

The other containment is also true ...can you do it?

Answer 3

Note $\,\ (2,1\!+\!w)^2 =\, (2)\ \ $ when $\ \ w^2 =\, \color{#c00}{4n}\!-\!1\,\,\ $ [e.g. $\ w^2=-5\ $ if $\ n=-1$]

since $\smash[b]{\,\ (2,1\!+\!w)^2 =\, (4,2\!+\!2w,\color{#c00}{4n}\!+\!2w)\, =\, 2\!\!\!\!\!\!\underbrace{(\color{#0a0}2,1\!+\!w,2n\!+\!w)}_{\large\quad \color{#0a0}{2n}+1+w-(2n\,+\,w)\:=\,1}\!\!\!\!\! \!\!\!= (2)}$

by using $\ \ (a,\,b)^2 =\, (a^2,\ ab,\ b^2)$