Why is $(2, 1+\sqrt{-5})$ not principal?

Why is $(2, 1+\sqrt{-5})$ not principal in $\mathbb{Z}[\sqrt{-5}]$?

Say $(2,1+\sqrt{-5})=(\alpha)$, then since $2\in(2,1+\sqrt{-5})$ we have $2\in (\alpha)$, so $\alpha\mid2$ in $\mathbb Z[\sqrt{-5}]$. Writing $2=\alpha\beta$ in $\mathbb Z[\sqrt{-5}]$ and taking norms, $4=N(\alpha)N(\beta)$ in $\mathbb Z$, so $N(\alpha)\mid4$ in $\mathbb Z$. Similarly, since $\sqrt{-5}\in(\alpha)$ we get $N(\alpha)\mid5$, thus $N(\alpha)$ is a common divisor of $4$ and $5$, therefore $N(\alpha)=(1)$, so $\alpha$ is a unit. But that means $1\in (2,1+\sqrt{-5})=(\alpha)$, so it must be the whole ring, but it cannot to reach a contradiction, so how can I find an element, which is not in $(2, 1+\sqrt{-5})$.

Or is there a simpler method ? (Maybe not UFD would imply not PID)

Solutions Collecting From Web of "Why is $(2, 1+\sqrt{-5})$ not principal?"

First if the ideal $(2, 1+\sqrt{-5})$ were principal, generated by $\alpha$, $N(\alpha)$ would divide $N(2)=4$ and $N(1+\sqrt{-5})=6$, hence would divide $\operatorname{gcd}(4,6)=2$.
There is no element with norm $2$, hence $N(\alpha)=1$, which means $\alpha$ would be a unit; in other words, we would have $$(2, 1+\sqrt{-5})=\mathbf Z[\sqrt{-5}].$$

Now $\mathbf Z[\sqrt{-5}]\simeq \mathbf Z[x]/(x^2+5)$. Hence
\begin{align*}\mathbf Z[\sqrt{-5}]/(2, 1+\sqrt{-5})&\simeq \mathbf Z[x]/(2,x+1,x^2+5)\simeq \mathbf Z_2[x]/(x+1,x^2+1)\\
&=\mathbf Z_2[x]/\bigl(x+1,(x+1)^2\bigr)=\mathbf Z_2[x]/(x+1)\simeq\mathbf Z_2.
\end{align*}

Hint $\,\ (2,1\!+\!w)\,=\, (\alpha) \,\Rightarrow\, (\alpha^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 $\,\ (a,\,b)^2 =\, (a^2,\ ab,\ b^2)$

We say
$$
a + b\sqrt{-5} \in \mathbb{Z}[\sqrt{-5}]
$$
is “good” if the parity of the integers $a, b$ is the same.

Claim: everything in your ideal is good. It’s obvious that good things are closed under taking sums, and it’s obvious that multiplying something arbitrary by 2 gives something good, so you just have to calculate that multiplying something arbitrary by $(1+\sqrt{-5})$ gives something good.

Indeed,
$$
(a + b\sqrt{-5})(1 + \sqrt{-5}) = (a – 5b) + (a + b)\sqrt{-5}.
$$
and $(a-5b)$ and $(a+b)$ have the same parity for any $a, b$.

On the other hand, there are bad elements in the ring, so your ideal can’t be the whole ring. Bonus: prove your ideal is exactly the set of good elements!

Yes, there is a simpler method, not that different from what you’ve done so far, but you have to keep in mind that there are still principal ideals even if the domain is not a principal ideal domain. For example, $\langle 5, \sqrt{-5} \rangle$ is a principal ideal; a little thought should reveal that it’s generated by a single element and is in fact $\langle \sqrt{-5} \rangle$.

So if $\langle 2, 1 + \sqrt{-5} \rangle$ is a principal ideal, we should be able to find a single element $x \in \mathbb{Z}[\sqrt{-5}]$ to generate it, an element that is not a unit. There isn’t one: $N(2) = 4$ and $N(1 + \sqrt{-5}) = 6$, as you already know, so we’d need for $x$ to satisfy $N(x) = 2$, which as you also already know, has no solutions.

It should be noted that $\sqrt{-5} \not\in \langle 2, 1 + \sqrt{-5} \rangle$. In fact, this ideal does not contain any numbers with odd norm, which means no purely real odd integers. There is no combination of $r, s \in \mathbb{Z}[\sqrt{-5}]$ that will give you $2r + s + s \sqrt{-5} = 3$, for example, because $N(2r + s + s \sqrt{-5})$ must be even. This confirms that $\langle 2, 1 + \sqrt{-5} \rangle$ is not the whole ring.


It might be helpful to compare a similar ideal in a similar domain: $\langle 2, 1 + \sqrt{5} \rangle$ in $\mathcal{O}_{\mathbb{Q}(\sqrt{5})}$. As before, $N(2) = 4$. But $N(1 + \sqrt{5}) = 4$ as well. This may or may not be in a PID (spoiler alert: it is), but this particular ideal is a principal ideal, and it is in fact $\langle 2 \rangle$ (verify that the familiar number $\frac{1}{2} + \frac{\sqrt{5}}{2}$ is an algebraic integer).