Show that $I_2 = (2, 1+ \sqrt{-5})$ is not a principal ideal

Question

Let $R$ be the quadratic integer ring $\mathbb{Z}[\sqrt{-5}]$, and $I_2 = (2, 1+ \sqrt{-5})$. Show that $I_2$ is not a principal ideal in $R$.

I still find such ring "alien", hence my original attempt was to show that these 2 are co-maximal thus no single element can generate them and still remain proper ideal. I want to do this by showing G.C.D is 1.

But according to Wikipedia this ring doesn't seem to be a euclidean domain, hence I can't apply euclidean algorithm here. How do I proceed from here ?

Any help or insight is deeply appreciated.

Answer 1

First we prove that $2$ is irreducible element of $\Bbb Z[\sqrt {-5}]=R.$

Suppose $2$ is reducible, then $2=xy$ where $x,y \in R$ and both are non-unit.

Then $N(2)=4=2 \times 2=N(x)N(y)$, where $N(a+b\sqrt {-5})=|a^2+5b^2|$ and $a,b \in \Bbb Z$.

This means $N(x)=2$. ($\because x$ is not unit.)

Let $x=a+b \sqrt {-5}$, then $|a^2+5b^2|=2$. But this is not possible for any $a,b \in \Bbb Z$.

Thus our assumption that $2$ is reducible is false $\Rightarrow 2$ is irreducible in $R$.

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Next we show that $I_2$ is not a principal ideal in $R$.

Suppose it is a principal ideal.

Then $(2,(1+ \sqrt {-5}))=(n)$ for some $n \neq 1 \in R$ (if $n=1$ then $R=I_2$ which is not possible$^\star$).

So the element $2=2 \times 1 + (1+ \sqrt {-5}) \times 0 \in (n)$

$\Rightarrow 2=nm$ for some $m \in R$.

But $2$ is irreducible in $R$, hence $n=2$ and $m=1$ (or $n=-2$ and $m=-1$).

Similarly $1+ \sqrt {-5} \in (n)=(2) \Rightarrow 1+\sqrt {-5}=2(x+y\sqrt {-5}) \Rightarrow x=\frac 12$ which is a contradiction ($\because x \in \Bbb Z$). (same process for $n=-2$)

Hence our assumption that $I_2$ is a principal ideal is wrong.

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To prove $^\star$,

Suppose $R=I_2$, then $2a+(1+\sqrt {-5})b=1$ for some $a,b \in R$.

Multiply both sides by $1-\sqrt {-5}$, we get $2(1-\sqrt {-5})a+6b=1-\sqrt {-5}$.

Since $2$ divides left hand side, $2$ divides $1-\sqrt {-5}$.

So $1-\sqrt {-5}=2(x+y\sqrt {-5})=2x+2y\sqrt {-5} \Rightarrow x=\frac 12$.

This is a contradiction since $x \in \Bbb Z$.

EDIT: I hope you know that $1$ and $-1$ are only units in $R$.

Answer 2

Assume $I_2$ is principal. Then there exists $f\in \mathbb{Z}[\sqrt{-5}]$ such that $I_2=(f)$. $Norm(2)=4$, $norm(1+\sqrt{-5})=6$ implies that norm of $f$ divide both of them which is 2 or 1. There are only a few possible $f$: $f=\pm 1$. We can easily eliminate all these possibilities. So $I_2$ is not a principal ideal.