$R=\mathbb{Z}[\sqrt{3}]=\{a+\sqrt{3}|a,b\in\mathbb{Z}\}$ and $I=(2)$ the from 2 generated ideal. Is (2) a prime ideal or a maximal ideal?
Hello,
I think, that $I$ is not a prime ideal, but a maximal ideal.
Prime ideal:
$x=(\sqrt{3}+1)$, $y=(\sqrt{3}-1)$, hence $xy=(\sqrt{3}+1)(\sqrt{3}-1)=2$, but neither $x$ nor $y$ are in $(2)$. Therefor it is not a prime ideal.
Maximal ideal:
I think, that $I$ is a maximal ideal. Consider the ideal $(2,\sqrt{3})\supset (2)$. This inclusion holds, since $\sqrt{3}\notin (2)$. I want to show, that $(2,\sqrt{3})=\mathbb{Z}[\sqrt{3}]$.
$(2,\sqrt{3})=\{2f+\sqrt{3}g|f,g\in\mathbb{Z}[\sqrt{3}]\}$
For $h\in (2,\sqrt{3})$ it is $h=2(a+\sqrt{3}b)+\sqrt{3}(c+\sqrt{3}d)=2a+3d+(2b+c)\sqrt{3}$.
For arbitrary $z=z'+z''\sqrt{3}\in\mathbb{Z}[\sqrt{3}]$, we can choose $c=0$ and $d=z''$. Hence we have to show, that for every $z'\in\mathbb{Z}$ there are $a,d\in\mathbb{Z}$, such that $z'=2a+3b$, or in other words:
The function $f:\mathbb{Z}\times\mathbb{Z}\to\mathbb{Z}$, $(a,d)\mapsto 2a+3d$ is surjective.
Which I could not proof yet, even tho it seems quite "obvious", but I can not really find a simple argument.
Any tips? Is there a simpler way to proof this, showing that $\mathbb{Z}[\sqrt{3}]/(2)$ is not a field seems out of range...
Thanks.
Just including this as you said showing it isn't a field is out of range.
$\mathbb{Z}[\sqrt{3}]/(2)\cong\mathbb{Z}_2[\sqrt{3}]$. Let's look at the inverse of a generic element: $$\frac{1}{a+b\sqrt{3}} = \frac{a-b\sqrt{3}}{a+3b}$$ For this to be integral, we need to have $a+3b\equiv 1\mod 2$. It follows that $(1+\sqrt{3})$ isn't invertible.
To see this explicitly: \begin{align*} (1+\sqrt{3})(a+b\sqrt{3}) &\equiv 1\mod 2 \\ \implies a+3b+(b+a)\sqrt{3}&\equiv 1\mod 2 \\ \implies (a+b)(1+\sqrt{3})&\equiv 1\mod 2 \end{align*} This can clearly never be true.
So, $\mathbb{Z}[\sqrt{3}]/(2)$ isn't a field, so $(2)$ isn't maximal.