Is $(1+\sqrt{-5})$ an ideal of $\mathbb{Z}[\sqrt{-5}]$?

Question

How to determine the ideal of $\mathbb{Z}[\sqrt{-5}]$? Is $(1+\sqrt{-5})$ is the whole ring? And is every $(a+b\sqrt{-5})$ an ideal, since it is generated by $a+b\sqrt{-5}$, consisting of the of $a+b\sqrt{-5}$ multiplied by every element in the ring?

Answer 1

If you have any commutative ring $R$ and any element $a \in R$, then the set $(a) = \{r\cdot a| r \in R\}$ is always an ideal.

Note that it is easy to show that $(a) = R \Leftrightarrow \exists b \in R:ab=1$

So in order to show that $(1+\sqrt{-5})$ is not the whole ring, we need to show that there is no $b \in \mathbb{Z}[\sqrt{-5}]$ such that $ab = 1$

Consider the so-called "norm function": $N:\mathbb{Z}[\sqrt{-5}] \to \mathbb{Z}, N(a+b\sqrt{-5})=a^2+5b^2$ It is easy to show that this is a multiplicative function. (You can do this by explicit calculation or by noting that is is the absolute value of the complex numbers squared) Note that if $ab=1$ for $a,b \in \mathbb{Z}[\sqrt{-5}]$, then applying $N$ and using multiplicativity, we have $N(a)N(b)=N(1)=1$. This is an equation over the integers, but the only integers with a multiplicative inverse are $\pm 1$. But $N(1+\sqrt{-5})=6 \neq \pm 1$, so there is no $b\in \mathbb{Z}[\sqrt{-5}]$ such that $(1+\sqrt{-5})b=1$, thus $(1+\sqrt{-5})$ is not the whole ring $\mathbb{Z}[\sqrt{-5}]$