Showing that an ideal is principal.

Question

I need to show that the ideal $(3 + i , 6)$ is principal in $\mathbb{Z}[i]$ and find its generator. So I know that I need to find an element such that $<t> = (3 + i , 6)$. My intuition tells me $t = i$. If this is right though I am not sure how to prove this. Maybe since $i(-3i + 1) = 3i + 1$ and $i(-6i) = 6i$ this shows it, can someone help me in what exactly I need to prove? thanks

Answer 1

Consider looking at the map $$\mathcal{N}\ : \mathbb{Z}[i]\longrightarrow\mathbb{N}\ :\ a+bi\mapsto a^2+b^2$$ This map has the property $\mathcal{N}({\alpha\beta})=\mathcal{N}({\alpha})\mathcal{N}({\beta})$ (you might like to prove this). What can you say about both $\mathcal{N}(3+i)$ and $\mathcal{N}(6)$? What might this tell you about a potential generator $t$?

Once you have found a candidate for the generator $t$, it simply becomes a matter of showing any $\mathbb{Z}[i]$-linear combination of $3+i$ and $6$ can be written as $\alpha t$ for some $\alpha\in\mathbb{Z}[i]$.

Answer 2

The ideal is closed under taking linear combinations (with coefficients in $\mathbb{Z}[i]$) of its elements. This means that i.e. $10=(3-i)(3+i)$ is in the ideal. Then $2=2\cdot6-10$ is in the ideal and so is $1-i=2\cdot2 - (3+i)$.
But $2=i(1-i)^2$.
Now you can try to reconstruct the elements $3+i,6$ from $1-i$