Simplifying Ideals to principal ideals

Question

I am given that $(6,2\sqrt{6}, 3\sqrt{6},6)$ is an ideal of $\mathbb{Q}(\sqrt{6})$ and I am trying to show that it is equal to $(\sqrt{6})$ (without the use of norms). Since there is closure under addition in the ideal $3\sqrt{6}-2\sqrt{6}\implies \sqrt{6}$ is in the ideal. I can also see trivially that I can generate everything in $(6,2\sqrt{6},3\sqrt{6},6)$ from $\sqrt{6}\mathbb{Q}(\sqrt{6})$ but I am not sure how this implies equality.

Answer 1

Let's call the ideal $I$. As you showed, $\sqrt{6} \in I$, which implies $(\sqrt{6}) \subset I$. But you also said that all the generators of $I$ are in $(\sqrt{6})$, which means $I \subset (\sqrt{6})$. Put those together and you have equality.