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I know what the ring of integers of $\mathbb{Q}(\sqrt{d})$ looks like when $d$ is square free, but what is the ring of integers for $\mathbb{Q}(\sqrt{d})$ for $d=18,45$ etc. Can I just remove the square factors?

Thank you

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    Yes. For instance $\mathbb{Q}(\sqrt{18}) = \mathbb{Q}(\sqrt{2})$2012-04-23

2 Answers 2

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What you are asking has nothing to do with rings of integers, but simply with the definition of $\mathbb{Q}(\sqrt{d})$: $\mathbb{Q}(\sqrt{d})=\{a+b\sqrt{d} : a,b\in\mathbb{Q}\}.$ Now suppose that $d$ is an arbitrary integer. Then, we can find integers $n\geq 1$ and $f\in\mathbb{Z}$ such that $d=n^2f$, and $f$ is square-free. Hence, $\mathbb{Q}(\sqrt{d})=\mathbb{Q}(\sqrt{f})$. Let's see this:

  • If $\alpha=a+b\sqrt{d}\in\mathbb{Q}(\sqrt{d})$, then $\alpha=a+b\sqrt{d}=a+b\sqrt{n^2f}=a+bn\sqrt{f}\in\mathbb{Q}(\sqrt{f})$.

  • Conversely, if $\beta=u+v\sqrt{f}\in\mathbb{Q}(\sqrt{f})$, then $\beta=u+\frac{v}{n}\cdot n \sqrt{f}= u+\frac{v}{n} \sqrt{d}\in \mathbb{Q}(\sqrt{d}).$

Notice that the argument above can be modified slightly to show that, in fact, for every rational number $d\in\mathbb{Q}$ there is a square-free integer $f\in\mathbb{Z}$ such that $\mathbb{Q}(\sqrt{d})=\mathbb{Q}(\sqrt{f})$.

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    Yes, that is correct, but I thought that at this level it would be clearer to show the equality of sets completely.2012-04-24
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Well it is quite obvious that say $\mathbb{Q}(\sqrt{18}) = \mathbb{Q}(\sqrt{2})$. This just boils down to the fact that $\sqrt{18} = 3\sqrt{2}$, which is a rational multiple of $\sqrt{2}$.

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    Was I wrong to assume that the original poster knows and understands the basics of fields?2019-02-08