Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
Yes. For instance $\mathbb{Q}(\sqrt{18}) = \mathbb{Q}(\sqrt{2})$ – user9413 Apr 23 '12 at 11:30
up vote 3 down vote accepted

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 $\mathbb{Q}(\sqrt{d})=\mathbb{Q}(\sqrt{f})$.

share|cite|improve this answer
You don't have to show this in general, it is enough to show that the generators $\sqrt{d}$ and $\sqrt{f}$ lie in the fields $\mathbb{Q}(\sqrt{f})$ and $\mathbb{Q}(\sqrt{d})$ respectively. – fretty Apr 24 '12 at 8:42
Yes, that is correct, but I thought that at this level it would be clearer to show the equality of sets completely. – Álvaro Lozano-Robledo Apr 24 '12 at 11:53

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 $\sqrt{2}$.

share|cite|improve this answer
This is not really an answer. It should have been posted as a comment. You should at least explain why it is obvious if you claim it to be so. – M Turgeon Apr 23 '12 at 13:00
This is exactly what I did in my second sentence... – fretty Apr 23 '12 at 13:29
Was I wrong to assume that the original poster knows and understands the basics of fields? – fretty Apr 23 '12 at 13:34

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.