Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

Suppose $x^2+mx+n\in\mathbb{Z}$ is irreducible with root $\alpha$. Now if $m^2-4n=D_0^2D$ with $D$ squarefree, then $\mathbb{Q}[\alpha]=\mathbb{Q}[\sqrt{D}]$.

I don't get how one gets the relationship $m^2-4n=D_0^2D$ to the equality $\mathbb{Q}[\alpha]=\mathbb{Q}[\sqrt{D}]$. I write $\beta=\alpha+m/2$, so $\mathbb{Q}[\alpha]=\mathbb{Q}[\beta]$. Also $$ \beta^2=\alpha^2+m\alpha+m^2/4\implies \beta^2+n-m^2/4=\alpha^2+m\alpha+n=0. $$ So $$ m^2-4n=4\beta^2=D_0^2D $$ for some $D_0$ and $D$? Can I say $2\beta=D_0\sqrt{D}$, so $\mathbb{Q}[\beta]=\mathbb{Q}[\sqrt{D}]$, or is that oversimplifying the problem?

share|cite|improve this question
up vote 2 down vote accepted

It's really just the quadratic formula. I think what you've written is fine. The basic idea is we know

$$\alpha= \frac{-m\pm \sqrt{m^2-4n}}{2}.$$

So $\mathbb Q(\alpha)=\mathbb (\sqrt{m^2-4n})$ and $D$ is simply the square free part of $m^2-4n$. Then it's a simple exercise to show that if $m$ is an integer and $m^\prime$ is its square free part then $\mathbb Q(\sqrt{m})=\mathbb Q(\sqrt{m^\prime})$.

share|cite|improve this answer
Oh, I didn't get the quadratic formula connection. Thanks Jacob! – Noomi Holloway Jan 14 '13 at 8:29

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.