Does the set of solutions to quadratics over $\mathbb{Q}$ form a subgroup of the additive group $\mathbb{R}$?

  • $\begingroup$ Why did you think it was an additive group? $\endgroup$ – lhf Jan 4 '12 at 3:03
  • $\begingroup$ On the other hand, the solutions to simple quadratic equations $X^2=a$ with $a\ne0$ do form a multiplicative subgroup of $\mathbb C^{\times}$ (and of $\mathbb R^{\times}$ for $a>0$). $\endgroup$ – lhf Jan 4 '12 at 3:06
  • $\begingroup$ @lhf: Only if $a^2=a$, i.e., $a=1$ or $a=0$: otherwise, if $x^2 =a$ and $y^2=a$, then $(xy)^2 = x^2y^2 = aa = a^2$. $\endgroup$ – Arturo Magidin Jan 4 '12 at 4:13
  • $\begingroup$ @ArturoMagidin, I meant: $x^2=a, y^2=b \implies (xy)^2=ab$. $\endgroup$ – lhf Jan 4 '12 at 9:57

No, because it is not closed under addition. For example, $\sqrt{2}$ and $\sqrt{3}$ are solutions to quadratics, but $\mathbb{Q}(\sqrt{2}+\sqrt{3})=\mathbb{Q}(\sqrt{2},\sqrt{3})$ is a degree 4 extension of $\mathbb{Q}$, and therefore $\sqrt{2}+\sqrt{3}$ is not a solution to a quadratic.

  • $\begingroup$ could you explain the equality $\mathbb{Q}(\sqrt{2}+\sqrt{3})=\mathbb{Q}(\sqrt{2},\sqrt{3})$? $\endgroup$ – user9352 Jun 20 '11 at 14:51
  • 2
    $\begingroup$ @user9352: $\sqrt{2}=((\sqrt{2}+\sqrt{3})^3-9(\sqrt{2}+\sqrt{3}))/2$ $\endgroup$ – Chris Eagle Jun 20 '11 at 15:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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