# Set of solutions to quadratics over $\mathbb{Q}$

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

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Why did you think it was an additive group? –  lhf Jan 4 '12 at 3:03
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 $a>0$). –  lhf Jan 4 '12 at 3:06
@lhf: Only if $a^2=a$, i.e., $a=1$ or $a=0$: otherwise, if $x^2 =a$ and $y^2=a$, then $(xy)^2 = x^2y^2 = aa = a^2$. –  Arturo Magidin Jan 4 '12 at 4:13
@ArturoMagidin, I meant: $x^2=a, y^2=b \implies (xy)^2=ab$. –  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.
could you explain the equality $\mathbb{Q}(\sqrt{2}+\sqrt{3})=\mathbb{Q}(\sqrt{2},\sqrt{3})$? –  user9352 Jun 20 '11 at 14:51
@user9352: $\sqrt{2}=((\sqrt{2}+\sqrt{3})^3-9(\sqrt{2}+\sqrt{3}))/2$ –  Chris Eagle Jun 20 '11 at 15:54