We will denote by $\mathbb{K}$ one of the fields $\mathbb{Q}, \mathbb{R}$ or $\mathbb{C}$. On $\mathbb{K}\times \mathbb{K}$ we define the following equivalence relation: $$(a,b)\equiv (a', b') \iff \exists (q,\alpha)\in \mathbb{K}^{*}\times \mathbb{K} \text{ such that } \begin{cases} a=q^2a'+\alpha^2-b\alpha \\ b=qb'+2\alpha \end{cases}.$$ We wish to determine the quotient set $\mathbb{K} \times \mathbb{K}/\equiv$ $\space$ for all $\mathbb{K}$s.
This problem was an extra problem in my abstract algebra class (don't worry, this isn't an attempt to cheat, I am posting this a week after the solution was due to be sent) and I kind of got stuck when it comes to $\mathbb{K}=\mathbb{Q}$.
For $\mathbb{K}=\mathbb{C}$, the things were nice and easy, because the original question asked me to prove that $\mathbb{C}\times \mathbb{C}/\equiv$ is equal to $\{\hat{(0,0)}, \hat{(0,1)}\}$ and this can be checked through (tedious) direct computations.
For $\mathbb{K}=\mathbb{R}$, a friend came up with the idea of expressing $\alpha$ from the second equation and then substituing it in the first one. This gives us the following equivalent characterisation of the equivalence relation: $$(a,b)\equiv (a', b') \iff \exists q\in \mathbb{K}^{*} \text{ such that } 4a+b^2=q^2(4a'+b'^2) \space (*).$$ (notice that this works for all $\mathbb{K}$s, the case $\mathbb{K}=\mathbb{C}$ can be solved much easier by using this, but I didn't really need to think that much for that one since direct computations worked in my context)
For real numbers, this rewrites as $(a,b)\equiv (a', b') \iff \operatorname{sgn}(4a+b^2)=\operatorname{sgn}(4a'+b'^2)$.
As a result, there will be three equivalence classes: the parabola $4x+y^2=0$, its interior and its exterior. A representative for each of these are, respectively, $(0,0), (-1,0)$ and $(0,1)$, so $\mathbb{R}\times \mathbb{R}/\equiv \space = \{\hat{(0,0)}, \hat{(-1,0)}, \hat{(0,1)}\}$.
For $\mathbb{K}=\mathbb{Q}$, the things get pretty nasty by this approach. In this case, $(*)$ rewrites as $(a,b)\equiv (a',b') \iff \sqrt{\frac{4a+b^2}{4a'+b'^2}}\in \mathbb{Q}$ and I haven't been able to make any further progress.
from Hot Weekly Questions - Mathematics Stack Exchange
Alexdanut
Post a Comment