Are two quadratic forms equivalent over $\mathbb{Q}$

Question

Are the quadratic forms $q_1$ and $q_2$ equivalent over $\mathbb{Q}$?

1. $q_1(x, y)=x^2+y^2 \text{ and } q_2(x, y)=x^2+3 y^2$
2. $q_1(x, y)=x^2+y^2 \text{ and } q_2(x, y)=-x^2-y^2$
3. $q_1(x, y)=x^2+y^2 \text{ and } q_2(x, y)=x^2+9 y^2$
4. $q_1(x, y)=x^2+y^2 \text{ and } q_2(x, y)=13 x^2+13 y^2$ 

I don't really know how to check this.

Denote the matrix of $q_i$ as $A_{q_i}$. So for the first example we have $ A_{q_1}= \left( {\begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} } \right)$ and $ A_{q_1}= \left( {\begin{array}{cc} 1 & 0 \\ 0 & 3 \\ \end{array} } \right)$, this $\det{(A_{q_1})}=1 = \text{discr}(A_{q_1})$ and $\det{(A_{q_2})}= 3.$ Thus $\text{discr}(A_{q_2}) = 1$. We conclude that $q_1$ and $q_2$ are equivalent over $\mathbb{Q}$?

Answer 1

So, first of all, all your quadratic forms have rank $2$ (they are all represented by a diagonal $2$-by-$2$ matrix with non-zero diagonal coefficients). In this case you can say that $q_1$ and $q_2$ are equivalent if and only if they have the same discriminant and they represent a common value, which means that there exists $a\in \mathbb{Q}^\times$ such that $a=q_1(x,y)=q_2(x',y')$ for some $x,y,x',y'\in \mathbb{Q}$. This works over any field of characteristic not $2$ (just replace $\mathbb{Q}$ by your base field).

Note that you don't seem to fully understand what the discriminant is: you say that $\det(A_{q_2})=3$ implies $\operatorname{disc}(q_2)=1$, but that is not the case. The discriminant of $q$ is the square class of $\det(A_q)$ in $\mathbb{Q}^\times$. So saying that $\operatorname{disc}(q)=1$ means that $\det(A_q)$ is a square in $\mathbb{Q}$, and obviously $3$ is not a square in $\mathbb{Q}$. Now on the other hand $\operatorname{disc}(q_2)=1$ in your example if you work over $\mathbb{R}$, since $3$ is of course a square in $\mathbb{R}$.

Let us look at your examples; in each case, $q_1$ is the same, and has discriminant $1$:

1. $q_2$ has discriminant $3$, so they are not equivalent over $\mathbb{Q}$;
2. $q_2$ has discriminant $1$, but it only represents negative values, while $q_1$ only represents positive values, so they are also not equivalent;
3. $q_2$ has discriminant $9$, which is a square, so it has disciminant $1$, and also $q_1$ and $q_2$ both represent the value $1$ (as $q_1(1,0)$ and $q_2(1,0)$ for instance), so they are equivalent;
4. $q_2$ has discriminant $13^2$, so discriminant $1$, and $q_1$ and $q_2$ both represent the value $13$ ($q_1(2,3)$ and $q_2(1,0)$ for instance), so they are equivalent.

Explicitly, in cases 3 and 4, one can find the base changes $(x',y')=(x,3y)$ and $(x',y')=(2x-3y,2y+3x)$ respectively, and then $q_1(x',y')=q_2(x,y)$.