Let $V$ be a vector space over field $K$ and $Q$ is the quadratic form on it, and $A$ be the matrix w.r.t. $e_1,e_2,...e_n$ of $V$. Now $discr(Q)$ is defined as $det(A)$ mod ${K^{*}}^{2}$.
Now my doubt is , If let $X^2+Y^2$ be the quadratic form on $\mathbb{Q}$(or take any $K$). Then $det (A)$ will be $1$, but $1$/ ${\mathbb{Q}^{*}}^{2}$ should be zero, as $1=1^2$. But why they sat it is $1$?.
The book says $discr(Q)=det (A)/{K^{*}}^{2}=0 \iff det(A)=0$ but It should be $0$ whenever $det(A)=a^2$ for any $a \in K$. No?
Llkewise $discr(2X^2-3Y^2)$ should be $6$ over $\mathbb{Q}$ but $0$ over $\mathbb{R}$ as $6$ is a square of a non zero real number i.e. $\sqrt{6}$. Please clear this doubt of mine. Thanks!
The unit element in $\mathbb{Q}^{\ast}/\mathbb{Q}^{\ast 2}$ is the class represented by $1$ and not by zero. The unit of the multiplicative group $\mathbb{Q}^{\ast}$ is $1$, not zero. In other words, we have still $\det(A)=1$ in $\mathbb{Q}^{\ast}/\mathbb{Q}^{\ast 2}$. This also answers the next questions.