Give a quadratic form $f=a_1X_1^2+a_2X_2^2+...+a_nX_n^2$ over a field $k$, with $d(f)$ being the discriminant of $f$.
I read in Serre's - A course in arithmetic that $d(f)\in k^*/(k^*)^2$.
(In page 35, he specifically refers to $k=\mathbb Q_p$)
Why is that?
Is it because $d(f)=a_1\cdot a_2\cdot ...\cdot a_n$ and he's assuming $a_i\in k^*/(k^*)^2$ $\forall 1\leq i\leq n$?
The point is that this formula for $f$ is not unique (up to isometry). You can do base changes that transform it into $f = b_1X_1^2+\dots +b_nX_n^2$. So if you want the discriminant to depend only on the isometry class of $f$, and not on a specific choice of basis, $d(f)=a_1\cdots a_n$ cannot work, because in general $a_1\cdots a_n\neq b_1\cdots b_n$.
On the other hand, it is not hard to check that $a_1\cdots a_n$ and $b_1\cdots b_n$ are the same up to a square, so they are congruent modulo $(k^*)^2$, and $d(f)=a_1\cdots a_n$ is well-defined, not as an element of $k$, but of $k^*/(k^*)^2$.