I am currently working on the proof of the Hasse-Minkowski Theorem for $n=3$ in Serre's Course on Arithmetic.
- First we start with a quadratic form (without cross terms) over $\mathbb{Q}$ \begin{equation} f = a_1X_1^2+a_2X_2^2+\ldots+a_nX_n^2 \end{equation} and raise the question when this form represents $0$ (meaning there is a non-trivial solution to $f(x)=0$) On page 41 he claims
Replacing $f$ by $a_1f$ one can moreover suppose $a_1=1$
Does he really mean replacing $f$ by $a_1^{-1}f$? This would absolutely make sense to me, so might this just be a small typo in his book? If not could somebody please explain why we yield a form $X_1^2+a_2X_2^2+\ldots+a_nX_n^2$ if we multiply the first coefficient by itself? Otherwise it is totally clear to me why we can do this.
However the main problem I have is on the following page. For the case $n=3$ we do exactly this and cancel the first coefficient to get \begin{equation} f = X_1^2 - aX_2^2 - bX_3^2 \end{equation} Now he sais we may assume that $a,b$ are square-free integers. If we only care about the question if $f$ represents $0$ we can obviously assume that the coefficients are square-free, as we might multiply any solution by the corresponding square. The trouble I'm facing is the claim that $a$ and $b$ are integers. If we have $a=\frac{x_1}{y_1}$ and $b=\frac{x_2}{y_2}$ we could use the same trick as before and look at $y_1y_2f$ to cancel the denominators, so we would have integer coefficients.
However in this case the leading coefficient would not be $1$ anymore and we would rather have a quadratic form like \begin{equation} y_1y_2X_1^2 - x_1X_2^2 - x_2 X_3^2 \end{equation} where $x_1,x_2$ are square-free integers as we wished. However we did not preserve our first coefficient to be $1$. In the rest of the proof we use exactly this property of having a negligible leading coefficient multiple times. If we were to try the trick from the beginning again to erase this leading coefficient, we would have rational coefficients $a,b$ again and we have the situation as we started.
So how exactly do we find the equivalent quadratic form \begin{equation} X_1^2 - aX_2^2 - bX_3^2 \end{equation} with $a,b$ (square-free) integers if we are given any arbitrary quadratic form (without cross terms) over $\mathbb{Q}$? Given this assumption the rest of the proof is clear to me. However for me the devil is in the detail and I would really wish to understand this basic assumption. If somebody could give a step-by-step explanation or some source where to find more explicit information I would be very thankful. Maybe it would also help doing this on some explicit example like \begin{equation} 2X_1^2 - \frac{1}{3}X_2^2 - \frac{1}{7}X_3^2 \end{equation} where I would first divide by $2$ and then multiply by the least common multiple of $6$ and $14$ which is $42$ to receive \begin{equation} 42X_1^2 - 7X_2^2 - 3X_3^2 \end{equation} by my understanding. How would I now get rid of the first coefficient $42$? Or did I completely get something wrong in my understanding?
- This is not as important to me as question 1. On page 34 in Theorem 1* he sais that any quadratic form is equivalent to one without cross terms as we have above. From the chapter it is clear to me that this has a strong connection to questions of diagonalization and Silvester's inertia theorem. However for my personal taste Serre is not thorough enough for me to fully understand his reasoning. So if anybody reading this has a good source in mind where this is explained more rigorously I would also be very thankful. I tried searching for a source already however I could not find a fully coherent source where this question is discussed fully and in detail.