Doubt about lemma for polynomial equivalence [closed]

Question

Multivariate polynomials $f,g$ are equivalent if there exists invertible linear transformation $A$ such that $f(X)=g(A\cdot X)$

From paper p.1:

Lemma 1.1. (Structure of quadratic polynomials). Let $F$ be an algebraically closed field of characteristic different from 2. For any homogeneous quadratic polynomial $f(X) \in F[X]$ there exists an invertible linear transformation $A \in F^{n \times n}$ and a natural number $1 \le r \le n$ such that $f (A \cdot X) = x_1^2 + x_2^2 + . . . + x_r^2$ Moreover, the linear transformation $A$ involved in this equivalence can be computed efficiently. Furthermore, two quadratic forms are equivalent if and only if they have the same number r of variables in the above canonical representation.

Let $F$ be the algebraic closure of the rationals.

Basically there are more than $n$ non-equivalent $f_i$ and only $n$ possible choices for $r$.

Fix $n$ and take $n+1$ non-equivalent $f_i$. By the pigeonhole principle there are at least two $r_i=r_j$, contradicting non equivalence of $f_i$.

Q1 What is wrong with this argument against the lemma?

Q2 Is homogeneous quadratic polynomial equivalence or isomorphism polynomial in $n$?

Answer 1

The lemma is correct, and there are indeed no more than $n$ non-equivalent $f_{i}$. Here is a sketch of a proof: We can do a transformation such that the coefficient of (without loss of generality) $a_{1}^2$ is non zero, and by scaling, 1. Now by replacing $a_1$ with $a_1 - c_2 \cdot a_2 - \dotsb - c_n \cdot a_n$, where the coefficients $c_i$ are chosen appropriately, we can ensure that $a_{1}^2$ is the only place where $a_1$ appears, and now we can induct on the statement with $n-1$ variables.