Let $q$ a $n$-variables indefinite quadratic form with integer coefficients, which has a non trivial isotropic vector, let say $z\in\mathbb{Z}$.
Easily this form is $\mathbb{Z}$-equivalent to a form of the shape:
$$
f(x_1,\ldots, x_n) = g(x_3,\ldots, x_n)+ \sum_{i=1}^n x_2\alpha_{i}x_i
$$
with $g$ having the same signature as $q$.
Is this result can be extended when in presence of a subspace on which the form is degenerated? (the previous case is then the case where the subspace is of dimension 1).
I was thinking of having a family $z_1, \ldots, z_i \in\mathbb{Z}^n$ so that the form has discriminant 0 on $\langle z_1, \ldots, z_i\rangle$, and being able to find a $\mathbb{Z}$-equivalent form of the shape:
$$
f(x_1,\ldots, x_n) = g(x_{i+2},\ldots, x_n)+ \sum_{i=1}^n x_{i+1}\alpha_{i}x_i
$$
A partial result consists in isolating isotropic vectors in $\langle z_1, \ldots, z_i\rangle$ and apply successively the construction for one vector cited above. But it's not necessarily giving such a result in the general case.
Best Answer
there is a simple procedure for finding all rational vectors in the null cone, given $q(z) = 0.$ Let $$ 2 q(x) = x^T H x, $$ where $H$ is the Hessian matrix, all integers. Next, take any primitive integer vector $v$ (we need to assume that $q(v) \neq 0$ and find the value of real $t$ such that $q(z + tv) = 0.$ It turns out that $$ t = \frac{-2z^T H v}{v^T H v} $$ which is the reason that we needed $q(v) \neq 0.$ Now we have a rational null vector $$ \frac{(v^THv)z -2(z^T Hv)v}{v^T Hv} $$
To get an integer null vector we may now just multiply by the denominator $q(v).$ The trouble is that the result will often be an integer vector that is not primitive, the GCD of the entries is something larger than one. Well, find the gcd and divide out by it.
So, take column vector $v,$ if $q(v) \neq 0$ find $$ (v^THv)z -2(z^T Hv)v $$ and divide through by its gcd of entries.
In low dimension there are ways to get around the gcd business. the familiar example would be Pythagorean triples, where primitive null vectors are found parametrized by three binary quadratic forms. I can be specific in dimension 3: given $H$ the Hessian of an integer isotropic form, and the matrix $W$ as the Hessian of $y^2 - zx,$ there is an invertible integer matrix $P$ such that $P^T HP = n W.$ Then recipes can be given for all null vectors.
Fricke and Klein (1897), in dimension 4, use $x^2 + y^2 + z^2 - w^2$ for signature $+++-,$ then $xy - zw$ for signature $++--.$