In a proof of a corollary in chapter 2, there is a step I don't understand.
Corollary 2: Suppose $p \neq 2$. Let $f(X) = \sum_j a_{ij}X_iX_j$ with $a_{ij} = a_{ji}$ be a quadratic form with coefficients in $\mathbb{Z}_p$ whose discriminant $\det (a_{ij})$ is invertible. Let $a \in \mathbb{Z}_p$. Every primitive solution of the equation $f(x) \equiv a \mod p$ lifts to a solution with coefficients in $\mathbb{Z}_p$.
Proof: (...), it suffices to show that $x$ does not annihilate all the partial derivatives of $f$ modulo $p$. Now $\frac{\partial f}{\partial X_i} = 2 \sum_j a_{ij} X_j$; since $\det (a_{ij}) \not\equiv 0 \mod p$ and $x$ is primitive, one of these partial deriviatives is $\not\equiv 0 \mod p$.
If I'm not mistaken, then one step I need to conclude: If $\det(a_{ij}) \not\equiv 0 \mod p$, then $a_{ij} \not\equiv 0 \mod p$.
However, I failed to show that.
I tried the following:
$\det(a_{ij}) = \det \begin{pmatrix} a_{11} & \ldots & a_{1n} \\ \vdots & \ddots & \vdots\\ a_{n1} & \ldots & a_{nn} \end{pmatrix} = \det \begin{pmatrix} (\ldots, a_{11}^{(k)}, \ldots, a_{11}^{(1)}) & \ldots & (\ldots, a_{1n}^{(k)}, \ldots, a_{1n}^{(1)}) \\ \vdots & \ddots & \vdots\\ (\ldots, a_{n1}^{(k)}, \ldots, a_{11}^{(1)}) & \ldots & (\ldots, a_{nn}^{(k)}, \ldots, a_{nn}^{(1)}) \end{pmatrix} = \left( \ldots, \det \begin{pmatrix} a_{11}^{(k)} & \ldots & a_{1n}^{(k)} \\ \vdots & \ddots & \vdots\\ a_{n1}^{(k)} & \ldots & a_{nn}^{(k)} \end{pmatrix}, \ldots, \det \begin{pmatrix} a_{11}^{(1)} & \ldots & a_{1n}^{(1)} \\ \vdots & \ddots & \vdots\\ a_{n1}^{(1)} & \ldots & a_{nn}^{(1)} \end{pmatrix} \right) $
Therefore, if $\det(a_{ij}) \not\equiv 0 \mod p$, then each component has to fulfill $\not\equiv 0 \mod p$.
Now, I could use Leibniz's formula [1] to calculate each determinant, but I don't see from here, why $a_{ij}^{(k)}$ has to be $\not\equiv 0 \mod p$.
Then I looked at the special case $n=2$. We have
$\det \begin{pmatrix} a_{11}^{(k)} & a_{12}^{(k)} \\ a_{21}^{(k)} & a_{22}^{(k)} \end{pmatrix} = a_{11}^{(k)}a_{22}^{(k)} - a_{21}^{(k)}a_{12}^{(k)} \not\equiv 0 \mod p$.
But for me this doesn't necessarily mean, that $a_{11}^{(k)} \not\equiv 0 \mod p$ and $a_{12}^{(k)} \not\equiv 0 \mod p$ and $a_{22}^{(k)} \not\equiv 0 \mod p$.
Did I miss something?
[1] https://en.wikipedia.org/wiki/Leibniz_formula_for_determinants
I think the point is that, if $M = (a_{ij})$ and $\det M\not\equiv 0\pmod{p}$, then $M$ has an inverse: there is some $N$ with $MN\equiv NM\equiv I \pmod{p}$.
Suppose $x$ is primitive and nonzero, and we have $\sum_{j} a_{ij} x_j \equiv 0 \pmod{p}$ for all $i$. Then $Mx \equiv 0\pmod{p}$. But multiplying on the left by $N$ gives us $x\equiv 0\pmod{p}$, which is impossible.