Two first integrals of an hamiltonian field $X_{H}$ are independent $\det \left[ \frac{\partial F_{i}}{\partial p_{k}} \right] \neq 0$ I want to understand how it is established if two first integrals of an hamiltonian field $X_{H}$ are independent. 
One hypothesis is:
Considering two first integrals $F(q^i,p_k)$ 
$$\det \left[ \frac{\partial F_{i}}{\partial p_{k}} \right] \neq 0.$$
I would like to have a confirmation.
 A: In this context, independent means that (this is the definition of independent)
if you fix any two values $f_1$ and $f_2$ in the range of the first integrals, the set $$S:= \{x \in M \:|\: F_i(x)=f_i, i=1,2 \}$$
in the $2n$-dimensional phase space $M$ is an embedded submanifold of dimension $2n-2$.
A necessary and sufficient condition for independence is that 
the Jacobian matrix (in any arbitrarily fixed coordinate system on $M$ not necessarily canonical) $\left[\frac{\partial F_i}{\partial x^k}\right]_{i=1,2\: k= 1,\ldots, 2n}$ has rank $2$ on $S$.
In other words
the $\mathbb R^{2n}$ vectors $(\frac{\partial F_1}{\partial x^{1}}, \cdots \frac{\partial F_1}{\partial x^{2n}})^t$ and $(\frac{\partial F_2}{\partial x^{1}}, \cdots \frac{\partial F_2}{\partial x^{2n}})^t$ must be linearly independent on $S$.
This last condition has not much to do with a determinant since the determinant function deals with $2n \times 2n $ matrices while here we have a $2 \times 2n$ matrix.
As @AccidentalFourierTransform suggested (in a discussion now removed), this condition is related with the other condition  $\{F_1, F_2\} \neq 0$ on $S$. 
In fact that condition implies the independence of $F_1$ and $F_2$. The proof is easy:
$$0 \neq \{F_1, F_2\} = Sym(dF_1,dF_2)$$
where $Sym(dF_1,dF_2)$ is the symplectic form which is bi-linear and anti-symmetric and thus it vanishes if $dF_1$, $dF_2$ are linearly dependent.
For $n>1$, it may happen that $dF_1$, $dF_2$ are linearly independent but $Sym(dF_1,dF_2)=0$, so the conditions are not equivalent and that of Poisson parentheses is stronger than the one about  the Jacobian matrix if $n>1$. 
They are equivalent however for $n=1$ and the result stated here holds in that case Proof of a property of the Poisson bracket.
