Degree of Jacobian of homogeneous polynomials What is the degree of the Jacobian (as a polynomial) of 3 homogeneous polynomials in 3 variables of degrees say $m_1, m_2$ and $m_3$ ?
I don't know how to prove that it is independent. In my case the degrees are 
$2,6$ and $10$. I have tried with some example and I found that the degree is $15$ in this case. 
 A: Having done an example or two should give you a feeling for what can be said about the degree of the Jacobian knowing only the homogeneous degrees of the three polynomials are $m_1,m_2,m_3$ in three variables (say) $x,y,z$.
The Jacobian is the determinant of the $3\times 3$ matrix of first partial derivatives of the three polynomials.
The first partial derivatives of a polynomial of homogeneous degree $m$ will be polynomials of homogeneous degree $m-1$.  So in the first row of the $3\times 3$ matrix the partials have degree $m_1-1$, in the second row $m_2-1$, and in the third row $m_3-1$.  The product of three polynomials of homogeneous degrees $m_1-1,m_2-1,m_3-1$ will be of homogeneous degree (their sum of degrees) $m_1+m_2+m_3-3$.  
Collecting like terms, any signed sum of polynomials of homogeneous degree $m_1+m_2+m_3-3$ will again be of that form, unless it happens to be identically zero.
An example of this with three polynomials of homogeneous degrees $2,6,10$ resp. is $p_1 = x^2, p_2 = y^3 z^3, p_3 = x^4 y^3 z^3$.  Their Jacobian is:
$$ \mathbf{J} = \det \begin{bmatrix} 2x & 0 & 0 \\ 0 & 3y^2z^3 & 3y^3z^2 \\ 4x^3y^3z^3 & 3x^4y^2z^3 & 3x^4y^3z^3 \end{bmatrix} $$
A brief if tedious calculation (expanding by minors) shows that $\mathbf{J}\equiv 0$.  This can be inferred from the functional relationship:
$$ p_3 = p_1^2 p_2 $$
which implies the polynomial system is not invertible (uniquely solvable for $x,y,z$) at any point in $\mathbb{R}^3$, and therefore vanishes at every point.
However if the Jacobian computed as above is nonzero, it will be homogeneous of degree $m_1+m_2+m_3-3$ because there exist nonzero terms of that degree.
