With $ x\equiv (x_1, x_2, x_3, x_4), $ consider the polynomial system, $P(x_{1}, x_{2}, x_{3}, x_{4}) = 0$, given by
$P_1(x) = x_1 * x_3,$
$P_2(x) = x_2 * x_4,$
$P_3(x) = x_1 + x_3 + \beta_1(x_1 * x4 + x2 * x3),$
$P_4(x) = x_2 + x_4 + \beta_2(x_1 * x4 + x2 * x3).$
I want to know if there is a simple way to show that the set of points where the Jacobian matrix is non-singular is an open dense set in $\mathbb{R} ^4 $. My present ideas go something like this:
The determinant of the Jacobian matrix is clearly a polynomial, say $g:\mathbb{R}^4 \rightarrow \mathbb{R}.$ I try to show $g$ is NOT an identically zero polynomial and hence its set of zeros, or equivalently the set of critical points, has measure zero. Hence its complement, the set of regular points, is dense. Lastly, since $g$ is continuous, the set of regular points is open.
