MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

## Regular Points of a polynomial system [closed]

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.

-
 This question is not appropriate for mathoverflow and it seems to me that you're be able to sort this out yourself. Perhaps you can try math.stackexchange.com if you need more help. – J.C. Ottem Dec 28 2011 at 3:42