I'm becoming desperate with a problem, which doesn't seem so difficult.
I have a real algebraic set $X$ in $\mathbb{R}^{15}$ defined by $$ x_1^2 + y_1^2 + z_1^2 - 1 = 0\\ x_2^2 + y_2^2 + z_2^2 - 1 = 0\\ x_3^2 + y_3^2 + z_3^2 -1 = 0 \\ c_1^2 + s_1^2 - 1 = 0 \\ c_2^2 + s_2^2 - 1 = 0 \\ c_3^2 + s_3^2 - 1 = 0 \\ l_1 = \ldots = l_6 = 0, $$ where $l_1, \ldots, l_6$ are linear in the $x_i,y_i,z_i,c_i,s_i$, (basically $X$ is the preimage of zero of a smooth map $(S^2)^3 \times (S^1)^3 \to \mathbb{R}^6$). I found 24 points, where all principal minors of the jacobian are zero. I want to show now, that these points are "singular", i.e. $X$ is not locally an embedded submanifold of $\mathbb{R}^{15}$ around these points.
I know that this set is irreducible of dimension $3$ and I think I can calculate the complex tangent cones $\mathcal{C}$ (tangent cone of the complex variety) in these points (with groebner bases) but I think it is not true in general, that $\mathcal{C} \cap \mathbb{R}^{15} = \mathcal{C}_\mathbb{R}$?
Does anybody have an idea how to proof this, or could maybe give a pointer to literature which considers this kind of problems or just vaguely related problems. Its quite difficult to find any research in this direction (or I don't know where to look).
Many Thanks!