Let $S_1$, $S_2$ be two surfaces in $\mathbb{R}^3$ implicitly defined by, respectively, $f_1(x,y,z)=0$ and $f_2(x,y,z)=0$. How do I check if the curve $S_1 \cap S_2$ has singular points without explicitly deriving a parametrization? Example:
$$S_1 = \{x^3 + 3(y^2+z^2)^2=2\} \\ S_2 = \{x^4 = y^3\}$$
I see no obvious way for a parametrization of $S_1 \cap S_2$, so how do I go about checking if it has singular points?
And, as a bonus more general question: in situations like this, how much information about the intersection curve can I derive just by looking at the surfaces?
Write down the Jacobian matrix $$ \big{(}\frac{\partial f_i}{\partial x_j}\big{)}. $$ A point will be singular iff the Jacobian matrix doesn't have full rank.
In this case, the matrix is $$ \begin{pmatrix} 3x^2 & 6y(y^2+z^2) & 6z(y^2+z^2) \\ 4x^3 & -3y^2 & 0 \end{pmatrix}. $$
Let's say $x = 0$. Then this matrix will fail to have full rank if $18y^2z(y^2+z^2) = 0$, i.e. if $y$ or $z$ is $0$. Staring at the equations for a minute, we see that the two points $(0, 0, \pm \sqrt[4]{\frac{2}{3}}$) will be singular points.
If $x \neq 0$, then column 3 is linearly independent of column $1$ unless $z = 0$. In that case, column 2 is $\begin{pmatrix} 6y^3 \\ -3y^2 \end{pmatrix}$, and so there could be a singular point where $-9x^2 -24x^3y^3 = 0,$ i.e. where $3+8xy^3 = 0$. Such a point is on $S_2$ iff $3 + 8x^5 = 0$, i.e. $x = \sqrt[5]{-\frac{3}{8}}$. Now we can solve for $y$ and see that such a point isn't on $S_1$, so there's no other singular points beyond the two we already found.