I'm having some trouble with the following problem:
Let $S_1$ and $S_2$ be two regular surfaces that intersect at a point $p$, such that $T_pS_1\neq T_pS_2$. Prove that there is an open neighborhood $U\subseteq\mathbb R^3$ of $p$ such that $S_1 \cap S_2\cap U$ is the trace of a regular curve.
My first idea was to use the fact that, because $S_1$ and $S_2$ are regular surfaces, there are open neighborhoods $W,V$ in $\mathbb R^3$ of $p$ and $f:W\to\mathbb R$, $g:V\to\mathbb R$ differentiable, such that:$$W\cap S_1=f^{-1}(\{0\})$$ $$V\cap S_2=g^{-1}(\{0\})$$ So with this, we can conclude, if we set $U=W\cap V$, that $S_1\cap S_2\cap U$ is defined by the following system of equations: $$f(x,y,z)=0$$ $$g(x,y,z)=0$$
This is a system with 2 equations and 3 variables, so my intuition tells me that the solution will have 1 free variable, and thus be a curve, but I have no idea how to formalize this.
My second idea was the following: If $T_pS_1$ and $T_pS_2$ intersect and are not the same, then $T_pS_1\cap T_pS_2$ is a one-dimensional linear subspace of $\mathbb R^3$, let it be the space generated by the vector $v$. Now we know a vector that will be tangent to the curve we want to find. Maybe with this, we can construct the curve?
Does any of these ideas work? If that's now the case, how can this be done?
Hint (for approach 1): Notice that $$F : U \mapsto \Bbb R^2, \qquad F : (x, y, z) \mapsto (f(x, y, z), g(x, y, z))$$ is a differentiable function with zero set $\Sigma := U \cap S_1 \cap S_2$, so to show that $\Sigma$ a smooth curve, it's sufficient to check that $F$ has constant rank $2$ along $\Sigma$, i.e., that $df \wedge dg$ vanishes nowhere on $\Sigma$. This property is not automatic for $f$ and $g$ as in the question statement, but notice that we can furthermore insist that our choice of defining functions $f, g$ satisfies $df \neq 0$, $dg \neq 0$ along $\Sigma$. Then, transversality guarantees that $df \wedge dg \neq 0$ there.