Let us fix a field $k$. Using Gauss Lemma for polynomial ring $k[X][Y]$ over the UFD $k[X]$, one can prove that for two polynomials $F, G \in k[X,Y]$ with no common factors we have $|V(F) \cap V(G)| < \infty$. This fact in turn shows that any irreducible plane curve $V(F)$ with $F \in k[X,Y]$ has only a finite number of singular points, i.e. the number of points $P$ such that $F(P) = \partial_X F (P) = \partial_Y F(P) = 0$ is finite.
I am curious if the result extends to general affine irreducible hypersurfaces cut out in $\mathbb{A}^n$ by $F \in k[X_1, \dots, X_n]$. Since the Gauss lemma argument does not pass through for $n > 2$, I think the result is false for general hypersurfaces. But I am having trouble finding a concrete example.
Is there an affine irreducible hypersurface $V(F)$ for $F \in k[X_1, \dots, X_n]$ where $n>2$, with infinitely many points at which all the partial derivatives vanish? If not, how do we prove the finiteness of the singular points? I can assume $k$ to be algebraically closed if necessary.
Take any irreducible algebraic curve $X\subset \mathbb A^2$ with finite non empty singular set $\emptyset\neq S\subset X$.
Then for $n\gt2$ the irreducible hypersurface $X\times \mathbb A^{n-2}\subset \mathbb A^2\times \mathbb A^{n-2}=\mathbb A^n$ has as infinite singular set $S\times \mathbb A^{n-2}$.
If $X$ has $f(x,y)=0$ as equation (for example $x_2^2-x_1^3=0$), then the hypersurface $X\times \mathbb A^{n-2}\subset\mathbb A^n$ has as equation the exact same equation $f(x_1,x_2)=0$, where we now see $f$ as a polynomial $f\in k[x_1,x_2,\cdots,x_n]$.