I am asking a follow up question to a previous question I posted here recently.
Let $f(x_1,...,x_n)$ be a (complex) irreducible polynomial in $n$ variables (you can assume that $n$ is very large, if that helps). Thus, the algebraic variety $$Z_f:=\{(x_1,...,x_n)\in \mathbb C^n:f(x_1,...,x_n)=0\}$$ is irreducible.
Let $S\subset \mathbb C^n$ be a given hyperplane, and let $V\subset S$ be a subvariety of codimension $1$ in $S$. Assume also that $Z_f\cap S\subset S$ is of codimension $1$ in $S$. I am interested to know under which condition $Z_f\cap V \subset S$ is of codimension $2$ in $S$. Of course, this should hold "generically", but since $S$ is already given, I cannot change it, so I am looking for another way to ensure this.
My original thought was to show that $Z_f\cap S$ is irreducible, since in this case it will not share a common factor with $V$, and so their intersection will be transversal (and thus of codimension $2$). But there is no reason for this to be true for our given $S$.
I also thought about considering the orthogonal projection $P:\mathbb C^n\rightarrow \mathbb C^n$ onto $S$, and looking at the set $Z'=P(Z_f)\subset S$. This is the image of an irreducible variety under a regular map, and is thus irreducible. This means that $Z'$ intersects $V$ transversally, and so their codimensions in $S$ add. Moreover, conveniently we have that $Z_f\cap S\subset Z'$, which helps relate all of this to our original question. If we knew that $Z'$ is of codimension $1$ in $S$, then once again we would be done - the intersection $Z'\cap V$ is of codimension $2$, giving the desired property for $Z_f\cap V$. But we do not know this (it could hypothetically be of codimension $0$).
This is where I am stuck. Are there any natural, yet relatively general assumptions I can make to still reach my desired result? I am open to different ideas.
Thanks in advance.
