Say $V$ is an irreducible variety in $\mathbb{C}^n$ of dimension $d$. I can view $\mathbb{C}^n$ as $\mathbb{R}^{2n}$ and inside $\mathbb{R}^{2n}$, $V$ is an irreducible (real) variety of dimension $2d$. If I pick a polynomial $f\in\mathbb{R}[x_1,\dots,x_{2n}]$, there are 2 options:
- $\overline{Z(f)\cap V}=V$ (this also includes the option $V\subset Z(f)$)
- $\dim\overline{Z(f)\cap V}<\dim V$.
Here, $\overline{Z(f)\cap V}$ denotes the Zariski closure inside $\mathbb{C}^n$. I guess that for a generic $f$, we have $\overline{Z(f)\cap V}=V$ since the intersection has dimension $2d-1$ (as a real variety) so its closure has dimension at least $d$.
I want to know for which polynomials $f$ the second option applies. Say I restrict to all polynomials of degree at most $k$. Do the polynomials satisfying the second option form a variety inside $P_k=\{f\mid \deg(f)\leq k\}$? If yes, what are the defining polynomials of this variety?
As an example, I consider $Z(x-y)\subset\mathbb{C}^2$. Viewed as a real variety, I have $V=Z(x_1-y_1,x_2-y_2)\subset\mathbb{R}^4$. If $f$ is a real polynomial in variables $x_1,x_2,y_1,y_2$, I have $\overline{Z(f)\cap V}\neq V$ if and only if $Z(f)\cap V$ is finite. And this happens if and only if $f(x_1,x_2,x_1,x_2)$ has only finitely many zeros in $\mathbb{R}^2$. For example $f$ might be of the form $$ f = x_1^2+x_2^2+y_1^2+y_2^2 $$ so $Z(f)\cap V=\{0\}$.
