I am trying to understand this statement, it is in "Principles of Algebraic Geometry" by Philip Griffiths and Joe Harris.
In page 13, third point, they are trying to prove that an analytic variety irreducible at $0$ is a analytic hypersurface. What I don't understand is this statement:
Last, let $V\subseteq U\subseteq\mathbb{C}^{n}$ be an analytic variety irreducible at $0\in V$ such that for arbitrary small neighbourhoods $\Delta$ of $0\in\mathbb{C}^{n}$, $\pi(V\cap\Delta)$ contains a neighbourhood of $0\in\mathbb{C}^{n}$. Write $V=\{f_{1}(z)=...=f_{k}(z)=0\}$ near $0$. Then the functions $f_{i}\in\mathcal{O}_{n}$ must have a common factor in $\mathcal{O}_{n}$, since otherwise $V$ would be contained in the common locus of two relatively prime functions, and by assertion $2$, $\pi(V\cap\Delta)$ would be a proper analytic subvariety of $\mathbb{C}^{n-1}$.
And so where is the contradiction here? The only one I can come up with is that if suppose $V$ is contained in $W=\{f_{i}=f_{j}=0\}$ (locally) where $f_{i}$ and $f_{j}$ are relatively prime, then we have the resultant $\gamma=\alpha f_{i}+\beta f_{j}$ where $\gamma\in\mathcal{O}_{n-1}$. But the locus $\{\gamma=0\}$ is $\pi(W)$ which contains an open set, so $\gamma$ is identically zero, which is a contradiction because we know that the resultant is not identically zero.
I am not sure if this argument works, but I would appreciate if someone can help me understand what is meant in the quote. (Particularly, what is wrong with $\pi(V\cap \Delta)$ being a proper analytic subvariety?)
Thanks!