TITLE: Algebraic varieties in $\mathbb{C}^n$ cannot have interior points
QUESTION [9 upvotes]: I know that the zero-set of a non-zero polynomial in $\mathbb{C}[x_1,...,x_n]$ can not have interior points, but I'm trying to find a proof that doesn't require a knowledge of complex analysis like holomorphic functions, etc...
So, suppose that $V=\{(x_1,\cdots, x_n)\in \mathbb{C}^n: p(x_1,\cdots,x_n)=0\}$. Now if there is an interior point in $\mathbf{y}=(y_1,\cdots,y_n) \in V$ then there exists $\epsilon>0$ such that $\displaystyle N_\epsilon(\mathbf{y}) \subseteq V$. But we know that in every metric space open balls are convex, so for all $\lambda \in \mathbb{R}$ the point $\mathbf{\bar{y}}=\lambda\cdot\mathbf{y}+(1-\lambda)\cdot \mathbf{y'} \in N_\epsilon(\mathbf{y})$ lies in $V$where $\mathbf{y'} \in N_\epsilon(\mathbf{y})$.
If I plug $\mathbf{\bar{y}}$ into $p(x_1, \cdots, x_n)=0$ I'll get a new non-zero polynomial $p_1(x_1,\cdots,x_n,\lambda)=0$. Now, if I fix $x_1,\cdots,x_n$ and consider $p_1(x_1,\cdots,x_n,\lambda)$ as a polynomial only in $\lambda$ then I see that $f(\lambda)=p_1(x_1,\cdots,x_n,\lambda)$ then because of convexity $f(\lambda)$ must vanish on every $\lambda \in [0,1]$, but from the fundamental theorem of algebra we know that every non-zero polynomial in $\mathbb{C}[\lambda]$ can have a finite number of roots, so, $f(\lambda)=0$. But I don't see how $f(\lambda)=0$ could lead me to a contradiction.

REPLY [2 votes]: One word: differentiate (the defining equations).
It's enough to prove the statement for varieties defined by one equation, because having no interior is a property inherited by subsets.  
If the zero set contains an interior point, all derivatives of the equation defining the variety are $0$ at that point (and thus in the a neighborhood of that point).  But then the variety defined by any derivative of the equation has the same interior point.  By a sequence of differentiations, the highest degree of any term in the defining equation can be reduced to $1$, and the question is reduced to whether a hyperplane has interior points.   
If you prefer, differentiate again to reduce to a variety of degree $0$ with no points, therefore no interior points.