Dimension of a variety defined by few non homogeneous polynomials Let $P_1$, $P_2$, $\ldots$ , $P_t$ be $t$ (possibly non homogeneous) polynomials in $n$ variables over the field of complex numbers. If the variety $V(P_1, P_2, \ldots, P_t)$ defined by these polynomials is non-empty, then is it true that the dimension of $V$ is at least $n-t$ ?
This is true if each $P_i$ is homogeneous, but what happens in the non-homogeneous case, assuming that the variety is non-empty? I would really appreciate any pointers.  

This is true. Consider the following lemma:
Let $X$ be an irreducible variety, $f\in k[X]$ a nonzero nonconstant function, and $Y$ an irreducible component of $V(f)$. $Y$ has codimension 1 in $X$.
Proof: We may restate as follows. Let $A$ be an an affine $k$-algebra which is a domain with $f\in A$ a nonzero nonunit. Let $P$ be a minimal prime ideal containin $f$- then $P$ must have height one. The height is at most one, by Krull's principal ideal theorem, and at least one, as $X$ irreducible.
I may then apply this lemma $t$ times to the chain $V(0)\supset V(P_1)\supset V(P_1,P_2)\supset\cdots\supset V(P_1,\cdots,P_t)$. This proves the claim. (The reason it must be at most instead of exactly is that a function $P_i$ may fail to be a nonzero nonunit on some irreducible component of $V(P_1,\cdots,P_j)$ where $i<j$. In this case, the final line of the lemma where it is shown that the height is at least one fails.)
