Dimension of $V(g_1(\mathbf{x})- g_1(\mathbf{y}), ...,g_s(\mathbf{x})- g_s(\mathbf{y}) )$ compared to $V(g_1(\mathbf{x}), ...,g_s(\mathbf{x}) )$? Let $g_1, ..., g_s$ be non-constant homogeneous polynomials in $n$ variables with coefficients in $\mathbb{C}$. Let $\mathbf{x}$ and $\mathbf{y}$ be two sets of $n$ variables. 
Let $V = V(g_1(\mathbf{x})- g_1(\mathbf{y}), ...,g_s(\mathbf{x})- g_s(\mathbf{y})   ) \subseteq \mathbb{A}_{\mathbb{C}}^{2n}$ and $W = V(g_1(\mathbf{x}), ...,g_s(\mathbf{x})   ) \subseteq \mathbb{A}_{\mathbb{C}}^{2n}$. 
Is it always true that
$\dim V \leq \dim W$? or maybe even that $\dim V = \dim W$? I would greatly appreciate any comments.  
 A: No to both questions.  Take $n=1, g_1(x)=17$. Then $W=V(17)=\emptyset, V=V(0)=\mathbb A^2_\mathbb C$.
Edit
This answers the original question which did not require homogeneity of the $g_i$'s. After seeing the counterexample above  the OP modified his question by requiring the $g_i$'s to be homogeneous and non-constant.
A: Here is a heuristic argument for the equality $\dim V=\dim W$ when the $g_i$'s are polynomials with constant term equal to zero (but not necessarily homogeneous).
1) Consider  the morphism $g=(g_1,\cdots, g_s):\mathbb A^n\to \mathbb A^s$  and let $$S=V_{\mathbb A^n}(g_1,\cdots, g_s)\subset \mathbb A^n, \;\Sigma=\overline {g(S)}\subset \mathbb A^s$$ Since $S=g^{-1}(0)$ is a non-empty fiber of $g$, we expect  $\dim S\stackrel {?}{=}n-\dim \Sigma$
2) Since the morphism  $p:\mathbb A^{2n}\to \Sigma:(x,y)\mapsto g(x)$ is dominant and since $W=p^{-1}(0)$ we expect $$\dim W\stackrel {?}{=}2n-\dim \Sigma     $$ 3) We have a dominant morphism $g\times g:\mathbb A^{2n}\to \Sigma\times \Sigma:(x;y)\mapsto (g(x),g(y))$.
 The variety $V\subset \mathbb A^{2n}$ is the inverse image under that morphism of the diagonal subvariety $\Delta =\{(\sigma,\sigma )\vert \sigma\in \Sigma \}\subset \Sigma\times \Sigma$. Thus we expect $$\dim V\stackrel {?}{=}2n-\dim \Delta=2n-\dim \Sigma$$ 
4) The expected equalities above show why it is reasonable to hope that $\dim V=\dim W$. 
