Dimension of affine variety $V(f+g)$ in terms of the dimensions of $V(f)$ and $V(g)$

Question

Suppose I have two affine varieties over $\mathbb{C}$, say $V_1 = V(f_1, f_2)$ and $V_2 = V(g_1, g_2)$. Is there anything we can say about the dimension of $V(f_1 + g_1, f_2 + g_2 )$ in terms of the dimensions of $V_1$ and $V_2$? Thank you very much!

Answer 1

Let $V(f) = V(f_1, \dots, f_m)$, $V(g) = V(g_1, \dots, g_m)$, and $V(f + g) = V(f_1 + g_1, \dots, f_m + g_m)$.

If $x \in V(f)\cap V(g)$, then $f_1(x) = \dots = f_m(x) = g_1(x) = \dots = g_m(x) = 0$ so for every $i$,

$$(f_i + g_i)(x) = f_i(x) + g_i(x) = 0 + 0 = 0$$

and hence $x \in V(f + g)$. Therefore $V(f)\cap V(g) \subseteq V(f+g)$ so

$$\dim V(f + g) \geq \dim(V(f)\cap V(g)).$$

This inequality is far from optimal in general. For example, take $f_1 = \dots = f_m = 1$ and $g_1 = \dots = g_m = -1$ on $\mathbb{C}^n$, then $V(f + g) = V(0, \dots, 0) = \mathbb{C}^n$ and $V(f) = V(g) = \emptyset$. The corresponding inequality $$n = \dim \mathbb{C}^n \geq \dim\emptyset = 0$$ is not that helpful.