# Algebraic Geometry Proof Explanation

In the last paragraph of the proof given in this SE question ($$I(V \times W ) =I(V) + I(W)$$), the OP alludes to an inductive / infinite descent argument, saying, "Continuing this process, we finally get an expression with zero terms... etc."

Whilst the proof seems to make perfect sense up to this point, I do not understand the inductive step.

Also, I just want to get some verification of the fact that the target identity, $$I(V \times W ) =I(V) + I(W)$$, holds for arbitrary sets $$V$$, $$W$$, not just algebraic sets.

Any help would go a long way!

You can phrase it another way. Suppose $$S = I(V\times W) \setminus (I(V) + I(W))$$ is nonempty. Among $$f\in S$$, pick $$f$$ so that it has an expression $$f = \sum_{i=1}^n f_i g_i$$ with $$f_i \in k[x_1,\dots,x_n]$$ and $$g_i \in k[x_{n+1},\dots,x_{n+m}]$$ for all $$1\le i \le n$$ so that $$n$$ is minimal, i.e. there is no other $$h\in S$$ with such an expression but with fewer terms.
Since $$f \notin I(W)$$, we must have that some $$g_i \notin I(W)$$. So there is some $$b\in W$$ with $$g_i(b) \ne 0$$, and we assume without loss of generality that $$i=1$$. Now, as in the post you refer to, there is $$p\in I(V)$$ so that $$f_1 = \frac{p}{g_1(b)} - \frac{g_2(b)}{g_1(b)} f_2 - \dots - \frac{g_n(b)}{g_1(b)}f_n.$$ Now we have $$f = \sum_{i=1}^n f_i g_i = \left(\frac{p}{g_1(b)} - \frac{g_2(b)}{g_1(b)} f_2 - \dots - \frac{g_n(b)}{g_1(b)}f_n\right)g_1+ \sum_{i=2}^n f_i g_i$$ $$= \frac{p\cdot g_1}{g_1(b)} + \sum_{i=2}^n f_i \left(g_i-\frac{g_i(b)}{g_1(b)}g_1\right).$$ Now, if $$f-\frac{p\cdot g_1}{g_1(b)} \in I(V) + I(W)$$, then $$f \in I(V) + I(W)$$ too, since $$\frac{p\cdot g_1}{g_1(b)} \in I(V)$$ (the expanded ideal). So $$h = f-\frac{p\cdot g_1}{g_1(b)} \notin I(V) + I(W)$$. But clearly $$h \in I(V+W)$$, and $$h$$ has the form in the first paragraph with fewer terms than $$f$$ had. This contradicts the minimality of the number of terms in the expression of $$f$$.