# Why is the following affine varieties equal? "Closure theorem"

I am very unsure about this "closure theorem" of ideals and varieties. I'm not sure if anyone here can answer this concisely as I think some notations may differ from what others know...

Say $$I$$ is an ideal and $$I=$$. I will state the extension theorem since it's tedious to explain the notation that follows.

Extension theorem

Let $$I=$$ be an ideal in $$\mathbb{C}[x_1,...,x_n]$$ and $$I_1$$ be the first elimination ideal of I. For each $$1 \leq i \leq s$$ write $$f_i$$ be written as $$f_i = g_i(x_2,...,x_n)x^{N_i}+\text{terms in which x_1 has degree less than N_i}$$. Where $$N_i \geq 0$$ and $$g_i \in \mathbb{C}[x_2,...,x_n]$$ is nonzero. Suppose that we have a partial solution $$(a_2,...,a_n) \in V(I_1)$$. If $$(a_2,...,a_n) \not\in V(g_i)$$ then there exists some $$a_1 \in \mathbb{C}$$ that extends to $$V(I)$$.

My main concern is the following though,

$$I$$ is an ideal as above and let us say $$J=$$. So $$J_l$$ and $$I_l$$(l-th elimination ideal) may differ but $$V(I_l)=V(J_l)$$.

This is (apparently) due to the following theorem

Closure theorem

$$V=V(f_1,...,f_s)$$ and $$I_l$$ is the l-th elimination ideal, then $$V(I_l)$$ is the smallest affine variety containing $$\pi_l(V) \subset \mathbb{C}^{n-l}$$

($$\pi_l$$ is the projection map $$\pi_l:\mathbb{C}^n \rightarrow \mathbb{C}^{n-l}$$)

I don't see why $$V(I_l)=V(J_l)$$ due to this closure theorem. It says in my lecture notes "they are both the smallest variety containing $$\pi_l(V)$$[by closure theorem] so they are equal"

But my argument is, isn't $$V$$ different in each $$J$$ and $$I$$? So for $$I$$, $$V=V(f_1,...,f_s)$$ but for $$J$$, shouldn't $$V=V(f_1,...,f_s,g_1,...,g_s)$$? They are different affine varieties, aren't they? So $$\pi_l(V)$$ should also be different.

So sure, $$V(I_l)$$ is the smallest variety that contains $$\pi_l(V=V(f_1,...,f_s) )$$ and $$V(J_l)$$ is the smallest variety containing $$\pi_l(V=V(f_1,...,f_s,g_1,...,g_s))$$. The notes seem to speak as if $$\pi_l(V(f_1,...,f_s))=\pi_l(V(f_1,...,f_s,g_1,...,g_s))$$ which I am not convinced with.

Does anyone know an explanation to this?

$V(f_1,\ldots,f_s) = V(f_1,\ldots,f_s,g_1,\ldots,g_s)$