1
$\begingroup$

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 $I=<f_1,...,f_s>$. I will state the extension theorem since it's tedious to explain the notation that follows.

Extension theorem

Let $I=<f_1,...,f_s>$ 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 $J=<f_1,...,f_s, g_1,...,g_s>$. 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?

$\endgroup$

1 Answer 1

0
$\begingroup$

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

See page 138 of Ideals, Varieties and Algorithms.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .