Why is $|V(I)| \leq d_1\cdots d_n$?

Question

If $I \subset K[x_1,\dots,x_n]$ is a zero dimensional ideal and $$V(I) = \{ (\alpha_1,\dots,\alpha_n) \in K^n: f((\alpha_1,\dots,\alpha_n)) = 0\ \forall f\in I\}$$ (the variety). Then if $G$ is a Groebner Basis for $I$ w.r.t. some monomial order $>$, then $\forall i$ $\exists g_i \in G$ such that $LT(g_i) = x_i^{d_i}$ for some $d_i > 0$. Why is $|V(I)| \leq d_1\cdots d_n$?

Answer 1

Suppose $x_1>\cdots>x_n$. Then $g_i=g_i(x_i,\dots,x_n)$ is a polynomial in $x_i,\dots,x_n$ for $i=1,\dots,n$.

In particular, $g_n$ is a polynomial only in $x_n$. Let $\alpha_n$ be a root of $g_n$. (There are at most $d_n$ roots.) Then consider $g_{n-1}(x_{n-1},\alpha_n)$. This has at most $d_{n-1}$ roots, and let $\alpha_{n-1}$ be one of them. Now consider $g_{n-2}(x_{n-2},\alpha_{n-1},\alpha_n)$, and so on. Finally we get at most $d_1\cdots d_n$ common roots for $G$.