Verify that a diagram commutes
by Hidenori
Proposition
Prove that the diagram (3) on P.134 (Ideals, Varieties, and Algorithms) commutes and $i(k^m) = V$.
Solution
For any $(t_1, \cdots, t_m) \in k^m$,
\[\begin{align*} (\pi_m \circ i)(t_1, \cdots, t_m) &= \pi_m(t_1, \cdots, t_m, f_1(t_1, \cdots, t_m), \cdots, f_n(t_1, \cdots, t_m)) \\ &= (f_1(t_1, \cdots, t_m), \cdots, f_n(t_1, \cdots, t_m)) \\ &= F(t_1, \cdots, t_m). \end{align*}\]$i(k^m) \subset V$ because each $x_i - f_i$ vanishes at $(f_1(t_1, \cdots, t_m), \cdots, f_n(t_1, \cdots, t_m))$ for any $(t_1, \cdots, t_m) \in k^m$. $V \subset i(k^m)$ because if $x_i - f_i$ vanishes at $(t_1, \cdots, t_{m + n})$, then $t_{m + i} = f_i(t_1, \cdots, t_m)$. Therefore, $i(k^m) = V$.
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