Proposition

Show that $I(\{a_1, \cdots, a_n\}) = \ev{x_1 - a_1, \cdots, x_n - a_n}$.

Solution

Let $I = I(\{a_1, \cdots, a_n\})$. Since each $x_i - a_i$ vanishes at $\{a_1, \cdots, a_n\}$, $x_i - a_i \in I$ for each $i$. Thus $\ev{x_1 - a_1, \cdots, x_n - a_n} \subset I$. $\ev{x_1 - a_1, \cdots, x_n - a_n}$ is a maximal ideal by Proposition 9 on P.210 (Ideals, Varieties, and Algorithms).

On the other hand, $I \ne k[x_1, \cdots, x_n]$ because $(x_1 - a_1) \cdots (x_n - a_n) + 1 \notin I$. Therefore, $I = \ev{x_1 - a_1, \cdots, x_n - a_n}$.