Proposition

Let $I \subset k[x_1, \cdots, x_n]$ be an ideal.

1. Prove that $I_l = I \cap k[x_{l + 1}, \cdots, x_n]$ is an ideal of $k[x_{l + 1}, \cdots, x_n]$.
2. Prove that the ideal $I_{l + 1} \subset k[x_{l + 2}, \cdots, x_n]$ is the first elimination ideal of $I_l \subset k[x_{l + 1}, \cdots, x_n]$.

Solution

1

• $0 \in I_l$, so $I_l$ is nonempty.
• Let $a, b \in I_l$. Then $a - b \in I$ because $I$ is an ideal, and $a - b \in k[x_{l + 1}, \cdots, x_n]$. Therefore, $a - b \in I_l$.
• $f \in I_l, g \in k[x_{l + 1}, \cdots, x_n]$. Then $gf \in k[x_{l + 1}, \cdots, x_n]$ and $gf \in I$. Thus $gf \in I_l$.

Hence, $I_l$ is an ideal of $k[x_{l + 1}, \cdots, x_n]$.

2