Proposition

Let $I, J, K \subset k[x_1, \cdots, x_n]$ be ideals. Prove the following:

$IJ \subset K$ if and only if $I \subset K:J$.
$(I:J):K = I:JK$.

Solution

1

Suppose $IJ \subset K$. Let $x \in I$. Since $IJ \subset K$, $xJ \subset K$. Therefore, $x \in K:J$. Therefore, $I \subset K:J$.

Suppose $I \subset K:J$. Let $x \in IJ$. Then $x = \sum_{i=1}^{n} a_ib_i$ for some $n \in \mathbb{N}$, $a_i \in I$, $b_i \in J$. Let $i$ be given. Since $a_i \in I \subset K:J$, $a_iJ \subset K$. Since $b_i \in J$, $a_ib_i \in K$. This implies that $x \in K$ because $K$ must be closed under finite addition. Therefore, $IJ \subset K$.

Hence, $IJ \subset K \iff I \subset K:J$.

2

Let $x \in k[x_1, \cdots, x_n]$.

\[\begin{align*} x \in (I:J):K &\iff xK \subset I:J \\ &\iff (xK)J \subset I & \text{(by part 1)}\\ &\iff x(KJ) \subset I \\ &\iff x(JK) \subset I \\ &\iff x \in I:(JK). \end{align*}\]

Therefore, $(I:J):K = I:(JK)$.