## Two basic properties of ideal quotients

Recall that if $R$ is a ring with ideals $I$ and $J$, the ideal quotient $(I:J)$ in $R$ is defined to be $(I:J) = \{r \in R \ |\ rJ \subseteq I \}$. This set is certainly an ideal of $R$.

1. Suppose $R$ is an integral domain, $d \neq 0$ is in $R$, and $I \subseteq R$ an ideal. Prove that if $G = \{g_i\}_{i=1}^k$ is a generating set for $I \cap (d)$ where $g_i = h_id$, then $H = \{h_i\}_{i=1}^k$ is a generating set for $(I:(d))$.
2. Let $R$ be a commutative ring, $I \subseteq R$ an ideal, and $D = \{d_i\}_{i=1}^k \subseteq R$ with $d_i \neq 0$. Prove that $(I : (D)) = \bigcap_{d_i \in D} (I : (d_i))$.

1. Suppose $r \in (I : (d))$. Then $r(d) \subseteq I$, so $rd \in I$. Certainly $rd \in (d)$, so that $rd \in I \cap (d) = (G)$. Say $rd = \sum s_ig_i = \sum s_ih_id$. Since $R$ is an integral domain, $r = \sum s_ih_i$, and thus $r \in (H)$. Conversely, suppose $r \in (H)$, with $r = \sum s_ih_i$. Then $rd = \sum s_ih_id = \sum s_ig_i$, and thus $rd \in I$. So $r(d) \subseteq I$, and thus $r \in (I : (d))$. Hence $(I : (d)) = (H)$.
2. Suppose $r \in (I : (D))$. Then $r(D) \subseteq I$, and in particular $rd_i \in I$ for all $i$. So $r(d_i) \subseteq I$ for all $i$, and we have $r \in \bigcap_{d_i \in D} (I : (d_i))$. Conversely, suppose $r \in \bigcap_{d_i \in D} (I : (d_i))$. Then $r(d_i) \subseteq I$ for all $i$, and $rd_i \in I$ for all $i$. Now $r(D) = (rD) \subseteq I$, so that $r \in (I : (D))$. Thus $(I : (D)) = \bigcap_{d_i \in D} (I : (d_i))$.