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)).
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