Facts about sums of ideals

Let R be a commutative ring with 1, and let A,B,C \subseteq R be ideals. Recall that A+B = \{a+b \ |\ a \in A, b \in B\}. Prove the following.

  1. A+B is an ideal
  2. A,B \subseteq A+B
  3. If A,B \subseteq C then A+B \subseteq C
  4. A+B = (A,B)
  5. If (A,B) = (1), then there exist \alpha \in A and \beta \in B such that \alpha+\beta = 1.
  6. If (A,B) = (1) and BC \subseteq A, then C \subseteq A.

Suppose a_1+b_1,a_2+b_2 \in A+B, and let r \in R. Then (a_1+b_1) + r(a_2+b_2) = (a_1+ra_2) + (b_1+rb_2) \in A+B since A and B are ideals. Moreover, 0 = 0+0 \in A+B. By the submodule criterion, A+B \subseteq R is an ideal.

For all a \in A, a = a+0 \in A+B. So A \subseteq A+B, and similarly B \subseteq A+B.

Suppose A, B \subseteq C for some ideal C. Since C is an ideal, it is closed under sums, so that a+b \in C for all a \in A and b \in B. Thus A+B \subseteq C.

Note that A, B \subseteq (A,B), so that A+B \subseteq (A,B) by the previous point. Now let x \in (A,B); by definition, x = \sum r_ic_i for some r_i \in R and c_i \in A \cup B. Collecting terms in A and B, we have x = \alpha + \beta for some \alpha \in A and \beta \in B. Thus A+B = (A,B).

Suppose (A,B) = (1). By the previous point, A+B = (1), and in particular 1 = \alpha+\beta for some \alpha \in A and \beta \in B.

Suppose (A,B) = (1). By the previous point, there exist \alpha \in A and \beta \in B such that \alpha+\beta = 1. Now let \gamma \in C. We have \gamma = \gamma(\alpha+\beta) = \gamma\alpha + \gamma\beta. Since BC \subseteq A, \gamma\beta \in A, and so \gamma \in A. Thus C \subseteq A.

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