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Homework Statement
Problem 35, Section 7.3 of Dummit and Foote:
Let I, J, and K be ideals of R.
(a) Prove that I(J+K) = IJ+IK and IJ+IK = I(J+K).
(b) Prove that if J \subseteq I then I \cap (J + K) = J + (I \cap K).
2. Concern/Question
Despite the problem statement specifically stating that I, J, and K are ideals, this information is irrelevant to the proofs (only that I, J, and K are subrings of R is needed). Am I missing something or is this really just unnecessary information?
3. The Proofs
(a) To prove that I(J+K) = IJ + IK, we will show that each set is a subset of the other. To see that IJ + IK \subseteq I(J+K), note that IJ, IK \subseteq I(J+K); since I(J+K) is a ring, IJ + IK \subseteq I(J+K) +I(J+K) = I(J+K).
To see the opposite inclusion, first consider x \in I(J+K); x = i_1(j_1+k_1) + ... + i_n(j_n+k_n) for i_* \in I, j_* \in J, k_* \in K, n \in \mathbb{Z}^+. Reorganizing this equation, we get that x=i_1j_1+...+i_nj_n + i_1k_1 + ... + i_nk_n \in IJ+IK. Since x was arbitrary, this gives us the opposite inclusion and, thus, the desired equality. The second proof is analogous to the above one.
(b) Once again, the equality will be proven by showing that each set is a subset of the other. Consider J+(I\cap K); since J \subseteq I and I is a ring, J+(I \cap K) \in I. Furthermore, every element of J + (I \cap K) can be written as the sum of an element of J and an element of K. Since I \cap (J+K) is the largest such subring of R that satisfies both these properties, J+(I \cap K) \subseteq I \cap (J+K).
Now consider x \in I \cap (J+K); we can rewrite x as j+k for j \in J and k \in K. Since J \subseteq I, j \in I as well, and so since rings are closed under subtraction, k = x-j \in I. Therefore, every x \in I \cap (J+K) can be written as j+k' for k' \in I \cap K. From this, I \cap (J+K) \subseteq J+(I \cap K) follows and, thus, so does the desired equality.
