**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.3 Exercise 7.3.35**

Solution:

(1) We show that $I(J+K) = IJ + IK$; the proof of the other equality is similar.

($\subseteq$) Let $\alpha \in I(J+K)$. Then $$\alpha = \sum a_i(b_i+c_i)$$ for some $a_i \in I$, $b_i \in J$, and $c_i \in K$. Then $$\alpha = \sum (a_ib_i + a_ic_i) = (\sum a_ib_i) + (\sum a_ic_i) \in IJ + IK.$$ ($\supseteq$) Note that since $J \subseteq J+K$, $IJ \subseteq I(J+K)$. Similarly, since $K \subseteq J+K$, $IK \subseteq I(J+K)$. By Exercise 7.3.34, $IJ+IK \subseteq I(J+K)$.

(2) ($\subseteq$) Let $x \in (J+K) \cap I$. Then $x \in I$ and $x = y+z$ for some $y \in J$ and $z \in K$. Since $J \subseteq I$, $x-y = z \in I$. Thus $z \in K \cap I$, and $x = y+z \in J+(K \cap I)$.

($\supseteq$) Let $x \in J+(K \cap I)$. Then $x = y+z$ where $y \in J$ and $z \in K \cap I$. Again because $J \subseteq I$, we have $x = y+z \in I$. Moreover, $$x = y+z \in J+K.$$ Thus $x \in (J+K) \cap I$.