## Some more properties of ideal arithmetic

Let $R$ be a ring and let $I,J,K \subseteq R$ be ideals.

1. Prove that $I(J+K) = IJ+IK$ and $(I+J)K = IK+JK$.
2. Prove that if $J \subseteq I$, then $(J+K) \cap I = J+(K \cap I)$.

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 this previous exercise, $IJ+IK \subseteq I(J+K)$.
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 $x \in (J+K) \cap I$.