**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.3 Exercise 7.3.34**

Solution:

(1) We wish to prove the following: (a) $I+J$ is an ideal of $R$ and (b) If $K \subseteq R$ is an ideal and $I,J \subseteq K$, then $I+J \subseteq K$.

(a) First we show that $I+J$ is an ideal of $R$. Let $a_1+b_1$, $a_2+b_2 \in I+J$ for some $a_i \in I$ and $b_i \in J$. Then $$(a_1+b_1)-(a_2+b_2) = (a_1-a_2) + (b_1-b_2) \in I +J$$ since $I$ and $J$ are closed under subtraction. Now let $r \in R$. We have $$r(a+b) = ra + rb \in I+J$$ since $I$ and $J$ absorb $R$ on the right; similarly, $(a+b)r \in I + J$. Thus $I+J$ is an ideal of $R$.

(b) Now suppose $K$ is an ideal of $R$ with $I,J \subseteq K$. If $a+b \in I+J$, then since $K$ is closed under addition, we have $a+b \in K$. Hence $I+J \subseteq K$.

(2) We wish to prove that $IJ$ is an ideal of $R$ and that $IJ \subseteq I \cap J$.

First, let $\alpha,\beta \in IJ$, with $\alpha = \sum a_ib_i$ and $\beta = \sum c_id_i$, where $a_i,c_i \in I$ and $b_i,d_i \in J$. Clearly $\alpha + \beta \in IJ$. Now let $r \in R$. Since $I$ is an ideal of $R$, $$r\alpha = \sum (ra_i)b_i \in IJ.$$ Similarly, $\alpha r \in IJ$. Thus $IJ$ is an ideal of $R$.

Now consider again $\alpha = \sum a_ib_i$. Since $a_i \in I$ and $I$ is an ideal, $a_ib_i \in I$, and thus $\alpha \in I$. Similarly, $\alpha \in J$. Thus $IJ \subseteq I \cap J$.

(3) Let $R = \mathbb{Z}$ and let $I = 2\mathbb{Z}$ and $J = 4\mathbb{Z}$. Evidently, $$(2\mathbb{Z})(4\mathbb{Z}) = 8\mathbb{Z},$$ while $$2\mathbb{Z} \cap 4\mathbb{Z} = 4\mathbb{Z}.$$ Thus it is not generally the case that $IJ = I \cap J$.

(4) Suppose $I+J = R$ and that $R$ is commutative. We have $IJ \subseteq I \cap J$ by part (3) above.

Note that $$(I \cap J)(I + J) = (I \cap J)R = I \cap J.$$ Thus, if $z \in I \cap J$, we have $z = w(x+y)$ for some $x \in I$ and $y \in J$ and $w \in I \cap J$. Then $$z = xw + wy \in IJ.$$ Thus $IJ = I \cap J$.