Some properties of ideal arithmetic

Let $R$ be a ring with 1, and let $I,J \subseteq R$ be ideals of $R$.

1. Prove that $I+J$ is the $\subseteq$-smallest ideal of $R$ containing both $I$ and $J$.
2. Prove that $IJ$ is an ideal contained in $I \cap J$.
3. Give an example where $IJ \neq I \cap J$.
4. Prove that if $R$ is commutative and if $I+J = R$, then $IJ = I \cap J$.

1. We wish to prove the following: (1) $I+J$ is an ideal of $R$ and (2) If $K \subseteq R$ is an ideal and $I,J \subseteq K$, then $I+J \subseteq K$.
1. 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 $R$.
2. 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 $IJ = I \cap J$.

• ysqure3John Magnum  On November 4, 2011 at 3:55 am

Observation: Dummit-Foote includes a note at the beginning of this section of problems that R is assumed to be a ring with identity. This fact is necessary for part 4, because otherwise (I \cap J)R =/= I \cap J

• nbloomf  On November 4, 2011 at 6:58 am

Thanks!