Given that $R$ is a commutative ring and $I,J$ are ideals in $R$, prove that $(I+J)(I\cap J) \subset IJ.$

Question

As stated in the title, it is given that $R$ is a commutative ring and $I,J$ are ideals in $R$. I want to prove that the following is true: $$(I+J)(I\cap J) \subset I J.$$

I have proven that both $I+J$ and $I\cap J$ are ideals, but not sure how to proceed further.

Any help would be appreciated.

Answer 1

I'll use $i$s to represent elements of $I$ and $j$s to represent elements of $J$.

Suppose that $a \in I \cap J$, and $b \in I + J$. Then we can write $b = i_1 +j_1$ and $a = i_2 = j_2$. Thus $ab = ai_1 + aj_1 = j_2i_1 + i_2j_1 \in IJ$.

Any element of $(I \cap J)(I+J)$ is thus a finite sum of elements in $IJ$, so being an ideal it is contained in $IJ$.

Note that this observation is usually used as the first step in the proof of the Chinese Remainder Theorem. We have just seen that $(I+J)(I \cap J) \subset IJ \subset I \cap J$ (the last inclusion is trivial).
Thus if $I$ and $J$ are comaximal, that is, $I +J = R$, then we have that $IJ = I \cap J$.

Answer 2

It's not difficult to show that, for ideals $A,B,C$, $(A+B)C=AC+BC$. Thus $$ (I+J)(I\cap J)=I(I\cap J)+J(I\cap J) $$ and all it's needed is to show that $$ I(I\cap J)\subseteq IJ \qquad\text{and}\qquad (I\cap J)J\subseteq IJ $$ Can you do it? There are some inclusion relations to be checked.

Answer 3

$I + J = \{ i + j \mid i \in I, j \in J \}; \tag 1$

$I \cap J = \{k \mid k \in I, k \in J \}; \tag 2$

for any two ideals $L$ and $K$,

$LK = \displaystyle \left \{ \sum_{p = 1}^n l_p k_p \mid n \in \Bbb N, l_p \in L, k_p \in K \right \}; \tag 3$

thus

$(I + J)(I \cap J) = \displaystyle \left \{ \sum_{p = 1}^n (i_p + j_p) \cdot k_p \mid n \in \Bbb N, i_p \in I, j_p \in J, k_p \in I \cap J \right \}; \tag 4$

we observe that the sum

$\displaystyle \sum_{p = 1}^n (i_p + j_p)k_p = \sum_{p = 1}^n i_p k_p + \sum_{p = 1}^n j_p k_p; \tag 5$

since $i_p \in I$ and $k_p \in I \cap J \subset J$,

$\displaystyle \sum_{p = 1}^n i_p k_p \in IJ; \tag 6$

as for the second sum on the right of (5), we note that $k_p \in I \cap J \subset I$, and then by the commutativity of $R$,

$\displaystyle \sum_{p = 1}^n j_p k_p = \sum_{p = 1}^n k_p j_p \in IJ; \tag 7$

it therefore follows that

$\displaystyle \sum_{p = 1}^n i_p k_p + \sum_{p = 1}^n j_p k_p = \sum_{p = 1}^n k_p j_p \in IJ; \tag 8$

we thus see that

$\displaystyle \left \{ \sum_{p = 1}^n (i_p + j_p) \cdot k_p \mid n \in \Bbb N, i_p \in I, j_p \in J, k_p \in I \cap J \right \} \subset IJ; \tag 9$

finally, returning to (4) we conclude

$(I + J)(I \cap J) \subset IJ, \tag{10}$

as required.