# If A and B are ideals of a commutative ring R with unity and A+B=R show that A nn B = AB

If A and B are ideals of a commutative ring R with unity and A+B=R show that $A\cap B=AB$
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avortarF

The set A and B should satisfy these two conditions,
1. $A\cap B\subseteq AB$
2. $AB\subseteq A\cap B$
For $x\in A\cap B,$
$x\in A\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}x\in B$
From problem, A and B are ideal of a commutative ring R with unity,
$1\in R$, and $A+B=R$
So,
For element $a\in A\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}b\in B,$
$a+b=1$
As $x\in A\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}x\in B,$
$x=1$
$\left(1\right)x=1$
$\left(a+b\right)x=1$
$ax+bx=1$
$ax+xb=1$ ($\because R$ is commutative)
For ax, $a\in A\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}x\in B\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}f\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}xb,x\in A\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}b\in B$, then
$ax+xb=1$ $\in B$
This indicates that,
$A\cap B\subseteq AB$
As A and B are ideal, then for any $y\in B\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}z\in A$, the product $yz\in A$.
For $x\in AB$,
$x=\sum _{i=1}^{n}{a}_{i}{b}_{i}$
For, $n>0$, here, ${a}_{i}\in A\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{b}_{1}\in B.$
So, the product ${a}_{i}{b}_{i}\in A$
This indicates that,
$x=\sum _{i=1}^{n}{a}_{i}{b}_{i}\in A\cap B$
So, $AB\subseteq A\cap B$
Here, both conditions are satisfied.