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I have proved one part: $AB \subset (A \cap B) \quad ...(1)$

To prove second part $(A \cap B) \subset AB$ my approach is as following:

Let $x \in A \cap B \Rightarrow x \in A \text{ and } x \in B \quad ...(2)$

$\because A$ and $B$ are ideals of ring $R$

$\Rightarrow 0 \in A \text{ and } 0 \in B$

$\because 1 \in R \text{ and } A+B = R$

$\Rightarrow 1 \in A \text{ and } 1 \in B$

Now $x.1 \in AB \quad (\because x \in A \text{ and } 1 \in B)$

$\Rightarrow x \in AB \quad ...(3)$

From (2) and (3) $\Rightarrow (A \cap B) \subset AB \quad ...(4)$

From (1) and (4) $\Rightarrow AB = (A \cap B)$ ... Hence Proved

I'm not able to figure out what mistake I made here as I've not used commutative property of ring $R$. Thanks for the help in advance.

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    Possible duplicate of [For a ring R, and ideals $A$, $B$, then $AB=A \cap B$ if $A + B = R$](http://math.stackexchange.com/questions/356684/for-a-ring-r-and-ideals-a-b-then-ab-a-cap-b-if-a-b-r) – 2017-01-31
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    Commutativity is necessary to define $AB$ (in a non-commutative ring $AB$ need not to be an ideal). Your proof is incorrect when you conclude that $1 \in A$ or $1 \in B$: this need not to hold, for example $A=2 \Bbb Z$, $B= 3 \Bbb Z$. – 2017-01-31

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Note that $1 \in R = A+B$ does not imply that $1 \in A$ and $1 \in B$, but that there are $a \in A$ and $b \in B$ such that $1 = a+b$. Multiplying this equation with $x$ gives $$ x = xa + xb = ax + xb $$ (we used commutativity for $xa=ax$). As $ax\in AB$ and $xb \in AB$, we have $x = ax + xb \in AB$.

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    I was writing the same proof (not really much of a coincidence), but I had even chosen the same letters, ie $x$ for an element in $A\cap B$ and $a+b=1$! – 2017-01-31