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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.

abhinav12369
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1 Answers1

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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 $x = ax + xb \in AB$.

martini
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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 $a+b=1$! – s.harp Jan 31 '17 at 11:44
  • @martini can you explain why $ax,xb\in AB$ implies $ax+xb\in AB$? – drfrankie Oct 27 '21 at 23:52