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Let $R$ a commutative ring and $I$, $J$ ideals of $R$ such that $I + J = R$. Prove that $IJ = I \cap J $

Is clear that $IJ \subseteq I$ and $IJ \subseteq J$ then $IJ \subseteq I \cap J$ this for any two ideals of a ring. But i not know as proof the other contains, because it is necessary that $I + J = R$?

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It's definitely necessary that $I+J=R$. For instance, take $R=\mathbb Z$ and $I=n\mathbb Z$ and $J=m\mathbb Z$. Then $I+J=(n,m)\mathbb Z$, $IJ=mn\mathbb Z$ and $I\cap J=\operatorname{lcm}(m,n)\mathbb Z$. Then $mn$ is the least common multiple iff $(m,n)=1$ –  Thomas Andrews Mar 30 '13 at 3:06
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1 Answer

Hint: Show that for any ideals $I$ and $J$, $$(I+J)(I\cap J)\subseteq IJ.$$ Then, because you are assuming that $I+J=R$, you have the other inclusion, $I\cap J\subseteq IJ$.

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Equivalently, write $1=i+j$ for $i\in I$ and $j\in J$. Then if $k\in I\cap J$ then $k=k\cdot 1\dots$. I'll leave the rest for OP to fill in. This is exactly the same proof, really, just at the element level. –  Thomas Andrews Mar 30 '13 at 3:10
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