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Im struggling to prove that if the function $\phi: R \to R/I \times R/J$ given by $\phi(x)=(x+I,x+J)$ is surjective then we have $R = I + J$. ($R$ is a commutative ring and $I$ and $J$ are ideals). I have proved that if $R=I+J$ then the map must be surjective but im not able to prove the converse. My attempts include trying to use the homomorphism theorem for rings, i.e we have $R/(I \cap J) \cong R/I \times R/J$, since the function is surjective and a homomorphism.

Although i cant make it...anyone with a solution or hint?

Loi
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1 Answers1

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If $\phi$ is surjective, then there is some $a\in R$ such that $\phi(a)=(1+I,0+J)$. In other words, there is some $a\in J$ such that $a-1\in I$. Then $$1=a-(a-1)\in I+J$$ so that $R=I+J$.

complexist
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