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Let $R$ be a ring (not necessary having "1"), and let $I,J$ be ideals of $R$ such that $I+J=R$. I want to prove that, for any $r, s \in R$, there is a $x\in R$ such that $$x\equiv r ({\rm mod} I) \quad x \equiv s ({\rm mod} J)$$

My proof: Since $I+J=R$, we can write $$r=r_i+r_j,\quad s=s_i+s_j \quad \mbox{for some}~~~r_i,s_i\in I,\quad r_j,s_j\in J$$ Let $x=r_j+s_i$. then $$x-r=r_i-s_i\in I, \quad x-s=r_j-s_j\in J$$ Thus, $$x\equiv r ({\rm mod} I) \quad x \equiv s ({\rm mod} J)$$

Is this proof wrong?? I can't look for a mistake.

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excellent proof doc very concise and explicit – user71994 Aug 27 '13 at 20:56
up vote 3 down vote accepted

You did everything correctly. The proof seems right to me.

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