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

I prove that since $I+J=R$ $$r=r_i+r_j, s=s_i+s_j \quad \mbox{for some}~~~r_i,s_i\in I, r_j,s_j\in J$$ 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

2 Answers 2

up vote 2 down vote accepted

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

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You're not in vector space. The hypothesis that $I$ and $J$ are stranger ideals i.e. they are such that $I+J=R$, does not imply that $\forall x \in R, \exists x_i \in I, x_j \in J$ such that $x=x_i+x_j$. What it [only] implies is $\exists a_i \in I, b_j \in J$ such that $a_i+b_j=1$. If you multiply both sides by $x$ you end up in $I$ or $J$ because of the very definition of what an ideal is and because of the property $I+J=R$.

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Let $a_i + b_j = 1$. Multiply both sides by $x$, you get $xa_i+xb_j=x$. Now $xa_i$ is in $I$ and $xb_j$ is in $J$, because $I$ and $J$ are ideals. That is, for every $x\in R$ you get $x_i\in I$ and $x_j\in J$ such that $x_i+x_j=x$. –  Gregor Bruns Sep 25 '12 at 10:38
my mistake, you are right. –  mak Sep 25 '12 at 10:40

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