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Let $I_1,...,I_m$ be ideals of a ring $R$ such that $I_j+\cap_{k\neq j}I_k=R$ for every $j\in\{1,...,m\}$. Then if $a_1,...,a_m\in R$ there exists $a\in R$ such that $a \equiv a_k \pmod{I_k}$ for every $k\in\{1,...,m\}$.

Fix $k\in\{1,...,m\}$. To show $a \equiv a_k \pmod{I_k}$ (for some $a$) is equivalenltly to show that $a_k\in I_k+a$. Since $a_k\in R=I_k+\cap_{t\neq j}I_t$, then $a_k=b_k+c_k$ for some $b_k\in I_k$ and $c_k\in\cap_{t\neq j}I_t$. However, clearly this $c_k$ is not the same for all of the $a_k's$.

Would anyone give me a hint? Thank you.

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  • $\begingroup$ However, I think in my problem I'm not assuming $R$ has $1$.@user26857 $\endgroup$
    – Talexius
    Jun 17, 2016 at 0:34
  • $\begingroup$ See here for the non-unitary case. $\endgroup$
    – user26857
    Jun 17, 2016 at 12:04

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