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Revision difference : Chinese remainder theorem |
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Let $R$ be a commutative ring with identity. If $I_1,\ldots,I_n$ are ideals of $R$ such that $I_i + I_j = R$ whenever $i\neq j$, then let $$I=\cap_{i=1}^n I_i = \prod_{i=1}^n I_i.$$ The sum of quotient maps $R/I\to R/I_i$ gives an isomorphism
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Let $R$ be a commutative ring with identity. If $I_1,\ldots,I_n$ are ideals of $R$ such that $I_i + I_j = R$ whenever $i\neq j$, then let $$I=\cap_{i=1}^n I_i = \prod_{i=1}^n I_i$$. The sum of quotient maps $R/I\to R/I_i$ gives an isomorphism
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| $$R/I\cong \prod_{i=1}^n {R}/{I_i}.$$ This has the slightly weaker consequence that given a system of congruences $x\cong a_i\pmod{I_i}$, there is a solution in $R$ which is unique mod $I$, as the theorem is usually stated for the integers. |
$$R/I\cong \prod_{i=1}^n {R}/{I_i}.$$ This has the slightly weaker consequence that given a system of congruences $x\cong a_i\pmod{I_i}$, there is a solution in $R$ which is unique mod $I$, as the theorem is usually stated for the integers. |
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