proof of Chinese remainder theorem

First we prove that 𝔞i+ji𝔞j=R for each i. Without loss of generality, assume that i=1. Then


since each factor 𝔞1+𝔞j is R. Expanding the productPlanetmathPlanetmath, each term will contain 𝔞1 as a factor, except the term 𝔞2𝔞2𝔞n. So we have


and hence the expression on the right hand side must equal R.

Now we can prove that 𝔞i=𝔞i, by inductionMathworldPlanetmath. The statement is trivial for n=1. For n=2, note that


and the reverse inclusion is obvious, since each 𝔞i is an ideal. Assume that the statement is proved for n-1, and condsider it for n. Then


using the induction hypothesis in the last step. But using the fact proved above and the n=2 case, we see that


Finally, we are ready to prove the . Consider the ring homomorphismMathworldPlanetmath RR/𝔞i defined by projection on each component of the product: x(𝔞1+x,𝔞2+x,,𝔞n+x). It is easy to see that the kernel of this map is 𝔞i, which is also 𝔞i by the earlier part of the proof. So it only remains to show that the map is surjectivePlanetmathPlanetmath.

Accordingly, take an arbitrary element (𝔞1+x1,𝔞2+x2,,𝔞n+xn) of R/𝔞i. Using the first part of the proof, for each i, we can find elements yi𝔞i and ziji𝔞j such that yi+zi=1. Put


Then for each i,


since xjzj𝔞i for all ji,


since xiyi𝔞i,


Thus the map is surjective as required, and induces the isomorphismPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath

Title proof of Chinese remainder theorem
Canonical name ProofOfChineseRemainderTheorem
Date of creation 2013-03-22 12:57:20
Last modified on 2013-03-22 12:57:20
Owner mclase (549)
Last modified by mclase (549)
Numerical id 8
Author mclase (549)
Entry type Proof
Classification msc 11A05
Classification msc 11N99
Classification msc 13A15