proof of Chinese remainder theorem
First we prove that for each . Without loss of generality, assume that . Then
since each factor is . Expanding the product, each term will contain as a factor, except the term . So we have
and hence the expression on the right hand side must equal .
and the reverse inclusion is obvious, since each is an ideal. Assume that the statement is proved for , and condsider it for . Then
using the induction hypothesis in the last step. But using the fact proved above and the case, we see that
Finally, we are ready to prove the . Consider the ring homomorphism defined by projection on each component of the product: . It is easy to see that the kernel of this map is , which is also by the earlier part of the proof. So it only remains to show that the map is surjective.
Accordingly, take an arbitrary element of . Using the first part of the proof, for each , we can find elements and such that . Put
Then for each ,
since for all ,
Thus the map is surjective as required, and induces the isomorphism
|Title||proof of Chinese remainder theorem|
|Date of creation||2013-03-22 12:57:20|
|Last modified on||2013-03-22 12:57:20|
|Last modified by||mclase (549)|