## The Chinese Remainder Theorem (Pt. II)

**Point of Post: **This is a continuation of this post.

Let’s see if we can give an example of the CRT which doesn’t involve . In particular, we shall show that the CRT can help identify mod certain ideals. In particular, we have the following

**Theorem: ***Let be a field and be -distinct non-zero numbers. Then, (where the denominator is the product of the ideals not the ideal generated by although, this is a completely artificial distinction, since, they are equal).*

**Proof: **What we claim first is that are coxmaximal for each distinct . Indeed, we merely note that and since we have that is a unit and so . It thus frollows from the CRT that . That said, recall that by the first isomorphism theorem and the epimorphism (we technically don’t know that has a zero at implies that is divisible by , but it’ll soon be proven, and not hard to believe). The conclusion follows.

**Corollary: ***Let be such that are non-zero and are distinct numbers. Then, .*

**Proof: **Merely note that .

*Number Theoretic Implications*

Not only does the CRT have important pure ring theoretic implications but it also implies a few classic number theoretic facts. Most obviously is the refinement of the one given in the motivation of this post:

**Theorem: ***Let be pairwise coprime numbers in . Then, the system of simultaneous linear equations*

*has a solution which is unique modulo .*

That said, it has much more satisfying, applicable number theoretic applications. For example:

**Theorem: ***Let be a natural number with prime factorization and . Then, the equation has a solution if and only if has a solution for each . Moreover, if (for some generic integer ) denotes the number of solutions then .*

Indeed, this follows immediately from the CRT used in conjunction with following lemma:

**Theorem: ***Let and be rings and then if is a monomorphism and is given by then the map (where and is defined similarly) given by is an injection.*

**Proof: **To see that is really well-defined (in the sense that maps into ) we merely note that if then , but since is a morphism, we know that and so . Since is injective evidently we have that is injective.

**Corollary: ***Since evidently if is an isomorphism the map given by is a two-sided inverse to we have that if is an isomorphism then .*

To see why this proves the stated theorem we first note that since equals (by definition) and so

But, by the above corollary since we have that

From where the second assertion of the theorem follows. The first assertion is immediate from the second since it clearly is equivalent to if and only if .

**References:**

1. Dummit, David Steven., and Richard M. Foote. *Abstract Algebra*. Hoboken, NJ: Wiley, 2004. Print.

2. Bhattacharya, P. B., S. K. Jain, and S. R. Nagpaul. *Basic Abstract Algebra*. Cambridge [Cambridgeshire: Cambridge UP, 1986. Print.

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