A few ways of stating the Chinese remainder theorem

I’m going to start off weird, with no motivation (because the reader will be able to recognize the statement almost immediately):

Theorem: Suppose p,p^\prime,p^{\prime\prime},\cdots,p^{\prime\cdots\prime} are relatively prime positive integers >1. Then the system x\equiv b_1\mod p, x\equiv b_2\mod p^{\prime},\cdots, x\equiv b_m\mod p^{\prime\cdots\prime}, has a unique solution mod p\times p^\prime\times p^{\prime\prime}\times\cdots\times p^{\prime\cdots\prime}.

That’s the Chinese remainder theorem. In this brief post, I’ll write down other ways of writing down the same statement, for comparison (and actually simply to have a collection of the statements). I refer the reader to Milne’s algebraic geometry notes for proofs, which I won’t be providing here (since that’s not the main purpose).

First Method: Let m=p\times q\times\cdots\times l. Then \mathbf{Z}/(m) is isomorphic as a ring to \mathbf{Z}/(p)\times\mathbf{Z}/(q)\times\cdots\times \mathbf{Z}/(l).

This is really just a homework exercise in basic algebra. We are going to generalize this. All rings are going to be commutative and unital.

Definition: An ideal of a ring R is a subset \mathfrak{p} of R such that:

  1. \mathfrak{p} is a subgroup of R as a group under addition.
  2. If a\in \mathfrak{p} and r\in R, then ra\in \mathfrak{p}.

Define \{p+q|p\in\mathfrak{p},q\in\mathfrak{q}\} to be \mathfrak{p+q}. This is an ideal (exercise). Two ideals \mathfrak{p,q} are coprime if \mathfrak{p+q} is isomorphic to R. Then (from Milne’s book):

Second Method: Suppose R is a commutative unital ring. Then for ideals \mathfrak{p}_1,\cdots,\mathfrak{p}_n such that if i\neq j, then \mathfrak{p}_i\neq\mathfrak{p}_j, the map

R\to\prod R/\mathfrak{p}_i

is surjective, and the kernel is \bigcap \mathfrak{p}_i.

One more method is using scheme theory. An ideal \mathfrak{p} is prime if R/\mathfrak{p} is an integral domain. Let R be a ring. \mathrm{Spec}(R) denotes the collection of all prime ideals of R. This can be equipped with a topology, called the Zariski topology. The basis of open sets for this topology are of the following form, for r\in R:

D(r)=\{\mathfrak{p}\text{ a prime ideal of }R|r\not\in\mathfrak{p}\}

We may define a presheaf \mathscr{O}_X of rings on X=\mathrm{Spec}(R) via \mathscr{O}_X(D(r))=R[x]/(xr-1). Then the Chinese remainder theorem follows from the following statement:

Third Method: The presheaf \mathscr{O}_X is actually a sheaf. (See here for a proof.)

The Chinese remainder theorem follows from applying the gluing property to disjoint open subsets. I’d love to know if there are more statements equivalent to the Chinese remainder theorem – these are just some that I can recall off the top of my head!

Cheers,

S.D.

Advertisements

2 thoughts on “A few ways of stating the Chinese remainder theorem

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s