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.