Another version of the Chinese Remainder Theorem

Question

Let $R$ be a ring, $I,J$ ideals such that $I+J=R$. The Chinese Remainder Theorem claims that if $I\cap J=(0)$, then $R\simeq R/I\times R/J$.

I remember seeing a version of the theorem for the case of principal ideals $I=(p),\ J=(q)$ which assumes instead of the comaximality of $I$ and $J$ (and probably of the condition on the intersection) that $I$ and $J$) that $p$ and $q$ are coprime elements in $R$. (I'm not sure that my statement is precise.) This version is easier to use sometimes, e.g. it looks easier for me to prove that $x-1$ and $x^2+x+1$ are coprime than to prove that $(x-1)+(x^2+x+1)=(1)$ in $\mathbb R[x]$.

So my questions are: whether my statement is precise, and how does it follow from the general Chinese remainder theorem?

Answer 1

The abstract Chinese remainder theorem that I know is this (stated for two ideals):

Let $R$ be a (commutative) ring, $I, J$ two coprime ideals in $R$, i.e. satisfying$I+J=R$. Then $IJ=I\cap J$, and we have an isomorphism \begin{align} R/I\cap J&\xrightarrow{\;\sim\;}R/I\times R/J\\ r+I\cap J&\longmapsto (r+ I, r+ J) \end{align}

Answer 2

The problem is what do you mean by the statement: "$x$ and $y$ are coprime elements in a ring $R$". Let's stick to integers for now. There are two different ways you can think about coprime integers. Two integers $x,y\in \mathbb{Z}$ are said to be coprime if one of the two equivalent conditions are satisfied.

1. The greatest common divisor of $x,y$ is equal to one.
2. There exists $a,b\in \mathbb{Z}$ such that $ax+by=1$ (Bezout's identity)

The problem with definition (1) is that there are many rings out there with no such thing as a greatest common divisor. Say $\mathbb{C}[x]/(x^2)$ is not even a domain. Or $\mathbb{Z}[-\sqrt{3}]$ which is a domain but the two elements $x=4=2\times 2=(1+\sqrt{-3})(1-\sqrt{-3})$ and $y=(1+\sqrt{-3})\times 2$ have no gcd. A commutative ring admitting a gcd for any pair of elements is called a GCD domain.

Your example of $\mathbb{R}[x]$ is a GCD domain (actually, more strongly, a principal ideal domain, PID). But even if a ring $R$ is a GCD domain, it does not mean $R$ admits a Bezout identity. An example would be $\mathbb{R}[x,y]$ which is a GCD domain (actually a unique factorization domain, UFD) which does not admit a Bezout identity. A ring that admits a Bezout identity is called a Bezout domain. All Bezout domains are GCD domains. Alternatively, a Bezout domain can be defined as follows:

An integral domain $R$ is called a Bezout domain if given any two elements $x,y$, the sum of ideals $(x)+(y)$ is again principal.

So you can see that in a Bezout domain, $x,y$ are coprime if and only if the ideals $(x), (y)$ are coprime (Two ideals $I,J\subset R$ are called coprime if $I+J=R$). However, one can define coprime ideals for all commutative rings, and coprime elements only for some rings. In a sense then, the way coprime ideals are defined is the natural generalization of how we expect "coprime" to be.

Now the Chinese Reminder theorem says

If the ideals $I,J\subset R$ are comaximal, then $R/I\cap J\simeq R/I\times R/J$.

If we assume $R$ is a Bezout domain, and $x,y\in R$ coprime, then $(x)\cap (y)=(xy)$ (which is not zero y the way). And Chinese remainder theorem, in this case, becomes $$ R/(xy)\simeq R/(x)\times R/(y) $$ So sure, if in your ring of interest it makes sense to talk about coprime elements, you can go with your special version of the theorem. But if there is no such thing, well... I hope, this sheds some light on the matter.