proof of Chinese remainder theorem

First we prove that $\mathfrak{{a}}_{i}+\prod_{j\neq i}\mathfrak{{a}}_{j}=R$ for each $i$ . Without loss of generality, assume that $i=1$ . Then

R=(\mathfrak{{a}}_{1}+\mathfrak{{a}}_{2})(\mathfrak{{a}}_{1}+\mathfrak{{a}}_{3% })\cdots(\mathfrak{{a}}_{1}+\mathfrak{{a}}_{n}),

since each factor $\mathfrak{{a}}_{1}+\mathfrak{{a}}_{j}$ is $R$ . Expanding the product, each term will contain $\mathfrak{{a}}_{1}$ as a factor, except the term $\mathfrak{{a}}_{2}\mathfrak{{a}}_{2}\cdots\mathfrak{{a}}_{n}$ . So we have

(\mathfrak{{a}}_{1}+\mathfrak{{a}}_{2})(\mathfrak{{a}}_{1}+\mathfrak{{a}}_{3})% \cdots(\mathfrak{{a}}_{1}+\mathfrak{{a}}_{n})\subseteq\mathfrak{{a}}_{1}+% \mathfrak{{a}}_{2}\mathfrak{{a}}_{2}\cdots\mathfrak{{a}}_{n},

and hence the expression on the right hand side must equal $R$ .

Now we can prove that $\prod\mathfrak{{a}}_{i}=\bigcap\mathfrak{{a}}_{i}$ , by induction. The statement is trivial for $n=1$ . For $n=2$ , note that

\mathfrak{{a}}_{1}\cap\mathfrak{{a}}_{2}=(\mathfrak{{a}}_{1}\cap\mathfrak{{a}}% _{2})R=(\mathfrak{{a}}_{1}\cap\mathfrak{{a}}_{2})(\mathfrak{{a}}_{1}+\mathfrak% {{a}}_{2})\subseteq\mathfrak{{a}}_{2}\mathfrak{{a}}_{1}+\mathfrak{{a}}_{1}% \mathfrak{{a}}_{2}=\mathfrak{{a}}_{1}\mathfrak{{a}}_{2},

and the reverse inclusion is obvious, since each $\mathfrak{{a}}_{i}$ is an ideal. Assume that the statement is proved for $n-1$ , and condsider it for $n$ . Then

\bigcap_{1}^{n}\mathfrak{{a}}_{i}=\mathfrak{{a}}_{1}\cap\bigcap_{2}^{n}% \mathfrak{{a}}_{i}=\mathfrak{{a}}_{1}\cap\prod_{2}^{n}\mathfrak{{a}}_{i},

using the induction hypothesis in the last step. But using the fact proved above and the $n=2$ case, we see that

\mathfrak{{a}}_{1}\cap\prod_{2}^{n}\mathfrak{{a}}_{i}=\mathfrak{{a}}_{1}\cdot% \prod_{2}^{n}\mathfrak{{a}}_{i}=\prod_{1}^{n}\mathfrak{{a}}_{i}.

Finally, we are ready to prove the . Consider the ring homomorphism $R\to\prod R/\mathfrak{{a}}_{i}$ defined by projection on each component of the product: $x\mapsto(\mathfrak{{a}}_{1}+x,\mathfrak{{a}}_{2}+x,\dots,\mathfrak{{a}}_{n}+x)$ . It is easy to see that the kernel of this map is $\bigcap\mathfrak{{a}}_{i}$ , which is also $\prod\mathfrak{{a}}_{i}$ by the earlier part of the proof. So it only remains to show that the map is surjective.

Accordingly, take an arbitrary element $(\mathfrak{{a}}_{1}+x_{1},\mathfrak{{a}}_{2}+x_{2},\dots,\mathfrak{{a}}_{n}+x_% {n})$ of $\prod R/\mathfrak{{a}}_{i}$ . Using the first part of the proof, for each $i$ , we can find elements $y_{i}\in\mathfrak{{a}}_{i}$ and $z_{i}\in\prod_{j\neq i}\mathfrak{{a}}_{j}$ such that $y_{i}+z_{i}=1$ . Put

x=x_{1}z_{1}+x_{2}z_{2}+\dots+x_{n}z_{n}.

Then for each $i$ ,

\mathfrak{{a}}_{i}+x=\mathfrak{{a}}_{i}+x_{i}z_{i},

since $x_{j}z_{j}\in\mathfrak{{a}}_{i}$ for all $j\neq i$ ,

=\mathfrak{{a}}_{i}+x_{i}y_{i}+x_{i}z_{i},

since $x_{i}y_{i}\in\mathfrak{{a}}_{i}$ ,

=\mathfrak{{a}}_{i}+x_{i}(y_{i}+z_{i})=\mathfrak{{a}}_{i}+x_{i}\cdot 1=% \mathfrak{{a}}_{i}+x_{i}.

Thus the map is surjective as required, and induces the isomorphism

\frac{R}{\prod\mathfrak{{a}}_{i}}\to\prod\frac{R}{\mathfrak{{a}}_{i}}.

Title	proof of Chinese remainder theorem
Canonical name	ProofOfChineseRemainderTheorem
Date of creation	2013-03-22 12:57:20
Last modified on	2013-03-22 12:57:20
Owner	mclase (549)
Last modified by	mclase (549)
Numerical id	8
Author	mclase (549)
Entry type	Proof
Classification	msc 11A05
Classification	msc 11N99
Classification	msc 13A15