proof of Chinese remainder theorem

First we prove that $\\mathfrak{{a}}_{i}+\\prod_{j\eq i}\\mathfrak{{a}}_{j}=R$ 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 $z_{i}\\in\\prod_{j\eq i}\\mathfrak{{a}}_{j}$ 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 $j\eq i$ ,

=\\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}}.