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Algebra-II: Companion notes

Abhay Soman

Section 4.10 Chinese Remainder Theorem

We generalize Theorem 4.1.6 to commutative rings.
First we show that \(\mfa_1\cdots\mfa_n=\mfa_1\cap\cdots\cap\mfa_n\text{.}\) We certainly have
\begin{equation*} \mfa_1\cdots\mfa_n\subseteq\mfa_1\cap\cdots\cap\mfa_n\text{.} \end{equation*}
We claim that \(\mfa_1\) and \(\mfb=\mfa_2\cdots\mfa_n\) are such that \(\mfa_1+\mfb=A\text{.}\) Indeed, since \(\mfa_1+\mfa_i=A\) for \(i\geq 2\text{,}\) there are \(x_i\in\mfa_1\) and \(y_i\in\mfa_i\) such that \(x_i+y_i=1\) for each \(i\geq 2\text{.}\) Thus, \(x_i+y_i\equiv y_i\mod\mfa_1\) and hence
\begin{equation*} 1=(x_2+y_2)\cdots(x_n+y_n)\in\mfa_1+\mfb. \end{equation*}
So, there are \(x\in\mfa_1\) and \(y\in\mfb\) such that \(1=x+y\text{.}\) Now for any \(a\in\mfa_1\cap\cdots\cap\mfa_n\) we get \(a=ax+ay\in\mfa_1\mfb=\mfa_1\cdots\mfa_n\text{.}\)
Consider the map \(\varphi\colon A\to A/\mfa_1\times\cdots\times A/\mfa_n\) given by
\begin{equation*} a\mapsto(a+\mfa_1,\ldots,a+\mfa_n) \end{equation*}
It is left to the reader to check that \(\varphi\) is a well-defined ring homomorphism. We compute its kernel.
\begin{align*} \ker(\varphi)\amp=\{a\in A:a\in\mfa_i\;\text{for all}\; i\}\\ \amp=\mfa_1\cap\cdots\cap\mfa_n \end{align*}
We now show that \(\varphi\) is surjective. It is enough to show that \((1+\mfa_1,0+\mfa_2,\ldots,0+\mfa_n)\) is in the image of \(\varphi\text{.}\) Since \(\mfa_1+\mfa_i=A\) for each \(i\geq 2\) there are \(x_i\in\mfa_1\) and \(y_i\in\mfa_i\) such that \(x_i+y_i=1\) for each \(i\geq 2\text{.}\) We put \(y=y_2y_3\cdots y_n\text{.}\) Then
\begin{equation*} y\equiv(1-x_2)(1-x_3)\cdots(1-x_n)\equiv 1\mod\mfa_1 \end{equation*}
and
\begin{equation*} y\equiv 0\mod\mfa_i\quad\text{for}\,i\geq 2. \end{equation*}
Hence \(\varphi(y)=(1+\mfa_1,0+\mfa_2,\cdots,0+\mfa_n)\text{.}\) By Theorem 3.2.1, we get the required result.
Let \(F[x]\) be the polynomial ring over a field \(F\text{.}\) Let \(f_1,f_2,\ldots,f_n\) be pairwise distinct irreducible polynomials such that \(f_i\) and \(f_j\) are not associates for any \(i\neq j\text{.}\) Then
\begin{equation*} F[X]\big/(f_1^{k_1}\cdots f_n^{k_n})\simeq F[x]/(f_1^{k_1})\times\cdots\times F[x]/(f_n^{k_n}). \end{equation*}
We first show that \((f_i^{k_i})+(f_j^{k_j})=F[x]\) for \(i\neq j\text{.}\) By Proposition 4.5.11 \(\gcd(f_i^{k_i},f_j^{k_j})=1\) and hence \((f_i^{k_i})+(f_j^{k_j})=F[x]\) for \(i\neq j\text{.}\) We have \((f_1^{k_1})\cap\cdots\cap(f_n^{k_n})=(f_1^{k_1}\cdots f_n^{k_n})\) (refer to the proof of Theorem 4.10.1). Thus by Chinese remainder theorem we get the result.
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