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Let $k\ge 2 $ and $m_{1},\ldots,m_{k}$ in $\mathbb{N}$ with $\gcd(m_{i},m_{j})= 1 \ (i \ne j)$.

  1. We show that $f(x) = x,\ldots,x)$ ist a ring isomorphism $f: \mathbb{Z}/ m\mathbb{Z} \rightarrow \mathbb{Z} / m_{1} \mathbb{Z} \times\cdots\times \mathbb{Z}/ m_{k}\mathbb{Z}$

  2. We show that for $y_{1},\ldots,y_{k} \in \mathbb{Z}$ it follows that there exists a y in $\mathbb{Z}$ so that $y\equiv y_{1} \pmod {m_{1}}, y\equiv y_{k} \pmod {m_{k}} \Leftrightarrow m_{1}|y-y_{1},\ldots, m_{k}| y- y_k$

    1). We know that $\mathbb{Z}/ m\mathbb{Z}$ is a Ring, therefore with $f(xy)=(xy,\ldots,xy)=(x,\ldots,x)(y,\ldots,y) = f(x)f(y)$ and $f(x+y)=(x+y,\ldots,x+y) = (x,\ldots,x)+(y,\ldots,y) = f(x)+f(y)$ f is a homomorphism . Suppose $x = y \pmod m$ . then with $m_{i}|m$ and $m|x-y \Rightarrow m_{i}| x-y$, it follows that} $x=y \pmod m_{i}$ and thus $f$ is a well defined ring homomorphism.

If $x \bmod n \in \operatorname{ker}(f)$, then it follows that $m_{j}|x$ for $j=1,\ldots,k$ and because $m_{k}$ ist relatively pairwise coprime, then also $x \bmod m = 0$. Thus f is injective. Because of $|\mathbb{Z} / m \mathbb{Z} | = | \mathbb{Z} / m\mathbb{Z}_{1}\times\cdots\times \mathbb{Z}/m \mathbb{Z}_{k}|$, $f$ is also surjective. Thus $f$ is an isomorphism.

2). The existence of such an y is equivalent to the surjectivity of $f$ , and the uniqueness in $0\le x \le m_1,\ldots, m_k-1$ follows from the injectivity.

Correct?

VVV

  • 0
    Technically, $f(x)\neq (x,x,\ldots,x)$; rather, $f(x) = (x\bmod m_1, x\bmod m_2,\ldots, x\bmod m_k)$. Also, the penultimate line of the paragraph before your (2) has a sequence of typos: it's not $\mathbb{Z}/m\mathbb{Z}_i$, it's $\mathbb{Z}/m_i\mathbb{Z}$. Otherwise, essentially correct.2011-10-27

1 Answers 1

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Begin by looking up the Chinese Remainder Theorem in any intro Number Theory textbook, or on a web search engine.

  • 0
    I edited my question, is my proof correct?2011-10-27