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대학생,일반 수학

Re:대수입니다. 증명 좀 부탁드립니다. 아시는 분 좀 까다라운 문제라서 도저히 모르겠네요.

작성자비는아픔|작성시간11.07.21|조회수127 목록 댓글 1

Let $D$ be an integral domain and $\phi :D\rightarrow D$ be a ring homomorphism.

Then $\phi (1)=\phi (1)^{2}$ so $\phi (1) = 0$ or $1$ .

If $\phi (1) = 0$ , trivially $\phi\equiv 0$ .

Now suppose $\phi (1) = 1$ and $D=\mathbb{Q}$ .

For any $x=\frac{p}{q}\in\mathbb{Q}$ ( $p,q\in\mathbb{Z}$ and $q$ is nonzero),

$q\phi (x) = \phi (qx) = \phi (p) = p\phi(1) = p$ so $\phi (x)=\frac{p}{q}=x$ , i.e., $\phi=\textrm{id}$ .

Finally assume that $\phi (1)=1$ and $D=\mathbb{R}$ .

For any nonnegative $x\in\mathbb{R}$ , $\phi (x) = \phi (\sqrt{x}^{2}) = \phi(\sqrt{x})^{2}\geq 0$ .

This implies for any $a,b\in\mathbb{R}$ with $a\geq b$ , $0\leq\phi (a-b)=\phi (a)-\phi (b)$ so $\phi$ is increasing.

Now noting that $\phi (q) = q$ if $q\in\mathbb{Q}$ as we’ve seen just before, suppose $x_{0} < \phi (x_{0})$ for some $x_{0}\in\mathbb{R}$ .

Since $\mathbb{Q}$ is dense in $\mathbb{R}$ , there exists $q_{0}\in\mathbb{Q}$ such that $x_{0}<q_{0}=\phi (q_{0})< \phi (x_{0})$ .

This contradicts the fact that $\phi$ is increasing.

On the other hand, the case that $x_{0}>\phi (x_{0})$ also derives a contradiction similarly.

Hence $\phi(x)=x$ for every $x\in\mathbb{R}$ , i.e., $\phi=\textrm{id}$ .

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작성자항상한상 | 작성시간 11.07.22 답변 달아주셔서 감사합니다.~~

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