青ㄟ
Lv 6
青ㄟ asked in 科學數學 · 1 decade ago

代數學-Prove or Disprove

〈U, . (一點)〉is isomorphic to〈R,+〉.

Update:

U是單位圓

1 Answer

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  • Anonymous
    1 decade ago
    Favorite Answer

    What's〈U, . (一點)〉? It's ambiguous!

    2007-12-13 21:26:38 補充:

    Answer:

    〈U, . 〉is not isomorphic to〈R,+〉,

    where U={e^ ix: x belongs to R} endowed with ordinary multiplication.

    Suppose on the contrary that they are isomorphic, then there exists an

    isomorphism from R onto U.

    (1)

    Note that f( the identity of R) = identity of U, so f(0) = 1 (= e^ i0 )---(#).

    (2)

    Since f is an onto mapping, so there exists an element x in R such that

    f(x) = - 1 ( = e^ i(pi) )---(*).

    (3)

    From the homomorphism property of f and (*), we got

    f(2 x ) = f(x + x) = f(x) . f(x) = (- 1) . (- 1) = 1.

    Now, both f(0) = 1, and f(2 x) = 1. Since f is one-to-one, this would

    force 2 x = 0, i.e., x = 0.

    Now, we have f(0) = f(x) = - 1 , by (*).

    This is a contradiction to f(0) = 1---(#) !

    Therefore, there doen't exist any isomophism between〈R,+〉and

    〈U, . 〉. Done!

    Remark:

    Although, 〈U, . 〉is not isomorphic to〈R,+〉, for every y in R,

    there does exists a unique continuous homomorphism from R onto U ,

    that is, x---> e^ ixy.

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