# 代數學-Prove or Disprove

〈U, . (一點)〉is isomorphic to〈R,+〉.

Update:

U是單位圓

Rating
• Anonymous

What's〈U, . (一點)〉? It's ambiguous!

2007-12-13 21:26:38 補充：

〈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.