PLEASE HELP

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# PLEASE HELP

Submitted by sleepy123 on Fri, 01/18/2008 - 06:27

Forums:

Prove that the fields real number and complex number are NOT isomorphic?

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## Re: PLEASE HELP

R^2 too!

## Re: PLEASE HELP

That's not an isomorphism between R and C, only between R and a subset of C. All you've shown is that R is isomorphic to the real line in C.

## Re: PLEASE HELP

An isomorphism needs to be bijective. You're function is into the set of complex numbers but not onto.

## Re: PLEASE HELP

If they are isomorphic and f:C -> R is an isomorphism... what's the image of i ?

Hope that helps,

Alvaro

## Re: PLEASE HELP

(C*,.) has an elements of order 4 (i,-i) while (R*,.) has none.

## Re: PLEASE HELP

there does exist an isomorphism between the field of real numbers and complex numbers. The elements of the form (x,0) in C is isomorphic to x in R.

## Re: PLEASE HELP

I just felt like opening a root level tree as I thought I could clarify some things.

alozano says :"If they are isomorphic and f:C -> R is an isomorphism... what's the image of i ?"

Suppose, there existed a isomorphism [;F:\mathbb{C}\to \mathbb{R};] and let [;f(i)=x;] for some [;x\in \mathbb{R};]. Then, [;f(1)=(f(i))^4=x^4;] which means that [;f(i)=-1;] (as [;f(1)=1;]). But [;f(i^2)=f(-1)=f(i)f(i)=1;] indicates no isomorphism.

shahab says: (C*,.) has an elements of order 4 (i,-i) while (R*,.) has none.

Note that orders of elements in a isomorphism are preserved.