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

R^2 too!

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.

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

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

Hope that helps,
Alvaro

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

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.

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.