Answer to Question #97174 in Abstract Algebra for Aakash

Question #97174
Prove or disprove that C~ R as fields.
1
Expert's answer
2019-10-23T15:12:31-0400

Let "f\\colon \\mathbb C\\to\\mathbb R" be a homomorphism of the fields. Since "i^2+1=0", we have "f(i^2+1)=f(0)".

Since "f(a)=f(0+a)=f(0)+f(a)" for all "a\\in\\mathbb C", we obtain that "f(0)=0".

Also, "f(a)=f(1\\cdot a)=f(1)f(a)" for all "a\\in\\mathbb C", so 1) "f(1)=1" or 2)"f(1)=0", and so "f(a)=0" for all "a\\in\\mathbb C"

So "0=f(0)=f(i\\cdot i+1)=f(i)f(i)+f(1)=f(i)^2+f(1)". Since the equation "x^2+1=0" has no solutions in "\\mathbb R", "f(1)" cannot be "1" (otherwise we obtain "0=f(i)^2+f(1)=f(i)^2+1"), so "f(1)=0", that is "f(a)=0" for all "a\\in\\mathbb C" .

So we obtain that there is only one homomorphism "f\\colon \\mathbb C\\to\\mathbb R", namely "f(a)=0" for all "a\\in\\mathbb C".

It is not a isomorphism because every isomorphism is a bijective homomorphism, but "f" is not a bijection, so there is no isomorphism between "\\mathbb C" and "\\mathbb R".

Answer: there is no isomorphism between "\\mathbb C" and "\\mathbb R".


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