Can $\Bbb R^+$ form a field

abstract-algebrafield-theorylinear algebravector-spaces

Consider $\Bbb R$ as a vector space. We have a vector space isomorphism $\exp:\Bbb R \to \Bbb R^+$ where we define new operations $a+b:=ab$ and $kb:=b^k.$

Now a natural inquiry at this point might be:

Does $\Bbb R^+$ also form a field?

For $\Bbb R^+$ to form a field we need to define addition, subtraction, multiplication, and division. Addition has already been defined above as $a+b:=ab.$

Best Answer

If I understand correctly, then the question asks whether it is possible to define a "multiplication" map $m: \Bbb R^+ \times \Bbb R^+\rightarrow \Bbb R^+$ such that, together with the "addition" map $a: \Bbb R^+ \times \Bbb R^+ \rightarrow \Bbb R^+$ sending $(a, b)$ to $a\cdot b$, makes $\Bbb R^+$ a field.

This is of course possible, in view that $\exp$ is an isomorphism of abelian groups from $(\Bbb R, +)$ to $(\Bbb R^+, \cdot)$.

We simply define the field structure on $\Bbb R^+$ by pushing out the field structure on $\Bbb R$ via the $\exp$ map.

More explicitly, we define $m: \Bbb R^+ \times \Bbb R^+ \rightarrow \Bbb R^+$ by sending $(a, b)$ to $\exp(\ln(a)\cdot \ln(b))$, which is equal to both $a^{\ln (b)}$ and $b^{\ln (a)}$.

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