Assume $(F,+,\cdot)$ is an arbitrary field. How to prove $(F,+)$ and $(F\setminus \{0\},\cdot)$ aren't isomorphic?

Thanks in advance.

  • $\begingroup$ @rschwieb:yeah as a group isomorphism $\endgroup$
    – M.H
    Dec 5 '13 at 21:39
  • $\begingroup$ And you want to prove this for every field or you just want a counterexample? $\endgroup$
    – rschwieb
    Dec 5 '13 at 21:39
  • $\begingroup$ i want prove for all field (arbitrary fields ) $\endgroup$
    – M.H
    Dec 5 '13 at 21:41
  • 2
    $\begingroup$ Consider elements of finite order. $\endgroup$
    – Cantlog
    Dec 5 '13 at 21:48
  • $\begingroup$ I think you can say something like that. For every field $(k,+)$ is divisible so is injective. This means that it can be true only for algebraic closed fields. If field has characteristic $0$ then its not true because $(k,+)$ dont have elements of finite order. I dont know what to do with infinite algebraic closed fields of finite characteristic. $\endgroup$
    – user52045
    Dec 5 '13 at 21:51

If $\operatorname{char}(F) \neq 2$ then $(-1)$ has order $2$ in $(F^{\times},\cdot)$, but there is no element of order $2$ in $(F,+)$.

If $\operatorname{char}(F)=2$ then any element has order $2$ in $(F,+)$ but no element has order $2$ in $(F^{\times}, \cdot)$ as

$$x^2=1 \Rightarrow (x-1)^2=0 \Rightarrow x-1=0$$


$\bullet$ if $k$ has not characteristic $2$, then if $f$ is an isomorphism from $(k^*,.)$ to $(k,+)$ we have: $0=f(1)=f((-1)^2)=2 f(-1)$ , then $f(-1)=0$ and $f(1)=f(-1)$, that is not possible since $f$ is injective.

$\bullet$ if $k$ has characteristic $2$, let $x \in k^*$; then $f(x^2)=2f(x)=0=f(1)$ that gives $x^2=1$ and then $x=1$. But $f$ is an isomorphism from $k \backslash\{0\}$ to $k$ gives $k$ is an infinite set.


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