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

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

• @rschwieb:yeah as a group isomorphism
– M.H
Dec 5 '13 at 21:39
• And you want to prove this for every field or you just want a counterexample? Dec 5 '13 at 21:39
• i want prove for all field (arbitrary fields )
– M.H
Dec 5 '13 at 21:41
• Consider elements of finite order. Dec 5 '13 at 21:48
• 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. 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.