1
$\begingroup$

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

Thanks in advance.

$\endgroup$
8
  • $\begingroup$ @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
9
$\begingroup$

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

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

$\endgroup$
3
$\begingroup$

$\bullet$ 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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.