$F$ be a field of non-zero prime characteristic $p$ , is it true that there is only one group homomorphism $f:(F,+) \to (F$ \ $\{0\},.)$?

Question

Suppose $F$ be a field of non-zero characteristic $p$ , is it true that there is only one group homomorphism $f:(F,+) \to (F$ \ $\{0\},.)$ ? I have tried taking $x \in F$ , then $px=0$ , so

$(f(x))^p=f(px)=f(0)=1$ , so $f(x)$ has finite order for every $x \in F$ , $f(x) \in (F$ \ $\{0\},.)$ , and

$o(f(x))|p$ . But then I am stuck . Please help

Oops. You don't assume the field to be finite. I was a bit trigger-happy there. Reopening. The related thread may still be of interest. — Jyrki Lahtonen, Feb 24 '15 at 14:20

score 2 · Accepted Answer · answered Feb 24 '15 at 14:18

2

Hint: $x^p-1=(x-1)^p$. This allows us to determine which elements have order $p$.

answered Feb 24 '15 at 14:18

Matt Samuel

58,164

1

The field is of characteristic $p$, not order $p$. The field may not even be finite. – Feb 24 '15 at 14:19
@Amudhan thanks, this hint should be more appropriate. – Matt Samuel Feb 24 '15 at 14:23

$F$ be a field of non-zero prime characteristic $p$ , is it true that there is only one group homomorphism $f:(F,+) \to (F$ \ $\{0\},.)$?

1 Answers1