If chatacteristic of $K$ is positive, show that every homomorphism from additive group to multiplicative group maps all elements of $K$ to $1$ Suppose $K$ is a field. Denote $(K,+)$ as the abelian group under addition operation and $(K,\times)$ as the abelian group under multiplication opearation. If the characteristic of $K$ is positive, show that every homomorphism $f:(K,+) \rightarrow (K,\times)$ maps all elements of $K$ to $1$.
I don't even know how to start this question. Any hint?

Suppose the field $K$ has characteristic $p>0$, and consider a group homomorphism
$$
f\colon (K,+)\to (K\setminus\{0\},\times).
$$
For any element $x\in K$, we must have $px=0$. Applying the group homomorphism to this equation,
$$
f(x)^p=1.
$$
On the other hand, by the Binomial Theorem
$$
\begin{align*}
(f(x)-1)^p&=f(x)^p+\{\text{terms that are a multiple of $p$}\}+(-1)^p\\
&=f(x)^p+(-1)^p\\
&=1+(-1)^p\\
&=0.
\end{align*}
$$
The last line follows by considering the cases when $p$ is odd and when $p=2$ separately. Hence we've shown that $(f(x)-1)^p=0$. Since fields lack zero divisors, this implies $f(x)=1$. Consequently the homomorphism maps all elements to $1$.
