If chatacteristic of $K$ is positive, show that every homomorphism from additive group to multiplicative group maps all elements of $K$ to $1$

Question

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?

Answer 1

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 $1$.

Answer 2

$f(p\cdot a)=f(0)=1$. And $f(p\cdot a)=f(a)^p$. So $f(a)^p=1$. So $0=f(a)^p-1=(f(a)-1)^p$. So $(f(a)-1)^p=0$. $K$ has no zero divisors so $f(a)-1=0$. So $f(a)=1$.