(Proof verification) there is no homomorphism between a finite field’s additive group to its multiplicative group.

Question

Given a finite field F with additive group $\text{F}^+$ and multiplicative group $\text{F}^{\times}$ Show that there doesn’t exist $f$:$\text{F}^+ \to \text{F}^{\times}$ s.t. $f(x+y)=f(x)f(y)$.
Proof by contradiction:
Assume $f$ exists.
Lemma 1: $\ker(f)$ has exactly two elements.
$\text{F}^{\times}$ has one less element,$0$, from $\text{F}^+$ thus, $$\exists !x,y,z \text{ s.t. } f(x)=f(y)=z$$ $$f(x)f(-x)=f(y)f(-x)=zf(-x)$$ $$f(x+-x)=f(y+-x)=zf(x)^{-1}$$ $$f(0)=f(x-y)=1$$ The kernel can’t have anymore elements because there is only one pair who gets mapped to the same value.
Because of lemma 1,$$\exists !i \text{ s.t. } f(0)=f(i) =1$$ $$f(i+i)=f(i)f(i)=1$$ Thus, $i+i$ can only equal $0$ or $i$.
If $i+i=0$ then $i=-i$ then $i=0.$
If $i +i=i$ then $i =0.$
But the kernel must have exactly $2$ distinct elements .
Contradiction! Can any of you verify my proof or help me improve it?

Answer 1

Let $\varphi:F^+\to F^\times$ be a homomorphism. Then, $|\varphi(F^+)|$ must divide both $|F^+|$ and $|F^\times|$.
Thus, it must divide their difference $|F^+| - |F^\times| = 1.$

This forces that $|\varphi(F^+)|$ must be $1$. As $\varphi(0)$ must necessarily be $1$, we see that $\varphi$ is forced to be the constant function $1$. Note that this is indeed a homomorphism. (And we have shown that this is the only one.)

Answer 2

The orders of $F^{+}$ and $F^{*}$ are coprime (they differ by $1$). Hence in a homomorphism every element of $F^{+}$ maps to $1$, the unique element whose order divides both group orders.