Projection onto projective space is injective iff $\Bbb K=\Bbb F_2$

Question

Consider $\Bbb K^{n+1}$ a vector space over a field $\Bbb K$ and $\Bbb P^n(\Bbb K)\colon\!=\dfrac{\Bbb K^{n+1}\setminus\{0\}}{\sim}$, where $\Bbb \sim $ is defined by: $v\sim w\Leftrightarrow \exists\,\lambda\in\Bbb K\setminus\{0\}:v=\lambda w\quad \forall v,w\in\Bbb K^{n+1}\setminus\{0\}$. I am trying to proof that the natural projection $\pi:\Bbb K^{n+1}\setminus\{0\}\to\Bbb P^n(\Bbb K)$ is injective if and only if $\Bbb K=\Bbb F_2$.

I think I have got the right to left implication, since $\pi(v)=\pi(w)\Leftrightarrow \exists\,\lambda\neq0:v=\lambda w$, and in the case $\Bbb K=\Bbb F_2$ we must have $\lambda=1$.

My problem lays with the left to right implication, any hints will be appreciated.

Thanks in advance!

Answer 1

If you know $\pi$ is injective then for all $v,w\in\Bbb K^{n+1}$ you have $\pi(v)=\pi(w)\Rightarrow v=w$.

You know $\pi(v)=\pi(w)\Leftrightarrow \exists\, \lambda\in\Bbb K\setminus\{0\}:v=\lambda w$, so $v=w=\lambda w$ and then (try to figure it out first!)

$w-\lambda w=0\Rightarrow (1-\lambda)w=0$, and since $w\neq0$ this means $\lambda=1$.

Therefore every nonzero element $\lambda$ of $\Bbb K$ is actually $1$, so $\Bbb K=\Bbb F_2$.