Characteristic of a field to make projective points the same In the resolution of a linear algebra exercise, they tell me that given a field $\Bbb K$ and the projective space $\Bbb P^n(\Bbb K)$, the projective points $(1:-1:2)$ and $(2:1:1)$ are the same if and only if $\text{char}(\Bbb K)=3$.
I understand the right to left implication, since if $\text{char}(\Bbb K)=3$ then $3=0$, hence $-1=-1+0=-1+3=2$ and $4=3+1=1$. Therefore, $(1:-1:2)=(1:2:2)=(2:4:4)=(2:1:1)$.
However, I have trouble understanding the left to right implication, I am not really sure of why any other value of the characteristic of the field would result in those two points being different.
Thanks in advance!
 A: For $P=(1:-1:2)$ to be the same as $Q=(2:1:1)$ we need the existence of some $\lambda\neq0$ such that $\lambda(1,-1,2)=(2,1,1)$. From the first component we get that $\lambda=2$, which means $P=Q$ if and only if $(2,-2,4)=(2,1,1)$ if and only if $-2=1\land 4=1$.
Both $-2=1$ and $4=1$ are true if and only if $3=0$ (add $2$ to the first equality and subtract $1$ to the first one), so we can conclude $P=Q$ if and only if $\text{char}(\Bbb K)=3$.
Another way to see this is thinking that you need $(1,-1,2)$ and $(2,1,1)$ to lie in the same vector line, since then they're projected onto the same projective point. This is equivalent to $(1,-1,2)$ and $(2,1,1)$ being linearly dependent,
which can be checked looking at the range of
$\begin{pmatrix}
         1 &      2  \\
          \llap{\text{-}}1&1\\
         2 & 1
    \end{pmatrix}$
Clearly, this matrix will have range greater or equal to 1, since it has entries with $1$ and the multiplicative identity of a field can't be zero (by convention). We need the range to be 1 in order to have the linear dependence, so we need to make sure the minors of order two are all 0:
$\left|\begin{matrix}
         \ 1 &      2  \\
          \ \llap{\text{-}}1&1
    \end{matrix}\right|=3,\quad\left|\begin{matrix}
         1 &      2  \\
         2 & 1
    \end{matrix}\right|=-3,\quad\left|\begin{matrix}
          \ \llap{\text{-}}1&1\\
         \ 2 & 1
    \end{matrix}\right|=-3$
and these are 0 if and only if $3=0$ if and only if $\text{char}(\Bbb K)=3$.
A: The simplest way is to rewrite $(1: -1: 2)$ as $(-1, 1, -2)$.
If we have $(-1, 1, -2) = \lambda(2, 1, 1)$, then by comparing the second coordinate, it's clear that $\lambda = 1$ and hence $-1 = 2$ and $-2 = 1$. This of course implies $3 = 0$.
