Check injectivity of the map $\mathbb{P}^1_{\mathbb{K}} \rightarrow \mathbb{P}^2_{\mathbb{K}}$ defined by $(t_0:t_1)\mapsto(t_0^3:t_0t_1^2:t_1^3)$

Question

The title of this question is explicit, I want to check the given map is injective, provided the field $\mathbb{K}$ is algebraically close; take for instance $\mathbb{K}=\mathbb{C}$ for clarification purposes.

My try: suppose $\phi(t_0:t_1)=\phi(s_0:s_1)$. Then, checking term by term, we obtain

$$t_0 = \lambda^{1/3}s_0^I$$ $$t_1=\lambda^{1/3}s_1^J$$

Where $I,J\in\{1,2,3\}$ are the superscrips of the distinct numbers in $\mathbb{K}$ for which the first and the third equation hold.

For simplicity, suppose $s_0=s_0^1$, $s_1 = s_1^1$. Substituting in the second equation:

$$t_0t_1^2=\lambda^{1/3}s_0^I \lambda^{2/3}{s_1^J}^2=\lambda s_0^I{s_1^J}^2=\lambda s_0^1{s_1^1}^2$$

In other words:

$$s_0^I{s_1^J}^2=s_0^1{s_1^1}^2$$

Which can only occur when $I=1$ and $J=1$, because taking square roots:

$$s_1^J=\pm s_1^1$$ And it would be impossible that $s_1^J=-s_1$, because then ${s_1^J}^3=-{s_1^1}^3$, contradicting the third equation, unless $s_1=0$, and in this case injectivity is clear.

Answer 1

Your proof is a bit confusing, so I'm not sure if it's correct. Here's the proof I would write:

Suppose $\phi(s_0:s_1) = \phi(t_0:t_1)$, and that $s_1 \neq 0$. It follows that (since we can divide by $s_1^2$) $$ (s_0s_1^2 :s_1^3) = (t_0t_1^2 :t_1^3) \implies\\ (s_0 : s_1) = (t_0t_1^2 :t_1^3) $$ Since $s_1 \neq 0$, conclude $t_1^3 \neq 0$, which means that $t_1 \neq 0$. Thus, we have $$ (s_0:s_1) = (t_0:t_1) $$ On the other hand, if $s_1 = 0$, then $\phi(s_0:s_1) = (s_0^3:0:0)$. If $\phi(t_0:t_1) = (s_0^3:0:0)$, then we have $t_0 = 0 \iff s_0 = 0$, which is to say that $(s_0:s_1) = (t_0:t_1)$ as desired.

Answer 2

It's slightly neater to work in the affine charts. Suppose $(t_0:t_1), (s_0:s_1)$ are two points in $\mathbb{P}^1_k$ having the same image under this map. Then clearly $$t_1 = 0 \Leftrightarrow s_1 = 0.$$ We distinguish to cases: either $t_1 = 0$, in which case $s_1 = 0$, hence $(t_0 : t_1) = (1:0) = (s_0:s_1)$. Otherwise $t_1 \neq 0$, hence $s_1 \neq 0$ hence we can write $(t_0:t_1) = (t:1)$ and $(s_0:s_1) = (s:1)$ and so $$(t^3 : t: 1) = (s^3:s: 1)$$ which shows that $s = t$, so we're done.

I chose to look at $t_1$ since in the chart $t_1 \neq 0$ the map looks like $t \mapsto (t^3,t)$ which is clearly injective.