Show that the map of graded rings $k[w, x, y, z] \rightarrow k[s, t]$ given by $(w, x, y, z) \mapsto (s^3, s^2t, st^2, t^3)$ induces a closed embedding $\mathbb{P}_k^1 \rightarrow \mathbb{P}_k^3$, which yields an isomorphism of $\mathbb{P}_k^1$ with the twisted cubic.
I think I need to use the following exercise: If $S \rightarrow R$ is a surjection of graded rings, then the domain of the induced morphism is $Proj (R)$, and the induced morphism $Proj (R) \rightarrow Proj (S)$ is a closed embedding.
However, I don't see how the map $(w, x, y, z) \mapsto (s^3, s^2t, st^2, t^3)$ is surjective.
Best Answer
You're correct that the map isn't surjective. The point is that this map has image in degrees divisible by $3$, and so factors through the associated Veronese subring of $k[s, t]$ composed of the pieces of degree divisible by $3$, onto which it is surjective. In precise language:
Let $S = k[x, y, z, w], R = k[s, t]$, and let $\varphi \colon S \to R$ be the morphism of graded rings described above. Let $R' := \bigoplus_{j = 0}^{\infty} R_{3j}$ be the subring of $R$ consisting of the components of $R$ with degree divisible by $3$. The morphism of graded rings $i \colon R' \to R$ induces an isomorphism $\tilde{i} \colon \mathrm{Proj}(R) \to \mathrm{Proj}(R')$; this is, for instance, Exercise 6.4H of Vakil's Foundations of Algebraic Geometry.
Note that $\varphi$ takes image in $R'$, so letting $\varphi'$ be the map $S \to R'$ defined exactly as $\varphi$ is, we have that $\varphi = i \circ \varphi'$. Since $\varphi'$ is surjective, the induced map $\tilde{\varphi'} \colon \mathrm{Proj}(R') \to \mathrm{Proj}(S)$ is a closed embedding. Hence, since $\tilde{\varphi} \colon \mathrm{Proj}(R) \to \mathrm{Proj}(S)$ is the composition of an isomorphism ($\tilde{i}$) with a closed embedding ($\tilde{\varphi'}$), it is itself a closed embedding, as desired.