The map of graded rings $k[w, x, y, z] \rightarrow k[s, t]$ induces a closed embedding $\mathbb{P}_k^1 \rightarrow \mathbb{P}_k^3$

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.

• Did you take a look at this post? – Liam Mar 29 at 2:36

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.
• When you say "Since $\varphi$ is surjective", should it be $\varphi'$? – Smash Mar 29 at 14:27