$\mathbb{P}^1$ isomorphic to conic in $\operatorname{Proj}(K[x,y,z])$ context I have recently started to study schemes and I found my self on the follow situation:
I want to prove that $\mathbb{P}^1$ is isomorphic to a conic.
I have used the morphisim $\mathbb{P}^1\longrightarrow \mathbb{P}^2 $ that in coordinates is bring by $[s:t]\longrightarrow [s^2-t^2:2st:s^2+t^2]$, the image of this morphisim is a conic $C$ and it is in fact a isomorphisim on the image so $\mathbb{P}^1\cong C$ as abstract varieties.
Now I want to express this situation on $\operatorname{Proj}(K[s,t])\longrightarrow \operatorname{Proj}(K[x,y,z])$
I am absolutly lost on this so I will acept all help, advice or trick.
 A: There are (at least) two approaches to define such maps between projective spaces. Let me start with the more difficult one which is the one I alluded to at first.
You can define maps on charts, $$D_+(s^2-t^2) \to D_+(x), \ \frac{y}{x} \mapsto \frac{2st}{s^2-t^2}, \ \frac{z}{x} \mapsto \frac{s^2+t^2}{s^2-t^2}$$ and similarly on the other charts. Such maps glue to a map $\mathbb{P}^1 \to \mathbb{P}^2$. This approach generalizes and maps $X \to \mathbb{P}^n$ are given exactly in this way once one introduces an invertible sheaf on $X$. For your particular case, you can also check out 6.3.M in Vakil's Rising Sea.
Probably easier here is the approach suggested by Andrea Marino. Your map corresponds to the map of graded rings $$\varphi: K[x,y,z] \to K[s,t], \ x \mapsto s^2-t^2, \ y \mapsto 2st, \ z \mapsto s^2 + t^2. $$ The map $\varphi$ is surjective onto the Veronese subring $K[s,t]^{(2)} = K[s^2,st,t^2]$. So, it induces $$\mathbb{P}_K^1 = \operatorname{Proj}{K[s,t]} \xrightarrow{\ \sim \ } \operatorname{Proj}{K[s,t]^{(2)}} \hookrightarrow \operatorname{Proj}{K[x,y,z]} = \mathbb{P}_K^2. $$ Such ideas are also explained in chapter 8.2 of Vakil's Rising Sea. Note also that this approach is only deceptively easier. The fact that the map of graded rings induces a map on $\mathrm{Proj}$'s certainly needs to be derived via gluing with which we return to our initial approach. But theorems are proven to be used and so we fearlessly do exactly that.
A: Hint: look for a ring map in the other direction. Then show that it is surjective on induced projective construction. Finally, remember that "closed subvarieties <---> ideals <---> surjective maps"
