# Coordinate rings of projective varieties as UFDs

I can't find the mistake in my logic and so I hope you can help me.

Let $$k$$ be an algebraically closed field. It is well known that the projective algebraic sets $$\mathbb{P}_{k}^1$$ and $$V=V(x^2+y^2-z^2)$$ are isomorphic. In an exercise problem, we were supposed to prove that coordinate rings of rational curves (i.e. curves which are birationally equivalent to $$\mathbb{P}^1$$) are UFDs. However, $$k[V] = k[x,y,z]/(x^2+y^2-z^2)$$ is not a UFD (see e.g. MSE/413506 ).

Where is my mistake?

• You are correct. Co-ordinate rings make sense only once you fix an embedding and so one could be a UFD and the other not. May 24, 2020 at 17:08
• @Mohan Thank you for your answer (again). I don‘t quite understand yet, unfortunately. Which embedding do you mean? May 24, 2020 at 17:43
• As an example in your case, $\mathbb{P}^1$ can be embedded just in itself or as a conic in $\mathbb{P}^2$. The two corresponding co-ordinate rings are different, one a UFD, the other not. May 24, 2020 at 18:03
• @Mohan I still fail to see how this answers my question. I know that the coordinate rings of isomorphic projective varieties must not be isomorphic but for me that contradicts the result that the coordinate ring of any rational curve is apparently a UFD. (I.e. doesn‘t that result imply that every embedding of $\mathbb{P}^1$ should have a UFD as its coordinate ring? Or is the result simply wrong?) May 24, 2020 at 18:03
• You are right and wherever you are quoting from is wrong in the sense that the co-ordinate rings of the same variety under different embeddings can exhibit different behaviors. May 24, 2020 at 18:06

I think you need to take the affine coordinate ring of the curve not the projective coordinate $$k[x, y, z]/(x^2+y^2-z^2)$$ . So in this example, the question is is $$k[x,y]/(x^2+y^2-1)$$ a UFD, and I think so since $$k$$ is algebraically closed: