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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?

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  • $\begingroup$ 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. $\endgroup$
    – Mohan
    May 24, 2020 at 17:08
  • $\begingroup$ @Mohan Thank you for your answer (again). I don‘t quite understand yet, unfortunately. Which embedding do you mean? $\endgroup$
    – Qi Zhu
    May 24, 2020 at 17:43
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    $\begingroup$ 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. $\endgroup$
    – Mohan
    May 24, 2020 at 18:03
  • $\begingroup$ @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?) $\endgroup$
    – Qi Zhu
    May 24, 2020 at 18:03
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    $\begingroup$ 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. $\endgroup$
    – Mohan
    May 24, 2020 at 18:06

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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:

Ring of trigonometric functions with real coefficients

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  • $\begingroup$ Yes, that was indeed what the instructor had intended when I asked him later. I guess I forgot to post it on MSE, thank you for your answer! $\endgroup$
    – Qi Zhu
    Apr 27, 2021 at 14:40

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