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So I have this question, and I'm confused at how exactly to show that they are isomorphic as varieties. Firstly, we have the "circle" $X^2 + Y^2 = Z^2$ in $\mathbb{P}^2$ (let's call it $C$), and we can consider the Zariski open sets of this variety $U = \{[X : Y : Z] \in C \mid Y \neq 0 \text{ or } X \neq Z\} = C \setminus \{[1:0:1]\}$ and $V = \{[X : Y : Z] \in C \mid Y \neq 0 \text{ or } X \neq -Z\} = C \setminus \{[1:0:-1]\}$. Then on $U$ we can define the morphism $f: C \to \mathbb{P}^1$ by $f([X:Y:Z]) = [X - Z : Y]$ and on $V$ we can define it by $f([X : Y : Z]) = [-Y, X + Z]$. This is well-defined as they agree on their intersection. Also, as sets, it isn't difficult to show that this is a bijection. It will also be helpful to see that it's inverse is given by $g([X : Y]) = [Y^2 - X^2 : 2XY : X^2 + Y^2]$.

Now, all the above is fine to me. Where I get a bit confused is showing that $f$ and it's inverse are morphisms of varieties. To do this, for each map we need to show that it is continuous on the Zariski topology and that the pullback of any regular function is again a regular function. The reason I get confused here is because I just haven't really seen any calculations or anything of this sort, just the definitions, so I'm a little confused about applying them.

I think I've made a little progress. My definition of a regular function is that it locally looks like the quotient of two homogenous polynomials of the same degree. So the second part is fairly clear as for $f$, $X - Z$, $Y$, and $X + Z$ are all homogenous degree 1 polynomials and for $g$, $Y^2 - X^2$, $2XY$, and $X^2 + Y^2$ are also all homogenous polynomials. Thus, the result comes as the composition of homogenous polynomials is homogenous.

So all we have left is to show that $f$ and $g$ are Zariski continuous. I get stuck here as to me, the Zariski topology seems to ugly to work with. On some level, $f$ and $g$ look very continuous as I mean, they're just polynomials really. But how do I actually show this? The open sets being defined by where polynomial equations are nonzero just seems so impossible to work with!

Maybe I'm overcomplicating this a little, but I'd love some help on this! Thanks so much!

Fnark Man
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    The $2$-uple of $(s:t)$ is $(x:y:z)=(s^2:st:t^2)$ Can you find a way from $xz-y^2=0$ to $x^2+y^2-z^2=0$ and back? – Jan-Magnus Økland Jan 30 '25 at 09:10
  • Hmm, so I can see that the isomorphism between these two sets comes from $(x + z)(z - x) - y^2 = 0$ and so there are natural maps two and from the two sets and they are in bijection. And because coordinate wise the maps are all homogenous polynomials, the pullback condition is fine.

    But I think I'm still confused on the Zariski continuity? I'm probably overcomplicating it all, but for some reason showing any of these maps are Zariski continuous seems very hard just as the Zariski topology feels very "unnatural", I guess.

     – Fnark Man Jan 30 '25 at 17:24
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    To show that they're continuous, you can show that the inverse image of a closed set is closed. A Zariski-closed set in $\mathbb{P^2}$ is of the form $\mathbb{V}(h)$ for some homogeneous polynomial $h \in k[X,Y,Z]$, so a closed subset of $C$ is of the form $C \cap \mathbb{V}(h)$. Can you show that $g^{-1}(\mathbb{V}(h)) = \mathbb{V}(h \circ g)$? It may be helpful to give your coordinates on $\mathbb{P}^1$ different names to avoid clashing with those on $\mathbb{P}^2$. – Viktor Vaughn Jan 30 '25 at 17:26
  • Oh, right! That $g^{-1}(V(h)) = V(h \circ g)$ essentially just falls out by definition, right? Ok, that's much nicer than I thought, I was overcomplicating it so much. Thanks! – Fnark Man Jan 30 '25 at 17:51
  • The consequence of that is that any equation of the form $f(P) = [f_0(P), \dots, f_n(P)]$ where each $f_i$ is a homogenous polynomial of the same degree will be Zariski continuous, right? As the same argument will always work? – Fnark Man Jan 30 '25 at 17:53
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    A very similar question already exists on MSE here. Does that answer your question? – KReiser Jan 30 '25 at 19:16
  • @KReiser I think my problem was a little different, but with that and the above comments I have a pretty good grasp on the situation, thanks! – Fnark Man Jan 31 '25 at 03:33

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