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I will denote by $X$ the set of rational points on the unit circle, i.e., $$X := \{ (x,y) \in \mathbb{Q}^2: x^2 + y^2 = 1 \}.$$ Viewing both $\mathbb{Q}$ and $X$ as an affine varieties, then every morphism $f: \mathbb{Q} \to X$ is given by two (global) regular functions on $\mathbb{Q}$. To be more specific, $$f(t) = \left(f_1(t),f_2(t)\right)$$ such that both $f_1$ and $f_2$ are rational functions of the form $\frac gh$, where $g$ and $h$ are polynomials on $\mathbb{Q}$, and $h$ never vanishes on $\mathbb{Q}$. My question is: does there exist an isomorphism $f: \mathbb{Q} \to X$?

On one hand, I know that the rational parametrization of the unit circle $$ t \mapsto \left( \frac{1-t^2}{1+t^2}, \frac{2t}{1+t^2} \right) $$ is a morphism from $\mathbb{Q}$ to $X$, but it's not an isomorphism. However, this doesn't mean there cannot be an isomorphism $f: \mathbb{Q} \to X$.

On the other hand, if we equip $\mathbb{Q}$ and $X$ with the usual topologies (i.e., the subspace topologies from $\mathbb{R}$ and $\mathbb{R}^2$, respectively), then $X$ is a countable metrizable space without isolated points, and hence homeomorphic to $\mathbb{Q}$, by a theorem of Sierpinski whose proof can be found here. So my question is indeed asking whether a homeomorphism $f: \mathbb{Q} \to X$ can be in the form of two rational functions $f_1$ and $f_2$ as mentioned above.

Yuhang Chen
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  • I don't know, but I hope not. – Matt Samuel Oct 25 '18 at 01:53
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    Key word for later users: Birational – AHusain Oct 25 '18 at 02:47
  • If they were isomorphic, wouldn’t they remain isomorphic to each other after you extend the scalars from $\Bbb Q$ to $\Bbb Q(i)$ ? But after such extension, your $X$ becomes the multiplicative group variety $\mathfrak X: {(\xi,\eta)\in \Bbb Q(i)^2: \xi\eta=1}$, definitely not the same as the affine line. – Lubin Oct 25 '18 at 05:14
  • Related: https://math.stackexchange.com/questions/1340461 ; if you take the projective closure, then you get a conic with a rational points, hence isomorphic to $\Bbb P^1$. – Watson Oct 25 '18 at 09:57
  • See also exercise 4.20.8 in Algebraic Geometry: A Problem Solving Approach by Garrity : « Show that the affine line $\Bbb A^1$ is birational to the circle » (they are birationally equivalent). – Watson Oct 25 '18 at 10:06
  • @Lubin By extending the scalars, did you mean the affine variety defined by the same equation(s) in a larger field? If so, then isomorphic varieties in the original space can become non-isomorphic after a change of field. For example, $x^2 + y^2 = 0$ and $x=0, y=0$ both define the origin $(0,0)$ in $\mathbb{Q}^2$, but not so in $\mathbb{Q}(i)^2$. – Yuhang Chen Oct 25 '18 at 22:09
  • @Watson Thank you for bringing up the relevant question. I understand that a circle is birational to an affine line over any field. But how does this help us determine whether they can be actually isomorphic over $\mathbb{Q}$? – Yuhang Chen Oct 25 '18 at 22:35
  • Sorry, but $x^2+y^2=0$ has only ore rational point, but is not same at all as the single point $x=0,y=0$. I guess, really, what we all need to know is what the objects are and what the morphisms are of your category. – Lubin Oct 25 '18 at 23:16
  • @Watson When I said "a circle is birational to an affine line over any field", I should have added that when the field is not of characteristic 2. – Yuhang Chen Oct 25 '18 at 23:24
  • @Lubin The objects are affine varieties as ringed spaces and the morphisms are morphisms between ringed spaces, which are continuous map pulling back regular functions to regular functions. If the target is an affine variety embedded in an $n$-dimensional affine space, then a morphism consists of $n$ global regular functions. I hope this clarifies why the morphism in my question is in the form I described. – Yuhang Chen Oct 25 '18 at 23:42
  • But for affine varieties $S_i=\text{Spec}A_i$, where $A_i$ are rings, isn’t a morphism $S_1\to S_2$ exactly equivalent to a ring morphism $A_2\to A_1$? (My apologies, it’s been so long…) Then in the case you asked about, the two affine spaces clearly are nonisomorphic. – Lubin Oct 26 '18 at 03:52
  • @Lubin If you think of an affine variety as an affine scheme, then yes. But as I described in this question, by affine varieties I really mean the points defined by polynomials. So a morphism between affine varieties can be more than a ring morphism between their coordinate rings. For example, the rational function $t \mapsto 1/(1+t^2)$ is a morphism from $\mathbb{Q}$ to itself (since the denominator is nonzero), although it is not a ring morphism from $\mathbb{Q}[t]$ to itself. – Yuhang Chen Oct 26 '18 at 04:16

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No, there is no isomorphism $\mathbb{Q}\to X$.

There is some risk of confusion here, because your notion of morphism for $\mathbb{Q}$-varieties is not the usual one (a morphism of $\mathbb{Q}$-varieties is usually required to be regular everywhere, not just on rational points). I will write $\mathbb{A}^1$ for the affine line over $\mathbb{Q}$, and $Y:=\mathbb{V}(x^2+y^2-1)\subset\mathbb{A}^2$. The $\mathbb{Q}$-rational points of $\mathbb{A}^1$ are $\mathbb{Q}$ and the $\mathbb{Q}$-rational points of $Y$ are your $X$.

In what follows, by "morphism" I mean morphism in the usual sense (so a morphism of affine varieties corresponds to a ring homomorphism going the other way). If I understand correctly, your question is whether there exists a birational morphism $f:\mathbb{A}^1\to Y$ that is bijective on $\mathbb{Q}$-points. Assume $f$ exists. Both $\mathbb{A}^1$ and $Y$ have $\mathbb{P}^1$ as a compactification, and $f$ extends to an isomorphism $f:\mathbb{P}^1\to \mathbb{P}^1$. Now $\mathbb{A}^1(\mathbb{Q})$ is a proper subset of $\mathbb{P}^1(\mathbb{Q})$, so $f(\mathbb{A}^1(\mathbb{Q}))$ must be a proper subset of $\mathbb{P}^1(\mathbb{Q})$. However, the inclusion $Y\subset\mathbb{P}^1$ is an equality on $\mathbb{Q}$-points, so we have $f(\mathbb{A}^1(\mathbb{Q}))\subsetneq Y(\mathbb{Q})$. In your language, this means $f(\mathbb{Q})\subsetneq X$.

Julian Rosen
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  • Thank you for your answer. But it's answering a different question because of your "usual" morphism for $\mathbb{Q}$-varieties. You can also see my discussion with @Lubin above, where I clarified what I meant by a morphism in my question. – Yuhang Chen Oct 29 '18 at 16:29
  • A isomorphism in your sense is a rational map that is regular on $\mathbb{Q}$-points, yes? I believe I have shown that such an isomorphism cannot exist. – Julian Rosen Oct 29 '18 at 23:00
  • I am not quite sure what you meant by "a rational map that is regular on $\mathbb{Q}$-points". But the morphism in my question doesn't correspond to ring homomorphisms. This was explained in my last comment with Lubin. – Yuhang Chen Oct 30 '18 at 01:38