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We say a $K$-valued point on a scheme is a map Spec$(K) \to S$, so in particular, a real valued point on the parabola $y = x^2$ should be a map Spec$(\mathbb{R}) \to $Spec$(\mathbb{C}[x,y]/(y-x^2))$. However, no such map exists because there are no ring homomorphisms $\mathbb{C}[x,y]/(y-x^2) \to \mathbb{R}$ (look at the image of $i$). But any high school student would say there are plenty of real points on the parabola: namely $(x,x^2)$ for $x \in \mathbb{R}$.

What is the benefit of allowing this type of behavior to occur with schemes? I read the question here ($\overline{\mathbb{Q}}$-valued points of a $\mathbb{C}$-scheme) which didn't seem satisfactory, as there was the distraction of transcendental elements in the defining equation of the scheme there. How can I tell a $\mathbb{C}$-scheme "really" has $\mathbb{R}$-valued points in an example like this? Perhaps there is a solution using Galois actions, but I'd like to have an idea of what to do in general.

Supersingularity
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2 Answers2

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This is a classical case of losing information after base change.

It seems obvious to you that $Y = Spec(\mathbb{C}[x,y]/(y-x^2))$ is $X_\mathbb{C}$ with $X = Spec(\mathbb{R}[x,y]/(y-x^2))$, because, well, it's obvious, it's the same equation, and it had coefficients in $\mathbb{R}$ all along !

But it's not obvious at all. Because there may be a lot of other $X$ over $\mathbb{R}$ with $Y\simeq X_\mathbb{C}$ that have very different $\mathbb{R}$-points. This is the basis of descent theory : you usually can't recover $X$ from $Y$.

For instance, $\mathbb{C}[x,y]\to \mathbb{C}[x]$ defined by $x\mapsto x$ and $y\mapsto x^2$ has kernel $(y-x^2)$, so your $Y$ is also isomorphic (quite "canonically" given the equation) to $\mathbb{A}_\mathbb{C}^1$. Then the $\mathbb{R}$-valued points of $Y$ are the point of $\mathbb{A}_\mathbb{R}^1$, right ? In this case this is not too different, since in fact $\mathbb{A}_\mathbb{R}^1\simeq Spec(\mathbb{R}[x,y]/(y-x^2))$.

But now take $Y = Spec(\mathbb{C}[x,y]/(x^2+y^2+1))$. It's obviously $Y\simeq X_\mathbb{C}$ with $X = Spec(\mathbb{R}[x,y]/(x^2+y^2+1))$ which has no real-valued points. But it's also $Y\simeq X'_\mathbb{C}$ with $X' = Spec(\mathbb{R}[x,y]/(x^2+y^2-1))$ which has an infinite number of real-valued points.

Captain Lama
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  • It might be worth noting (for the OP) that the actual input to descend (as in 'descent theory') to something over $\mathbb{R}$ is, in this case, an action of $\text{Gal}(\mathbb{C}/\mathbb{R})$. But, if you have that Galois action, you can actually get at $\mathbb{R}$-points as Galois invariant $\mathbb{C}$-points. – Alex Youcis Apr 20 '16 at 13:59
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The real points on the parabola are the real points of $\text{Spec } \mathbb{R}[x, y]/(y - x^2)$, which is a scheme over $\text{Spec } \mathbb{R}$. Your scheme is the base change of this one to $\text{Spec } \mathbb{C}$.

Qiaochu Yuan
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