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Let $X$ be an affine smooth variety over the complex numbers, $X^{an}$ its associated smooth complex analytic space, and $\mathcal{O}$, resp. $\mathcal{O}^{an}$ the respective structure sheaves.

Is the natural map

$$\mathcal{O}(X)\to \mathcal{O}^{an}(X^{an})$$

injective?

Denoting by $f : X^{an}\to X_{Zar}$ the morphism of sites, this map is given as the composition

$$\Gamma(X_{Zar}, \mathcal{O})\to \Gamma(X^{an},f^{-1}\mathcal{O})\to \Gamma(X^{an},\mathcal{O}^{an})$$ and in this composition we know the last map is injective (see Serre’s GAGA paper, Prop. 10(b)).

What about the first map?

Morally, this map should send a regular function on $X$ to the holomorphic function it induces on $X^{an}$ in an evident way, so I would expect the answer to this question to be trivially yes, but I want to make sure I’m not missing anything.

If we call $\alpha$ the map $\Gamma(X_{Zar}, \mathcal{O})\to \Gamma(X^{an},f^{-1}\mathcal{O})$, then for any closed point $x\in X$, $\alpha(s)_x = \alpha_x(s_x)$, where $s\in \Gamma(X_{Zar}, \mathcal{O})$, $\alpha_x$ is the map induced on stalks, and $x$ is also regarded as a point of $X^{an}$, since analytification identifies closed points.

It is known that $\alpha_x$ is an isomorphism for all $x$ (see the discussion before Prop. 10 in loc cit), so this should be it. Am I right?

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  • $\begingroup$ By the strong Nullstellensatz, the zero ideal in $\mathcal{O}_X(X)$ equals the intersection of all kernels of all evaluation $\mathbb{C}$-algebra homomorphism, $$\text{ev}_x:\mathcal{O}_X(X)\to \mathbb{C}.$$ Each of these homomorphisms factors through $\mathcal{O}_{X^\text{an}}(X^{\text{an}})$. $\endgroup$ – Jason Starr 3 hours ago

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