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I would like to compute the divisor class group of the projective quadric cone $$ Q=\mathrm{Proj}(\mathbb{C}[X_0,X_1,X_2,X_3]/(X_1X_2-X_3^2)). $$ It has as an open subset the quadric cone $U$ in $\mathbb{C}^3$, which, according to Example II.6.5.2 of Hartshorne's, satisfies $\mathrm{Cl}(U)\simeq \mathbb{Z}/2\mathbb{Z}$. Therefore we have an exact sequence: $$ \mathbb{Z}\rightarrow \mathrm{Cl}(Q)\rightarrow \mathbb{Z}/2\mathbb{Z}\rightarrow 0, $$ where the first map sends $1$ to the class of the divisor $Q\setminus U$, and the second map is restriction.

I am not sure if this helps, but it is all I have been able to do so far.

By the way, what would be the canonical class of $Q$? Could we use the adjunction formula to compute it?

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  • $\begingroup$ I don't think you can use the adjunction formula because this is a singular cone. If you can show the first map is injective then there are only two possibilities for the middle group. To see that the first map is injective, you can compute the Euler characteristic (to do this you can first push it forward to P^3 since the inclusion is finite). $\endgroup$ – Eoin Jun 11 '18 at 9:05
  • $\begingroup$ should it be $\operatorname{Proj}(\mathbb{C}[x_0,x_1,x_2]/(x_0x_1-x_2^2))$? $\endgroup$ – user347489 Jun 11 '18 at 9:12
  • $\begingroup$ I think you can get the rest by showing there is an element that squares to the divisor Q\U. Conceptually, I'm thinking of this divisor as a conic, which is degree 2. Something that would square to this would be degree 1, so just a line. So I would try to find a line in the cone. I think all of this should imply CL(Q) is $\Bbb{Z}$. $\endgroup$ – Eoin Jun 11 '18 at 9:23
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Let $X = Proj(\mathbb{C}[X_1,X_2,X_3]/(X_1 X_2 - X_3^2))$. Now, observe that $Q = \overline{C(X)}$, where $C(X)$ is the affine cone over $X$ and bar above it denotes the projective closure. Now use the exercise 6.3(a) of Hartshorne's Algebraic Geometry, Chapter 2 to get

$$Cl(Q) \cong Cl(X)$$

Now clearly $X \cong \mathbb{P}^1$ via the veronese embedding. Thus $Cl(X) \cong Cl(\mathbb{P}^1) \cong \mathbb{Z}$.

The exact sequence written above has an interpretation in the following form

$$\mathbb{Z} \rightarrow Cl(Q) \rightarrow Cl(C(X))$$

This is well-described in exercise 6.3(b).

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