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I've been turning this exercise over for a while, and I appear to be stuck in particular on part (c). The question is:

Let $V$ be a projective variety in $\mathbb{P}^n$ of dimension $\geq 1$ and nonsingular in codimension $1$. Let $X$ be the affine cone over $V$ in $\mathbb{A}^{n+1}$, and $\bar{X}$ it's projective closure. Let $S(V)$ be the homogeneous coordinate ring of $V$. Show $S(V)$ is a UFD if and only if $V$ is projectively normal and $\text{Cl } V \simeq \mathbb{Z}$, generated by the class of $V.H$

I can do one direction. Namely, if $S(V)$ is a UFD, then the following exact sequence

$$\mathbb{Z} \to \text{Cl } V \to \text{Cl }X \to 0$$

can be modified to $$0 \to \mathbb{Z} \to \text{Cl }V \to 0 $$ since $\text{Cl }X$ is $0$, as $S(V)$ is also the coordinate ring of the cone $X$, and it is proven in the previous part that the map from $\mathbb{Z}$ is injective. This proves the divisor condition. However it is also true that a UFD is integrally closed, so projective normality follows.

What I can't do is the other direction. This should be just a statement about commutative algebra. I would like to use the equivalent condition that all prime ideals are principal, for example, but I don't know how to relate this to the divisor class group. Another possible approach that might be promising is the also equivalent condition that the ring satisfy the ascending chain condition on principal ideals (which is trivial, since the ring we're dealing with is already Noetherian, as a variety is a finite type scheme over a field and so is Noetherian) and every irreducible is prime. So it suffices to check that projective normality and having class group the integers is enough to prove every irreducible is prime.

Is this the right way to go? Perhaps there is a better way?

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We have the following three statements; note that the third is probably the commutative algebraic result you were looking for.

  1. $V$ is projectively normal if and only if $S(V)$ is an integrally closed domain by definition (Exercise II.5.14).

  2. $\operatorname{Cl} V \cong \mathbf{Z} \cdot V.H$ if and only if $\operatorname{Cl} X = 0$ by the short exact sequence (b): $$0 \longrightarrow \mathbf{Z} \longrightarrow \operatorname{Cl} V \longrightarrow \operatorname{Cl} X \longrightarrow 0.$$

  3. If $A$ is a noetherian domain, then $A$ is a UFD if and only if $A$ is integrally closed and $\operatorname{Cl} X = 0$, where $X = \operatorname{Spec} A$ (Proposition II.6.2).

Now we can prove (c). By (1) and (2) above, $V$ is projectively normal with $\operatorname{Cl} V \cong \mathbf{Z} \cdot V.H$ if and only if $S(V)$ is an integrally closed domain and $\operatorname{Cl} X = 0$. But by (3), this occurs if and only if $S(V)$ is a UFD.

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  • $\begingroup$ Thank you. I had forgotten that 6.2 is in fact an if and only if. For a while what I was doing was trying to reprove the theorem but for a homogeneous coordinate ring, trying to find an analogue of that fact. What I should have done is true and use the exact sequence. $\endgroup$ – Alfred Yerger Apr 30 '18 at 22:07

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