# Divisors on cone variety not passing through cone point are principal

We work over an algebraically closed field $$k$$, and we use "divisor" to mean Weil divisor.

Let $$V$$ be a projective algebraic variety of dimension $$\geq 1$$ which is nonsingular in codimension one. Let $$X$$ be the affine cone over $$V$$ (the affine variety whose coordinate ring is the homogeneous coordinate ring of $$V$$).

Claim: Any divisor on $$X$$ not passing through the cone point must be principal.

This question comes from Exercise II.6.3(d) in Hartshorne's Algebraic Geometry. I can see that the above claim is equivalent to injectivity of the map of divisor class groups $$\mathrm{Cl}~X\rightarrow\mathrm{Cl}(\mathrm{Spec}~\mathcal{O}_P)$$ where $$\mathcal{O}_P$$ is the local ring of the cone point $$P$$. (The exercise in Hartshorne asks us to show $$\mathrm{Cl}~X\rightarrow\mathrm{Cl}(\mathrm{Spec}~\mathcal{O}_P)$$ is an isomorphism; surjectivity is clear.)

The corresponding algebraic statement is the following: if $$A=k[x_0,\ldots,x_n]/I$$ is the homogeneous coordinate ring of $$V$$, any height one prime $$\mathfrak{p}\subseteq A$$ not contained in $$\mathfrak{m}=(x_0,\ldots,x_n)$$ is principal. Here, $$\mathfrak{m}$$ corresponds to the cone point. Note that such a prime $$\mathfrak{p}$$ is necessarily non-homogeneous.

How should one see this? I have been unsuccessful in verifying the claim.

I solved the problem after posting this. My solution is included below.

Solution: Let $$A=k[x_0,\ldots,x_n]/I$$ be the homogeneous coordinate ring of $$V$$. Here $$I$$ is a homogeneous ideal. We may assume $$x_i\neq 0$$ in $$A$$ for all $$i$$.

The height one primes of $$A$$ are either homogeneous or non-homogeneous. We say that the homogeneous primes correspond to "type one" prime divisors and the non-homogeneous primes are "type two" prime divisors. (This is the terminology of Hartshorne's Proposition II.6.6.)

Let $$\mathfrak{p}\subseteq A$$ be a height one prime with $$\mathfrak{p}\not\subseteq (x_0,\ldots,x_n)=\mathfrak{m}$$. This corresponds to a prime divisor $$Y$$ of the cone $$X$$ which does not pass through the cone point $$P$$. Note this prime divisor is type two, because every type one divisor passes through $$P$$. Select $$h\in \mathfrak{p}$$ such that $$h\equiv 1 \pmod{m}$$. Pick any index $$i$$ and consider the successive localizations

$$A \hookrightarrow A_{x_i}=A_{(x_i)}[x_i,{x_i}^{-1}] \hookrightarrow K[x_i,{x_i}^{-1}] \subseteq \mathrm{Frac}(A)$$

where $$A_{(x_i)}$$ is the degree zero elements in the localization $$A_{x_i}=A[x_i^{-1}]$$ and $$K=\mathrm{Frac}\left(A_{(x_i)}\right)$$ is the function field of $$V$$.

Note that the generic point of any type two divisor is contained in the chart $$D(x_i)=\mathrm{Spec} ~ A_{x_i}$$ (i.e. the complement of $$x_i=0$$), because the hyperplane section $$x_i=0$$ contains only type one prime divisors (every minimal prime of a homogeneous ideal is homogeneous). The same argument shows that $$\mathfrak{p}$$ cannot contain any nonzero homogeneous elements so $$(\mathfrak{p}A_{(x_i)}[x_i,x_i^{-1}])\cap A_{(x_i)} = 0$$. Indeed, the type two prime divisors are precisely the pre-images in $$A$$ of nonzero primes in $$K[x_i,x_i^{-1}]$$.

Since $$K[x_i,x_i^{-1}]$$ is a principal ideal domain, the extended prime $$\mathfrak{p}K[x_i,x_i^{-1}] \subseteq K[x_i,x_i^{-1}]$$ is principal, generated by some $$f \in K[x_i,x_i^{-1}]$$. Scaling by $$K$$, $$x_i$$, and $$x_i^{-1}$$, we may assume that the lowest degree term of $$f$$ is equal to $$1$$.

If $$Y'\neq Y$$ is another type two prime divisor on $$X$$ with associated valuation $$v_{Y'}$$ on $$\mathrm{Frac}(A)$$, we have $$v_{Y'}(Y)=0$$. This is because the irreducible polynomial $$f$$ is contained in exactly one height one prime of $$K[x_i,x_i^{-1}]$$, which is the one corresponding to $$Y$$. We have $${v_{Y}} (f)=1$$, where $$v_Y$$ is the valuation on $$\mathrm{Frac}(A)$$ associated to $$Y$$.

Write $$h=fg$$ for some $$g \in K[x_i,x_i^{-1}]$$. Note that $$g$$ must also have lowest degree term equal to $$1$$.

Suppose $$Z$$ is a type one prime divisor whose generic point lies in $$D(x_i)$$, and write $$v_Z$$ for the associated valuation on $$\mathrm{Frac}(A)$$. If we write $$f=1+\sum_{\ell=1}^r a_{\ell}x_i^{\ell}$$ then $$v_Z(f)=\min\limits_{\ell\in\{0,\ldots,r\}} v_Z(a_\ell)$$ where $$a_0=1$$. Note $$v_Z(f)\leq 0$$ for all such $$Z$$. The same statements hold when $$f$$ is replaced with $$g$$. Since $$h$$ has smallest degree term equal to $$1$$, we know $$h\not \in \mathfrak{q}$$ for any homogeneous prime $$\mathfrak{q}\subseteq A$$. This implies $$v_Z(h)=v_Z(fg)=0$$ for all such $$Z$$. The non-positivity from above then implies $$v_Z(f)=v_Z(g)=0$$ for all such $$Z$$.

The chart $$D(x_i)$$ may not contain all type one divisors however, so pick another index $$j$$ and write $$f=1+\sum_{\ell=1}^r a_{\ell}'x_j^{\ell}\in K[x_j]\subseteq K[x_j,x_j^{-1}]$$ where $$a_{\ell}'=a_{\ell}(x_i/x_j)^\ell$$. As an element of $$\mathrm{Frac}(A)$$, this $$f$$ is the same one as before. We may do the same for $$g$$. The previous argument applied to this situation then shows that $$v_Z(f)=v_Z(g)=0$$ for all type one divisors $$Z$$ whose generic point is contained in $$D(x_j)$$. Varying over the index $$j$$ so that we cover $$X$$, we see that $$v_Z(f)=0$$ for all type one divisors $$Z$$.

Thus the divisor on $$X$$ associated to the rational function $$f$$ is simply $$Y$$, i.e. $$Y$$ is a principal divisor.

Remark: If $$A$$ is not normal, I am not sure whether $$f$$ has to lie in $$A$$, i.e. $$A$$ may not be a Krull domain. So the algebraic reformulation in my original question may be slightly stronger than necessary.

Corollary: Let $$V\subseteq \mathbb{P}^n_k$$ be a projectively normal variety (i.e. the homogeneous coordinate ring is normal) and let $$X$$ be the affine cone over $$V$$. Then the Cartier class group of $$X$$ is trivial.

Proof: For normal varieties, the Cartier class group may be identified with the subgroup of the Weil class group given by locally principal divisors. A Weil divisor which is principal at the cone point $$P$$ is linearly equivalent to a divisor not passing through the cone point. By the above problem, such a divisor is principal.

Example: The cone $$\mathrm{Spec} ~k[x,y,z]/(xy-z^2)$$ is normal. It has trivial Cartier class group but a nontrivial Weil class group.

• Could you please elaborate more on “Write $h=fg$ for some $g\in K[x_i,x_i^{-1}]$.”?Why such $g$(with lowest degree term equal to 1) exists? – XiaYu Feb 4 '20 at 12:20
• @XiaYu We know the ideal $\mathfrak{p}\cdot K[x_i,x_i^{-1}]$ is principal and generated by $f$, and also contains $h$, so $h$ is divisible by $f$. Next, we check that $f$ and $h$ have lowest degree term equal to $1$. For $f$, this is in the construction. For $h$, we required $h\equiv 1\pmod{\mathfrak{m}}$. Select a lift $\tilde{h} \in k[x_0,\ldots,x_n]$ of $h$. It's of the form $(1+\text{higher degree terms})$. Thus $h\in K[x_i,x_i^{-1}]$ is also $(1+\text{higher degree terms})$. For example if $\tilde{h}=1+x_0$ then $h\in K[x_i,x_i^{-1}]$ is $1+(x_0/x_i) \cdot x_i$, where $x_0/x_i \in K$. – 351910953 Apr 27 '20 at 2:26
• Yes, $f$ may not lie in $A$. But you can still conclude for the type I case that each summand in $f$ and $g$ has a valuation $\ge 0$. Indeed, if $Z$ is codimension variety associated to type one prime ideals of height 1 (i.e projects onto generic point of $V$) and if $\eta_Z$ is the generic point of $Z$, then the valuation $v_Z$ of any element of $\mathcal O_X,\eta_Z$ is non-negative. – quantum Nov 7 '20 at 11:56

I would give a simpler answer. I haven't read OP's answer cuz it's too long. However I guess these two are essentially the same. It's just a imitation of II.6.6 in the Hartshorne.

Let $$Y$$ be such a divisor not passing through $$P$$. Let the $$\mathbb{P}^n = \mathrm{Proj}\;k[x_1, \dots , x_{n+1}]$$ be where $$V$$ lives in. Let $$H_i = \{x_i = 0\} \subseteq \mathbb{P}^{n+1}=\mathrm{Proj}\;k[x_0, \dots , x_{n+1}]$$, where $$i>0$$, and $$H_i$$ not containing $$Y$$. Since $$Y$$ doesn't passing through $$P$$, it's prime ideal in $$X$$ containing $$1+f(x_j / x_0)$$ ($$f(x_j/x_0)$$ means $$j$$ runs over what it should run.). Then $$\mathcal{O}_Y$$, the stalk at the generic point, contains $$x_o^{deg f}/ x_i^{deg f}(1+f)$$.

Look at $$\bar{X} \backslash H_i = U_i \times \mathbb{A}^1$$, where $$U_i = X \backslash H_i$$. $$x_o^{deg f}/ x_i^{deg f}(1+f) \in \mathcal{O}_Y$$ implies it is a divisor of type two, since type one divisors shouldn't contain this element. As $$K[x_0/x_i]$$ is a UFD where $$K$$ is the function field of $$V$$, we can choose a prime polynomial $$x_0^m/x_i^m + \cdots +k_{0} \in \mathcal{O}_Y$$ where the coefficients $$k_j$$ are all in $$K$$.

Now consider the principal divisor $$(\frac{1}{x_0^m/x_i^m + \cdots +k_{0}}) = (\frac{x_i^m}{x_0^m + \cdots + k_0 x_i^m})$$. Its type 2 part is just $$-Y$$, and type one part is just $$mH\cdot V$$. The last one is zero as a divisor in $$X$$, which is proved in part (c) of this problem. So we've done.