Zac Maeder-Wolland: Some Daily AG Problems

Hartshorne Exercise III.5.2.

NB: This problem was pretty difficult for me. Please note that this problem also requires a good level of comfort with properties of pullbacks and pushforwards of coherent sheaves.

(a) We follow the hint. First note that $\op{Supp} \mathcal{F}(n)= \op{Supp}\mathcal{F}$ for any $n\in\Z$. Denote by $f_r:\Z\ra\Z$ the function $\chi (\mathcal{F}(\cdot))$, where $r=\dim\op{Supp}\mathcal{F}$. Suppose that $\dim \op{Supp}\mathcal{F}=0$, so the support of $\mathcal{F}(n)$ is just finitely many closed points. $\mathcal{F}(n)$ is the skyscraper sheaf $\displaystyle\bigoplus_{x\in\op{Supp}\mathcal{F}}\iota_{x,*}(\mathcal{F}(n)_x)$ where $\iota _x$ is the inclusion of ${x}\ra X.$ Then for $i>0$,

$$H^i(X,\mathcal{F}(n))\cong \displaystyle\bigoplus_{x\in\op{Supp}\mathcal{F}}H^i(\{x\},\mathcal{F}(n)_x)=0,$$

where we have used that $H^i(Y,\mathcal{E})=0\:\:\forall i>\dim X$ for a quasicoherent sheaf $\mathcal{E}$ on a noetherian separated scheme $Y$ and that cohomology commutes with direct sums and plays nicely with the pushforward. For $i=0$, we see that

$$H^0(X,\mathcal{F}(n))=\displaystyle\bigoplus_{x\in\op{Supp}\mathcal{F}}\mathcal{F}_x\otimes_{\OO_{\PP{k}{n},x}} \OO_{\PP{k}{n}}(n)_x\cong\displaystyle\bigoplus_{x\in\op{Supp}\mathcal{F}}\mathcal{F}_x.$$

Hence $\chi (\mathcal{F}(n))=\displaystyle\prod_{\op{Supp}\mathcal{F}}\dim_k\mathcal{F}_x$ is independent of $n$ and $P_0(z):=f_0\in\Z\subset \Q[z]$ is a constant polynomial. This solves the base case.

Let $r>0$ and suppose the claim true for coherent sheaves $\mathcal{F}$ with $\dim\op{Supp}\mathcal{F}< r$. Proposition I.7.3 says this: (b) If $f:\Z\ra\Z$ is any function, and if there exists a numerical polynomial $Q(z)\in\Q[z]$ such that the difference function $\Delta f=f(n+1)-f(n)$ is equal to $Q(n)$ for all $n>>0$, then there exists a numerical polynomial $P(z)$ such that $f(n)=P(n)$ for all $n>>0$. We will use this proposition. Assume that $\dim\op{Supp}\mathcal{F}=r$. Then,

$$ f_r(n):=\chi(\mathcal{F}(n))=\displaystyle\sum_{i=0}^r(-1)^ih^i(X,\mathcal{F}(n))\ra h^0 (\mathcal{F}(n))\:\:\text{ as }\:\:n\ra\infty,$$

by Serre vanishing. For all $n\in \Z$

$$\Delta f_r(n)=\chi(\mathcal{F}(n))-\chi(\mathcal{F}(n-1)).$$

Moreover, if

$$0\ra\mathcal{K}\ra\mathcal{F}(-1)\ra\mathcal{F}\ra\mathcal{R}\ra0$$

is any exact sequence of coherent sheaves on $X$, then tensoring with $\OO_X(n)$ for and splitting this into two SES’s yields,

$$0\ra \mathcal{K}(n)\ra\mathcal{F}(n-1)\ra\mathcal{M}(n)\ra 0$$

$$0\ra \mathcal{M}(n)\ra\mathcal{F}(n)\ra\mathcal{R}(n)\ra 0,$$

where $\mathcal{M}:=\ker (\mathcal{F}\ra\mathcal{R})$. Considering their LES’s in cohomology yields that,

$$ \chi (\mathcal{F}(n))-\chi (\mathcal{F}(n-1))=\chi (\mathcal{R}(n))-\chi (\mathcal{K}(n))\:\:\forall n\in Z.$$

This will allow us to apply the proposition about numerical polynomials. Now we seek a suitable exact sequence.

I claim that there is a hyperplane $H\subset\PP{k}{n}$ that does not contain any one of the irreducible components of $\op{Supp}\mathcal{F}$. Indeed, $\op{Supp}\mathcal{F}$ is closed, so it has finitely many irreducible components; choosing a point in each one, we can always find a hyperplane $H$ that does not contain their union. This is clear when we consider the projective space dual $\mathbb{P}H^0(\PP{k}{n},\OO(1))$; for each point, the subspace of sections vanishing on that point has dimension strictly less than $n$, so a finite union of such subspaces is a proper subspace of $\mathbb{P}H^0(\PP{k}{n},\OO(1))$. By the projective dimension theorem, all of the irreducible components of $\op{Supp}\mathcal{F}$ are varieties of dimension $\le r$ that have nonempty intersection with $H$. Moreover, this intersection $H\cap \op{Supp}\mathcal{F}$ has dimension strictly less than $r$ because because the intersection of $H$ with each irreducible component is a closed proper subset of that component. Write $\iota:H \cap X\ra X$ for the closed immersion of $H\cap X$. Note that $H$ is given by a section $s\in H^0(\PP{k}{n},\OO(1))$. Putting $\mathcal{K}:=\op{ker}(\mathcal{F}(-1)\ra\mathcal{F})$ and $R:=\mathcal{F}|_{H\cap X}:=\iota_*\iota^*\mathcal{F}$ gives us an exact sequence,

$$0\ra\mathcal{K}\ra\mathcal{F}(-1)\xrightarrow{\cdot s}\mathcal{F}\ra\mathcal{F}|_{H\cap X}\ra0.$$

Let’s see why this is exact. Let $j:X\ra \PP{k}{n}$ be the immersion of $X$, and $l:H\ra\PP{k}{n}$ the closed immersion of $H$. Starting with the exact sequence $0\ra\OO_{\PP{k}{n}}(-1)\ra \OO_{\PP{k}{n}}\ra l_*\OO_H\ra0$, we apply $j^*$ and then $\_\_\otimes_{\OO_X}\mathcal{F}$, both of which are only exact on the right. We get the sequence above because $j^*l_*\OO_H\otimes_{\OO_X}\mathcal{F}$ is canonically isomorphic to $\mathcal{F}|_{X\cap H}$ (this is easily checked on stalks). Therefore, for all $n\in\Z$,

$$\Delta f_r(n)=\chi (\mathcal{F}(n))-\chi (\mathcal{F}(n-1))=\chi(\mathcal{F}|_{H\cap X}(n))-\chi(\mathcal{K}(n)).$$

For large $n$, the same equality holds with $\chi(\cdot)$ replaced by $h^0(\cdot)$. Now, by construction, $\dim \op{Supp}\mathcal{F}|_{H\cap X}\lt r$ so by induction there is a polynomial $S(z)\in\Q[z]$ such that $S(n)=\chi(\mathcal{F}|_{H\cap X}(n))$ for all $n\in\Z$. I claim that the support of $\mathcal{K}$ is also lower dimensional. Indeed, looking at stalks, the support of $\mathcal{K}$ is exactly where $\mathcal{F}(-1)_x\ra\mathcal{F}_x$ is not injective. For one, we certainly have $\op{Supp}\mathcal{K}\subset\op{Supp}\mathcal{F}$. But after a closer look, I claim that $\op{Supp}\mathcal{K}\subset\op{Supp}\mathcal{F}\cap H$. To see this, we need get our hands dirty; we must give in cash the map that is “multiplication by s”.

At every stalk, this map is actually multiplication by $(j^*s)_x$. We see that $(j^*s)_x$ is just the image of $s_x$ under the ring homomorphism $\OO_{\PP{k}{n},j(x)}\ra\OO_{X,x}$. Hence, if $x\notin H=\{s=0\}$ then $x\in \{s_x\ne 0\}$, so $s_x$ must be a unit in both rings. Therefore $(j^*s)_x$ is a unit, and multiplication by a unit is injective. If $x\in H$, multiplication by $(j^*s)_x$ is injective if and only if $(j^*s)_x$ is not a zero divisor. This is what we wanted to see: $\op{Supp}\mathcal{K}\subset\op{Supp}\mathcal{F}\cap H$.

Thus again by induction, there is a polynomial $K(z)\in\Q[z]$ such that $K(n)=\chi(\mathcal{K}(n))$ for all $n\in\Z$. This shows that there is a numerical polynomial, $S(z)-K(z)\in\Q[z]$, such that

$$\Delta f_r(n)=S(n)-K(n)$$

for all $n\in\Z$. By the proposition, there is a numerical polynomial $P_r(z)\in\Q[z]$ such that $P_r(n)=f_r(n)$ for $n>>0$. Unfortunately this equality only holds for large $n\ge n_0$ right now. However I claim that we do indeed have $P_r(n)=f_r(n)$ for every $n\in\Z$. Suppose ab absurdo that $n_0$ above were minimal, i.e., $P_r(n_0-1)\neq f_r(n_0-1)$. Notice that $f_r(n_0-1)=P_r(n_0)-S(n_0)+K(n_0)$. The key here is to use the proof of the proposition on numerical polynomials: the construction involved that $\Delta P_r(z)=S(z)-K(z)\in \Q[z].$ It follows that,

$$f_r(n_0-1)=P_r(n_0)-\Delta P_r(n_0)=P_r(n_0-1),$$

which contradicts the minimality of $n_0$. We thus conclude the induction: the numerical polynomial $P_r(z)\in \Q[z]$ has $P_r(n)=\chi(\mathcal{F}(n))$ for all $n\in\Z$. This completes part (a).

(b) Now we have $X=\PP{k}{r}$ and $M:=\Gamma_*(\mathcal{F})=\displaystyle\bigoplus_{n\in\Z}H^0(X,\mathcal{F}(n))$. As defined in section I.7, the Hilbert polynomial of $M$ is the unique polynomial $P_M(z)\in\Q[z]$ with $P_M(l)=\varphi_M(l)$ for all $l>>0$, where $\varphi_M(l)=\dim_k M_l$, and $M_l$ means the degree-$l$ part of $M$. This polynomial also satisfies $\deg P_M=\dim \op{Supp}(\OO_{\PP{k}{r}}/\mathcal{I}_M)$, where $\mathcal{I}_M$ is the ideal sheaf for $\op{Ann}_S(M)$ (equivalently $\op{Supp}(\OO_{\PP{k}{r}}/\mathcal{I}_M)=\op{Proj}(k[x_0,…,x_r]/I)$ with $I=\Gamma_*(\mathcal{I}_M)$), however the uniqueness follows from the first specification only. We defined our Hilbert polynomial above as

$$P(n)=\chi(\mathcal{F}(n))=\displaystyle\sum_{i=0}^r(-1)^ih^i(\PP{k}{r},\mathcal{F}(n)).$$

Since the Hilbert polynomial $P_M$ from I.7 was unique, it suffices to show that $P(l)=\varphi_M(l)\:\:\forall l>>0$. It is immediate by Serre vanishing that, for $l>>0$,

$$P(l)=h^0(\PP{k}{r},\mathcal{F}(l))=\dim_kM_l=\varphi_M(l).$$

We are done.