Why is the sequence $0\rightarrow\mathscr{O}_C(C)^\ast\rightarrow k(C)^\ast\rightarrow\rm Div(C) \rightarrow Pic(C)\rightarrow 0$ exact?

Question

Definition

• A curve over a field $k$ is a separated scheme $C$ of finite type over $k$ which is integral of dimension 1.
• Let $C$ be a normal curve over a field $k$. A divisor is an element of the free abelian group $\operatorname{Div}(C):=\bigoplus_{p\in C\text{,p closed}}\mathbb{Z}[p]$
• Let $D=\Sigma_{p}n_p[p]\in \operatorname{Div}(C)$. Define $\Gamma(U,\mathscr{O}_C(D)):=\{f\in k(C)|\forall p\in U:v_p(f)\geq -n_p\}$. Here $v_p$ is the valuation function.
• $D=\Sigma_pn_p[p]\in \operatorname{Div}(C)$, $\operatorname{Supp}(D):=\{p:n_p\neq 0\}$

Let $C$ be a normal curve over $k$ then we have the following theorem

Theorem The sequence $$0\rightarrow \Gamma(C,\mathscr{O}_C)^*\rightarrow k(C)^*\rightarrow \operatorname{Div}(C)\rightarrow \operatorname{Pic}(C)\rightarrow 0$$is exact.

I have read the proof but I have some issues in understanding

• the well-definedness and surjectivity of the map $\operatorname{Div}(C)\rightarrow\operatorname{Pic}(C)$
• $\ker(\operatorname{Div}(C)\rightarrow \operatorname{Pic}(C))\subset\operatorname{im}(k(C)^*\rightarrow \operatorname{Div}(C))$

Lemma For any $D\in \operatorname{Div}(C)$, $\mathscr{O}_C(D)$ is a line bundle.

Proof: Let $D=\Sigma_pn_p[p]$. For any closed point $p$, pick $f\in \mathscr{O}_{C,p}$ with $v_p(f)=|n_p|$. Pick $U\subset C$ open with $p\in U$ and $\operatorname{Supp}(D)\cap U=\operatorname{Supp}(\operatorname{div}(f))\cap U=\{p\}$. Define $e:=\frac{v_p(f)}{n_p}=\operatorname{sign}(n_p)\in \{+1,-1\}$. Then for any $V\subset U$ we have $$t\in \mathscr{O}_{C}(D)(V)\Leftrightarrow \forall q\in V: v_q(t)\geq -n_q=e\cdot v_q(f)\\ \Leftrightarrow \forall q\in V: v_q(t/f^e)\geq 0\\ \Leftrightarrow s:=t/f^e\in \mathscr{O}_{C}(V)$$ Then $$\mathscr{O}_C|_U\rightarrow \mathscr{O}_C(D)|_U\\ s\mapsto f^e\cdot s$$ is an isomorphism.

I think I understand the proof for this lemma. Basic idea is to locally replace $D$ by some principal divisor.

Lemma For any $D$, $D' \in \operatorname{Div}(C)$, there exists an isomorphism $\mathscr{O}_C(D)\otimes _{\mathscr{O}_C}\mathscr{O}_C(D')\sim \mathscr{O}_C(D+ D')$

Proof: For any $U\subset C$ open, we have a homomorpshism $$\Gamma(U,\mathscr{O}_C(D))\otimes \Gamma(U, \mathscr{O}_C(D'))\rightarrow \Gamma(U,\mathscr{O}_C(D+D'))\\ f\otimes g\mapsto f\cdot g$$ since $v_p(fg)=v_p(f)+v_p(g)$ for all closed points $p\in C$. These homomorphisms are compatible with restriction, hence pass to sheafification. Being an iso can be checked on stalks $$\mathscr{O}_C(D)_p\otimes _{\mathscr{O}_{C,p}}\mathscr{O}_C(D')_p\sim \mathscr{O}_C(D+ D')_p.$$

I understand everything but I do not know how to give explicit isomorphism $\mathscr{O}_C(D)_p\otimes _{\mathscr{O}_{C,p}}\mathscr{O}_C(D')_p\sim \mathscr{O}_C(D+ D')_p$. I assume it follows from the first lemma.

Proposition $\operatorname{Div}(C)\rightarrow \operatorname{Pic}(C)$ is surjective.

Proof: Let $\eta\in C$ be the generic point. Let $\mathscr{L}\in \operatorname{Pic}(C)\rightarrow \mathscr{L}_{\eta}$ is a 1-dim vector space over $K:=\mathscr{O}_{C,\eta}=k(C)$. For any closed $p\in C$, $\mathscr{L}_p \hookrightarrow \mathscr{L}_{\eta}=K\cdot s$ is free $\mathscr{O}_{C,p}$-submodule of rank 1. Say $\mathscr{L}_p=\mathscr{O}_{C,p}\cdot fs \text{ for some $f\in K$}=\{gs|g\in K, v_p(g)\geq -n_p\}$ with $n_p=-v_p(f)$. Then by the claim $D:=\Sigma_pn_p[p]$ is a divisor and by construction $$\mathscr{O}_C(D)\rightarrow \mathscr{L} \\ f\mapsto f\cdot s$$ is an isomorphism.
Claim: $n_p=0$ for almost all $p$.
Proof of claim: Let $\emptyset\neq U\subset C$ be open with $s\in \mathscr{L}(U)\subset \mathscr{L}_{\eta}$. Shrinking $U$ we may assume any closed point $p\in U$ has the property $s(p)\neq 0\in \mathscr{L}_p\otimes_{\mathscr{O}_{C,p}}\kappa (p)$. $\Rightarrow \forall p\in U:n_p=0$.

I have trouble understanding the claim. Why can we assume every closed $p\in U$ has the property $s(p)\neq 0\in \mathscr{L}_p\otimes_{\mathscr{O}_{C,p}}\kappa (p)$. And why this implies $\forall p \in U: n_p=0$? I guess this is a typo, maybe he means $n_p\neq 0$?

To show $\ker(\operatorname{Div}(C)\rightarrow\operatorname{Pic}(C))\subset\operatorname{im}(k(C)^\ast\rightarrow \operatorname{Div(C)}))$. It suffices to show $\mathscr{O}_C(D)\sim \mathscr{O}_C \Rightarrow \exists f\in k(C)^* $ such that $D=\operatorname{div}(f)$. Assume $\mathscr{O}_C(D)\sim \mathscr{O}_C$, then there exists isomorphism on generic stalk $\mathscr{O}_C(D)_{\eta}\xrightarrow[1]{\sim} \mathscr{O}_{C,\eta}$. Since $\eta \in C\setminus \operatorname{Supp}(D)$, there is a canonical isomorphism $\mathscr{O}_{C,\eta}\xrightarrow[2]{\sim}\mathscr{O}_C(D)_{\eta}$. Then $1\circ 2$ is an automorphism 1-dim $\mathscr{O}_{C,\eta}$-vector space and it is multiplication by some element $f\in \mathscr{O}_{C,\eta}^*$. Then $D=\operatorname{div}(f)$.

For the final part, my question is what is the canonical isomorpism $(2)$? Is it because $\mathscr{O}_C{D}$ and element of the Picard group? In the end why can we claim $\operatorname{div(f)}=D$?

I understand my questions are bit long and I have asked many. So any help is appreciated.

Answer 1

Just to start with:

Why is the algebra structure map $\mu:f\otimes g\mapsto f\cdot g$ (a priori a map $\mathscr{K}\otimes\mathscr{K}\to\mathscr{K}$ which restricts to a map $\mu:\mathscr{O}(D)\otimes\mathscr{O}(D')\to\mathscr{O}(D+D'))$ an isomorphism at the stalk of every $p\in C$ (and therefore an isomorphism)?

One way: use your first lemma. Choosing a suitably small neighbourhood $U$ of $p$ and $f\in\mathscr{O}_p,g\in\mathscr{O}_p$ with $\nu_p(f)=|n_p|,\nu_p(g)=|m_p|$ - where $D=n_p[p]+\cdots,D'=m_p[p]+\cdots$ - we have, for some fixed signs $e,e'\in\{\pm1\}$, that $\mathscr{O}_U\cong\mathscr{O}(D)_U,\mathscr{O}_U\cong\mathscr{O}(D')_U$ via $s\mapsto f^es,g^{e'}s$ respectively.

Therefore, $\mathscr{O}_U\cong\mathscr{O}_U\otimes\mathscr{O}_U\cong(\mathscr{O}(D)\otimes\mathscr{O}(D'))_U$ via $s\mapsto (f^e\cdot s)\otimes(g^{e'}\cdot s)$.

Your map $\mu$ composes with this to give $\mathscr{O}_U\to\mathscr{O}(D+D')_U,\,s\mapsto f^eg^{e'}\cdot s$. For some choice of signs $\sigma,\sigma'$ we have that $h:=f^{\sigma}g^{\sigma'}\in\mathscr{O}(U)$ satisfies $\nu_p(h)=|n_p+m_p|$. If $\tilde{e}$ is the sign of $n_p+m_p$, I claim $\sigma,\sigma'$ may be chosen such that: $\tilde{e}\sigma=e,\tilde{e}\sigma'=e'$. Indeed, $\nu_p(h)=\sigma|n_p|+\sigma'|m_p|$ and $n_p+m_p=\tilde{e}\nu_p(h)=(\tilde{e}\sigma)|n_p|+(\tilde{e}\sigma')|m_p|$ so we can clearly see we may define $\sigma:=e/\tilde{e},\sigma'=e'/\tilde{e}$ and everything works.

Hence, the composite map is exactly the isomorphism $\mathscr{O}_U\cong\mathscr{O}(D+D')_U$ associated to the function $h\in\mathscr{O}_p$, since $h^{\tilde{e}}=f^e\cdot g^{e'}$. Note the lemma is still applicable since $\operatorname{div}(h)=\pm\operatorname{div}(f)\pm\operatorname{div}(g)$ has support - in $U$ - only at $\{p\}$, if at all, since that is true of both $f$ and $g$. Since $\mu$ composed with an isomorphism is again an isomorphism, $\mu$ itself must be an isomorphism!

$$\mathscr{O}_U\cong\mathscr{O}(D)_U\otimes\mathscr{O}(D')_U\overset{\mu}{\longrightarrow}\mathscr{O}(D+D')_U\text{ is an isomorphism}\\\therefore\,\mu\text{ is an isomorphism over $U$}$$

Now:

If $s$ is nonvanishing and regular over $U$ then $s\in\mathscr{L}(U)$ restricts to a generator of $\mathscr{L}_p$; if we choose an isomorphism $\mathscr{L}_p\cong\mathscr{O}_p$, $s$ would be identified with some $g\notin\mathfrak{m}_p$ i.e. a unit, and: $\mathscr{O}_p\cdot g=\mathscr{O}_p\iff\mathscr{O}_p\cdot s=\mathscr{L}_p$. Then the embedding $\mathscr{O}_p\cong\mathscr{L}_p\hookrightarrow\mathscr{L}_\eta$ identifies $1=g^{-1}\cdot g$ in the left hand side with $g^{-1}\cdot s$, so if we made any choice of $f$ making the image of the embedding $\mathscr{O}_p\cdot fs$, it would have to be the case that $hf=g^{-1}$ for some $h$ and, in particular, that $f$ is also a unit of $\mathscr{O}_p$ - hence, with valuation $0$: $n_p=0$.

As we consider $\mathscr{O}(D)$ a subsheaf of the sheaf of rational functions $\mathscr{K}$ (which is just $\underline{K}$ in your case), when we localise at the generic point we have $\mathscr{O}(D)_\eta\le K=\mathscr{O}_\eta$ is a necessarily $0$ or $1$-dimensional subspace; clearly, it is $1$. The canonical inclusion $\mathscr{O}(D)\hookrightarrow\mathscr{K}$ therefore reduces to an equality $\mathscr{O}(D)_\eta=K=\mathscr{O}_\eta$, explaining $(2)$. Explicitly: let $V:=C\setminus\operatorname{Supp}(D)$, which is the nonempty open complement of a finite set of points. Any nonzero $f\in K$ may be represented by a regular function $f'$ over some nonempty open $U\subseteq V$, and $f\mapsto f'\mapsto (f')_\eta$ in $\mathscr{O}(D)(U)$ and then $\mathscr{O}(D)_\eta$ gives a well-defined function $K=\mathscr{O}_\eta\to\mathscr{O}(D)_\eta$, and this map is the isomorphism $(2)$.

If $\varphi:\mathscr{O}(D)\cong\mathscr{O}$ is an isomorphism, the point is then that $\varphi_\eta$ composed with $(2)$ gives an automorphism $f\mapsto\varphi(f')_\eta,\,K\to K$ of $K$-vector spaces so must be multiplication by some nonzero scalar $\lambda$. That is, if $f\in\mathscr{O}(U)=\mathscr{O}(D)(U)$ is regular, with $U\subseteq V$ a nonempty open subset, we have that $\lambda\cdot f$ and $g=\varphi(f)\in\mathscr{O}(U)$ have the same stalk at the generic point which, as your scheme is integral, forces them to be equal. Thus, $\varphi_V:\mathscr{O}(D)_V=\mathscr{O}_V\to\mathscr{O}_V$ is the multiplication by $\lambda$ isomorphism.

Let $p\in C$ be an arbitrary closed point, $n_p$ the associated coefficient in $D$ and let $f$ be a rational function over some neighbourhood $W$ of $p$ with $\nu_p(f)=-n_p$ and $W\cap\operatorname{Supp}(D)\subseteq\{p\}$, and such that $\nu_q(f)=0$ at every other point $q\in W$; $\mathscr{O}(D)_W=\mathscr{O}_W\cdot f$ (your first lemma) and therefore $\mathscr{O}_W=\mathscr{O}_W\cdot\varphi(f)$ holds i.e. $h=\varphi(f)\in\mathscr{O}_W^\ast$. On $W\setminus\{p\}=W\cap V$, $h=\lambda\cdot f$, so by integrality $h=\lambda\cdot f$ on all of $W$; $\lambda\cdot f$ is a nowhere vanishing regular function over $W$. We find for $q\in W\setminus\{p\}$ that $\nu_q(\lambda)=0$, as $\nu_q(f)=0$ too, and we find $\nu_p(\lambda)+\nu_p(f)=\nu_p(h)=0\implies\nu_p(\lambda)=n_p$. Hence, on the open neighbourhood $W$ of $p$, we see $\operatorname{div}(\lambda)_W=D_W$.

We conclude $\lambda\in K^\ast=\mathscr{O}_\eta^\ast$ has $\operatorname{div}(\lambda)=D$.

Answer 2

The map $\text{Div}(C) \to \text{Pic}(C)$ sends $D$ to the line bundle $\mathscr{O}_C(D)$. The lemma proves $\mathscr{O}_C(D)$ is locally free of rank $1$. This makes the map well-defined.

For surjectivity, take any line bundle $\mathscr{L}$. Choose a nonzero rational section $s \in \mathscr{L}_\eta$. For each closed point $p$, write $\mathscr{L}_p = \mathscr{O}_{C,p} \cdot (f_p s)$ with $f_p \in k(C)^*$. Set $n_p = -v_p(f_p)$.

The claim $n_p = 0$ for almost all $p$ holds because we can shrink to an open $U$ where $s$ is regular and nonvanishing at every closed point. Such a $U$ exists since the vanishing set of $s$ is finite on a curve. For $p \in U$, $s$ generates $\mathscr{L}_p$, so $f_p = 1$ and $n_p = 0$.

Thus $D = \sum n_p [p]$ is a divisor and $\mathscr{O}_C(D) \cong \mathscr{L}$ via $1 \mapsto s$. This proves surjectivity.

Assume $\mathscr{O}_C(D) \cong \mathscr{O}_C$. At the generic point $\eta$, $\mathscr{O}_C(D)_\eta = k(C) = \mathscr{O}_{C,\eta}$. The isomorphism on stalks at $\eta$ is multiplication by some $f \in k(C)^*$.

For a closed point $p$, $\mathscr{O}_C(D)_p = \{ g \in k(C) \mid v_p(g) \geq -n_p \}$ where $D = \sum n_p [p]$. The isomorphism forces $v_p(f) = n_p$. Hence $D = \text{div}(f)$.

The canonical isomorphism $\mathscr{O}_{C,\eta} \to \mathscr{O}_C(D)_\eta$ is the identity map. This shows $\ker(\text{Div}(C) \to \text{Pic}(C))$ equals the image of $k(C)^* \to \text{Div}(C)$. The sequence is exact.