# Notes on divisors-2

###### Weil divisors v.s. Cartier divisors

Let’s continue the discussion of the last post. In this post, we shall establish a correspondence between Weil divisors and Cartier divisors for special schemes.

First let’s repeat what this correspondence is:

Suppose that $X$ is an integral, separated noetherian scheme such that all of its local rings are UFDs. Then the group $Div(X)$ of Weil divisors on $X$ is isomorphic to the group $\Gamma(X,\mathfrak{K}^*/\mathfrak{O}_X^*)$ of Cartier divisors. Besides, the principal Weil divisors correspond to the principal Cartier divisors.

This correspondence and the following proof are taken from Hartshorne’s book, ‘algebraic geometry’.

We shall not give a complete and rigorous proof to this statement. We only give a rough idea of what this correspondence is like. First note that $X$ is integral(reduced and irreducible), thus the sheaf $\mathfrak{K}$ is the constant sheaf of $K$ on $X$ where $K$ is the function field on $X$, which is also the local ring $\mathfrak{O}_{X,x_0}$ with $x_0$ the generic point of $X$. So, a Cartier divisor is given by a family $\{U_i,f_i\}$ where $U_i$ is an open cover of $X$ and $f_i\in \Gamma(U_i,\mathfrak{O}_X)=K^*$. We now construct a Weil divisor: for each prime divisor $Y$ of $X$, by definition, this means that $Y$ is an integral closed subscheme of $X$, now for each $f_i$, we set the coefficient of $Y$ to be $v_Y(f_i)$. Note that for any two $U_i,U_j$ such that $U_i\bigcap U_j$, then by definition, $f_i/f_j\in \Gamma(U_i\bigcap U_j,\mathfrak{O}_X)$, that is to say, $f_i/f_j$ is invertible on $U_i\bigcap U_j$, thus $v_Y(f_i)=v_Y(f_j)$, so the coefficient of $Y$ is well-defined. And we set the Weil divisor to be $D=\sum_Yv_Y(f_i)$. Note here we should guarantee that this is a finite sum. This can be deduced easily from the assumption that $X$ is noetherian(we will say about this later).

Now conversely, if $D$ is a Weil divisor on $X$, then for each point $x\in X$, $D$ gives a divisor on $Spec(\mathfrak{O}_{X,x})$. Now by assumption, $\mathfrak{O}_{X,x}$ is a UFD, we can show that $D_x$ is a principal divisor(we shall not prove this point for the present), that is, $D_x=(f_x)$ for some $f_x\in K^*$. This means that there is an open subset $x\in U$ of $X$ such that $f_x$ and $D$ agree on $U$. In this way, varying $x$, we get an open cover $U_i$ of $X$ with $f_i$ such that $f_i$ and $D$ agree on $U_i$. And we set $\{U_i,f_i\}$ to be the corresponding Cartier divisor. So, we finish the construction of the correspondence between Weil divisors and Cartier divisors.

It is not too hard to see that these two construction are inverse one to another.

###### Relation to invertible sheaves

Next we want to relate divisors to invertible sheaves.

Suppose that $X$ is a scheme and $\mathfrak{O}_X$ its structure sheaf. Then we define an invertible sheaf to be an $\mathfrak{O}_X$-module locally free of rank $1$.

Now let $D$ be a Cartier divisor on $X$ given by $\{U_i,f_i\}$, then we define the sheaf $\mathfrak{L}(D)$ to be the submodule of $\mathfrak{K}$ generated by $1/f_i$ on each $U_i$ for all $i$. We call $\mathfrak{L}(D)$ to be the sheaf associated to $D$.

The important result is that, this gives a one-to-one correspondence between the set of Cartier divisors and the set of invertible submodules of $\mathfrak{K}$. This can be seen like this: given a invertible submodule $\mathfrak{L}$ of $\mathfrak{K}$, for an open subset $U$ of $X$, $\mathfrak{L}(U)$ is a free module of $\mathfrak{O}_X(U)$ of rank $1$, which is also a submodule of $\mathfrak{K}(U)$. So, we take a generator $f_U$ of $\mathfrak{L}(U)$(it is not zero since $\mathfrak{L}(U)$ is of rank $1$, not $0$), then we get an open covering $U_i$ of $X$ such that on each $U_i$ there is an $f_i$. We can verify that these $f_i$ can glue together, and thus we can set $\{u_i,f_i\}$ to be the Cartier divisor. It can be verified that this process is converse to the above one, thus we showed that there is a correspondence between the set of Cartier divisors and the set of invertible submodules of $\mathfrak{K}$.

In fact, this is more than a bijection, this is a group homomorphism. Recall that we give a group structure to the set of invertible sheaves on $X$ by tensor product: suppose that $\mathfrak{L},\mathfrak{L}'$ are two invertible sheaves on $X$, then by definition, $\mathfrak{L}\bigotimes_{\mathfrak{O}_X}\mathfrak{L}'$ is again an invertible sheaf on $X$. Besides, $\mathfrak{L}^*=Hom_{\mathfrak{O}_X}(\mathfrak{L},\mathfrak{O}_X)$ is also an invertible sheaf on $X$, and $\mathfrak{L}\bigotimes \mathfrak{L}^*\simeq\mathfrak{O}_X$. We call this group the Picard group of $X$,

Charles Emile Picard(photo taken from here)

and denote it by $Pic(X)$. So the set of invertible submodules of $\mathfrak{K}$ is a subgroup of $Pic(X)$.

It is not hard to see that $\mathfrak{L}(D/D')=\mathfrak{L}(D)\bigotimes\mathfrak{L}(D')^*$, so this gives a group homomorphism between the group of Cartier divisors and the group of invertible submodules of $\mathfrak{K}$.

In general, the group of Cartier divisors is not equal to the Picard group of $X$. Yet, under some mild condition, we can show that, these two groups are isomorphic under the above morphism:

Suppose that $X$ is an integral scheme, then the group of Cartier divisors on $X$ is isomorphic to $Pic(X)$.

We will establish this isomorphism in the next post.