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

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.

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