I am reading the recent book by Kawamata, Algebraic Varieties: Minimal Models and Finite Generation. There is an English translation here .

In the bottom of page 16 he says that an $\mathbb{R}$-Cartier divisor can be written as $D = \sum d_i D_i$ with $D_i$ Cartier and $d_i \in \mathbb{R}$. Then for $f: X \rightarrow Y$ between normal varieties we can define $f^* D $ as $\sum d_i f^*D_i$. He says that this doesn't depends on the expression of $D$ (There are many ways to decompose $D$ as linear combination of Cartier divisors). How can I see that?

A similar fact(?) mentioned in this book is that an $\mathbb{R}$-Cartier divisor that is a $\mathbb{Q}$-Weil divisor is a $\mathbb{Q}$-Cartier divisor. I don't think this is obvious also, since we may write an $\mathbb{R}$-Cartier divisor as $\sum d_i \sum k_jD_{ij}$ with $d_i \in \mathbb{R}$ but after summing up the coefficient it becomes a $\mathbb{Q}$ divisor.

  • 1
    $\begingroup$ Maybe this is false. Could you please give an example? $\endgroup$
    – XT Chen
    Apr 21, 2021 at 14:46

1 Answer 1

  1. If you have morphism of free abelian groups $g : A \to B$ then $g \otimes Id_{\mathbb R} : A \otimes \mathbb R \to B \otimes \mathbb R$ which sends $a \otimes r \mapsto g(a) \otimes r$ is well-defined (in the sense that it does not depend on how you present the element $\sum a_i \otimes r_i \in A \otimes \mathbb Q$ in coordinates), this is just about the construction of tensor products, just check that $g \otimes Id_{\mathbb R}$ respects relations which are part of tensor product construction. To apply to your case you have to take $A = CartierDiv(Y), B=CartierDiv(X), g = f^*$.

  2. You have to check that if $A \subset B$ is embedding of free abelian groups then $(A \otimes \mathbb R) \cap (B \otimes \mathbb Q) = A \otimes \mathbb Q$ (as $\mathbb Q$-vector spaces inside $B \otimes \mathbb R$). We have obvious embedding $A \otimes \mathbb Q \subset (A \otimes \mathbb R) \cap (B \otimes \mathbb Q)$ of $\mathbb Q$-vector spaces, tensoring this embedding by $\mathbb R$ we would obtain $A \otimes \mathbb R = (A \otimes \mathbb R) \cap (B \otimes \mathbb R)$ (we use the fact that intersections commutes with tensor products in flat case), but for any two $\mathbb Q$-vector spaces $W$ and $V \subset W$, we have that $V \otimes \mathbb R = W \otimes \mathbb R$ implies $V = W$ (you can check this simply by choosing the basis and considering everything in coordinates). To apply in your case just take $A = CartierDiv(X), B = WeilDiv(X)$.


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