Pullback of $\mathbb{R}$-Cartier divisors

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.

• Maybe this is false. Could you please give an example? Apr 21 at 14:46

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 $$A = CartierDiv(X), B = WeilDiv(X)$$.