I am studying the proof of Lemma 7.2. on page 108 in Dolgachev's "Lectures on invariant theory". It states (everything is done over the field $k = \overline{k}$):

Let $X$ be a normal affine variety (for example, nonsingluar) and $G$ is a connected affine algebraic group. Let $L$ be a line bundle on $G \times X$. Then there exist line bundles $L_1 \in Pic(G)$, $L_2 \in Pic(X)$ such that $L \cong pr_1^*(L_1) \otimes pr_2^*(L_2)$ (where $pr_i$ is the $i$-th projection of $G \times X$).

The proof uses the following facts about algebraic groups:

  • $G$ containes a Zariski open set $U$, which is isomorphic to $(\mathbb{A}^1 \setminus \{0\})^N$.
  • $pr_2^* \colon Pic(X) \to Pic((\mathbb{A}^1 \setminus \{0\}) \times X)$ is an isomorphism.

Then it is clear that $L$ restricted to $U \times X$ is isomorphic to $pr_2^*(L_2)$ for some $L_2 \in Pic(X)$.

The following is said afterwards: Let $D$ be a Cartier divisor on $G \times X$ representing $L$.

Question 1: Why does such a $D$ exist? $X$ is not projective, so I don't quite see why $D$ should exist.

The proof continues as follows:

Then the preceding isomorphism implies that there exists a Cartier Divisor $D_2$ on $X$ such that $D' = D-pr_2^*(D_2)|_{U \times X} = 0$.

Question 2: Why exactly does $D_2$ exist?

Now, take any irreducibly component $D_i'$ of $D'$ and consider the projection $D_i' \to G$. The image is contained in $G \setminus U =: Z$. By the theorem on the dimension of fibres, the fibres of this projection have dimension equal to $\dim X$. Then the last part which I don't understand follows:

This easily implies that $D_i' = pr_1^*(D_i)$, where $D_i \subset Z$.

Question 3: Why?

Thanks in advance for any help.


Question 1: Maybe there's some subtlety I'm missing here, but isn't there a Cartier divisor attached to any line bundle? You just pick an affine where it's trivial, and let the trivialization be your rational section (using the definition here).

Question 2: Since $pr_2^*$ is an isomorphism to $\operatorname{Pic}(U\times X)$, it's surjective.

Question 3: Because $Z\times X$ is already a divisor. Thus, the only divisor that could be supported on it is a sum of the components of $Z\times X$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy