The Stacks project

Lemma 27.16.4. In Situation 27.15.1. Let $T$ be a scheme. Let $(d, f, \mathcal{L}, \psi )$, $(d', f', \mathcal{L}', \psi ')$ be two quadruples over $T$. The following are equivalent:

  1. Let $m = \text{lcm}(d, d')$. Write $m = ad = a'd'$. We have $f = f'$ and there exists an isomorphism $\beta : \mathcal{L}^{\otimes a} \to (\mathcal{L}')^{\otimes a'}$ with the property that $\beta \circ \psi |_{f^*\mathcal{A}^{(m)}}$ and $\psi '|_{f^*\mathcal{A}^{(m)}}$ agree as graded ring maps $f^*\mathcal{A}^{(m)} \to \bigoplus _{n \geq 0} (\mathcal{L}')^{\otimes mn}$.

  2. The quadruples $(d, f, \mathcal{L}, \psi )$ and $(d', f', \mathcal{L}', \psi ')$ are equivalent.

  3. We have $f = f'$ and for some positive integer $m = ad = a'd'$ there exists an isomorphism $\beta : \mathcal{L}^{\otimes a} \to (\mathcal{L}')^{\otimes a'}$ with the property that $\beta \circ \psi |_{f^*\mathcal{A}^{(m)}}$ and $\psi '|_{f^*\mathcal{A}^{(m)}}$ agree as graded ring maps $f^*\mathcal{A}^{(m)} \to \bigoplus _{n \geq 0} (\mathcal{L}')^{\otimes mn}$.

Proof. Clearly (1) implies (2) and (2) implies (3) by restricting to more divisible degrees and powers of invertible sheaves. Assume (3) for some integer $m = ad = a'd'$. Let $m_0 = \text{lcm}(d, d')$ and write it as $m_0 = a_0d = a'_0d'$. We are given an isomorphism $\beta : \mathcal{L}^{\otimes a} \to (\mathcal{L}')^{\otimes a'}$ with the property described in (3). We want to find an isomorphism $\beta _0 : \mathcal{L}^{\otimes a_0} \to (\mathcal{L}')^{\otimes a'_0}$ having that property as well. Since by assumption the maps $\psi : f^*\mathcal{A}_ d \to \mathcal{L}$ and $\psi ' : (f')^*\mathcal{A}_{d'} \to \mathcal{L}'$ are surjective the same is true for the maps $\psi : f^*\mathcal{A}_{m_0} \to \mathcal{L}^{\otimes a_0}$ and $\psi ' : (f')^*\mathcal{A}_{m_0} \to (\mathcal{L}')^{\otimes a_0}$. Hence if $\beta _0$ exists it is uniquely determined by the condition that $\beta _0 \circ \psi = \psi '$. This means that we may work locally on $T$. Hence we may assume that $f = f' : T \to S$ maps into an affine open, in other words we may assume that $S$ is affine. In this case the result follows from the corresponding result for triples (see Lemma 27.12.4) and the fact that triples and quadruples correspond in the affine base case (see proof of Lemma 27.16.2). $\square$

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 01NW. Beware of the difference between the letter 'O' and the digit '0'.