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). $\square$

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