Can I cancel the vector bundle $E$ in an equation $E \otimes L_1 \cong E \otimes L_2$ to conclude $L_1 \cong L_2$?

Question

Suppose $E$ is a vector bundle and $L_1, L_2$ are line bundles on a scheme $X$, such that $$E \otimes L_1 \cong E \otimes L_2.$$ Can I conclude $L_1 \cong L_2$?

I know that taking determinants gives me $$\det E \otimes L_1^{\otimes r} \cong \det E \otimes L_2^{\otimes r},$$ where $r$ is the rank of $E$. Since I can cancel line bundles, I certainly get $L_1^{\otimes r} \cong L_2^{\otimes r}$.

Also if $E$ admits a line bundle quotient $E \to L \to 0$, I think I can conclude $L \otimes L_1 \cong L \otimes L_2$, which yields $L_1 \cong L_2$. That is not true, see Sasha's answer for a counter example.

Answer 1

No. Assume, for instance, that $L$ is a line bundle of order 2, i.e., $$ L^2 \cong \mathcal{O}, \qquad L \not\cong \mathcal{O}. $$ Let $E = \mathcal{O} \oplus L$. Then $$ E \otimes \mathcal{O} \cong E \cong E \otimes L, $$ but $L \not\cong \mathcal{O}$.

Answer 2

No, if $ X $ is an elliptic curve then it has exactly four $ 2 $-torsion points, $ O, P, Q, P+_EQ $. Let $ E = \mathcal{O}_X(P+Q-2O) \oplus \mathcal{O}_X $ with $ L_1 = \mathcal{O}_X(P) $ and $ L_2 = \mathcal{O}_X(Q) $. Then the two line bundles are not isomorphic but $ E \otimes L_1 $ and $ E \otimes L_2 $ are both $ \mathcal{O}_X(P) \oplus \mathcal{O}_X(Q) $ because the degree zero part of the Picard group of $ X $ is just the elliptic curve $ X $.

Edit: I was late in posting this but I’m going to leave it as a concrete example of the above answer.