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Let $X=\mathbb{P}_{\mathbb{P^1}}(\mathcal O(-1)+\mathcal O(-1)+\mathcal O)$, $Y=\mathbb{P}_{\mathbb{P^1}}(\mathcal O(-1)+\mathcal O+\mathcal O+\mathcal O)$.

Can we embed $X$ into $Y$ as a hypersurface? If so, which divisor class of $Y$ would correspond to $X$?

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Yes, there is an exact sequence $$ 0 \to \mathcal{O}(-1) \oplus \mathcal{O}(-1) \oplus \mathcal{O} \to \mathcal{O}(-1) \oplus \mathcal{O} \oplus \mathcal{O} \oplus \mathcal{O} \to \mathcal{O}(1) \to 0, $$ obtained as the direct sum of the standard exact sequence $$ 0 \to \mathcal{O}(-1) \to \mathcal{O} \oplus \mathcal{O} \to \mathcal{O}(1) \to 0 $$ with $\mathcal{O}(-1)$ and $\mathcal{O}$. It induces the required embedding of projective bundles, and since the quotient is $\mathcal{O}(1)$, the image is linearly equivalent to $H + h$, where $H$ is the relative hyperplane class and $h$ is the pullback of the hyperplane class on $\mathbb{P}^1$.

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