For the <closed immersion> $i:Y\hookrightarrow X$, the <ideal sheaf of a closed subscheme> gives
$$
0\longrightarrow\mathcal I_{Y/X}\longrightarrow\mathcal O_X
\longrightarrow i_*\mathcal O_Y\longrightarrow0.
$$
Because the <projective hypersurface> is cut out by one <homogeneous polynomial> $f$ of degree $d$, multiplication by $f$ identifies
$$
\boxed{\mathcal I_{Y/X}\cong\mathcal O_X(-d)=\mathcal O_X(1)^{\otimes(-d)}.}
$$
This is the <ideal-sheaf sequence of a projective hypersurface>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 113 / 4 / c / i / Solution

{title2=$\mathcal I_{Z/X}$}

For a <closed immersion> $i:Z\hookrightarrow X$, the ideal sheaf is the kernel $\mathcal I_{Z/X}=\ker(\mathcal O_X\to i_*\mathcal O_Z)$. It therefore fits into the <short exact sequence of sheaves>
$$
0\longrightarrow\mathcal I_{Z/X}\longrightarrow\mathcal O_X\longrightarrow i_*\mathcal O_Z\longrightarrow0.
$$

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 Incoming links: Ideal sheaf of a closed subscheme