Lemma 33.35.2. Let $k$ be a field. Let $n \geq 1$. Let $i : H \to \mathbf{P}^ n_ k$ be a hyperplane. Then there exists an isomorphism

such that $i^*\mathcal{O}(1)$ pulls back to $\mathcal{O}(1)$.

Lemma 33.35.2. Let $k$ be a field. Let $n \geq 1$. Let $i : H \to \mathbf{P}^ n_ k$ be a hyperplane. Then there exists an isomorphism

\[ \varphi : \mathbf{P}^{n - 1}_ k \longrightarrow H \]

such that $i^*\mathcal{O}(1)$ pulls back to $\mathcal{O}(1)$.

**Proof.**
We have $\mathbf{P}^ n_ k = \text{Proj}(k[T_0, \ldots , T_ n])$. The section $s$ corresponds to a homogeneous form in $T_0, \ldots , T_ n$ of degree $1$, see Cohomology of Schemes, Section 30.8. Say $s = \sum a_ i T_ i$. Constructions, Lemma 27.13.7 gives that $H = \text{Proj}(k[T_0, \ldots , T_ n]/I)$ for the graded ideal $I$ defined by setting $I_ d$ equal to the kernel of the map $\Gamma (\mathbf{P}^ n_ k, \mathcal{O}(d)) \to \Gamma (H, i^*\mathcal{O}(d))$. By our construction of $Z(s)$ in Divisors, Definition 31.14.8 we see that on $D_{+}(T_ j)$ the ideal of $H$ is generated by $\sum a_ i T_ i/T_ j$ in the polynomial ring $k[T_0/T_ j, \ldots , T_ n/T_ j]$. Thus it is clear that $I$ is the ideal generated by $\sum a_ i T_ i$. Note that

\[ k[T_0, \ldots , T_ n]/I = k[T_0, \ldots , T_ n]/(\sum a_ i T_ i) \cong k[S_0, \ldots , S_{n - 1}] \]

as graded rings. For example, if $a_ n \not= 0$, then mapping $S_ i$ equal to the class of $T_ i$ works. We obtain the desired isomorphism by functoriality of $\text{Proj}$. Equality of twists of structure sheaves follows for example from Constructions, Lemma 27.11.5. $\square$

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