Exercise 10.4

Proof. Define the hyperplane $\bar{K}$ by

where $(a_0,\dots ,a_n)\ne (0,\dots ,0)$ (if $(a_0,\dots ,a_n)=(0,\dots ,0)$, then $\bar{K}=P_n(F)$ is not a hyperplane). Note that, if $(x_0,\dots ,x_n)\sim (y_0,\dots ,y_n)$, there is $\lambda \in F^{\text{∗}}$ such that $y_i=\lambda x_i,i=0,\dots ,n$, thus $a_0{x}_0+\cdots +a_n{x}_n⟺0=a_0{y}_0+\cdots +a_n{y}_n=0$, so that the condition doesn’t depend on the choice of the projective point representative.

Since $(a_0,\dots ,a_n)\ne (0,\dots ,0)$, suppose, without loss of generality, that $a_0\ne 0$. Consider

Then $\chi$ is well defined. Indeed, if $(x_0,\dots ,x_n)\sim (y_0,\dots ,y_n)$, there is some $\lambda \in F^{\text{∗}}$ such that $(y_0,\dots ,y_n)=\lambda (x_0,\dots ,x_n)$. In particular, $(y_1,\dots ,y_n)=\lambda (x_1,\dots ,x_n)$, thus $[x_1,\dots ,x_n]=[y_1,\dots ,y_n]$.

If $\chi ([x_0,\dots ,x_n])=\chi ([y_0,\dots ,y_n])$, where $[x_0,\dots ,x_n]$ and $[y_0,\dots ,y_n]$ are in $\bar{K}$, then $[x_1,\dots ,x_n]=[y_1,\dots ,y_n]$, thus there is $\lambda \in F^{\text{∗}}$ such that $(y_1,\dots ,y_n)=\lambda (x_1,\dots ,x_n)$. Since $a_0\ne 0$,

therefore $[x_0,\dots ,x_n]=[y_0,\dots ,y_n]$. So $\phi$ is injective.

At last, let $[x_1,\dots ,x_n]$ be any point of $P_{n-1}(F)$. Define $x_0=-\frac{1}{a_0}(a_1{x}_1+\cdots +a_n{x}_n)$. Then $a_0{x}_0+\cdots +a_n{x}_n=0$, so that $[x_0,\dots ,x_n]\in \bar{K}$, and $\chi ([x_0,\dots ,x_n])=[x_1,\dots ,x_n]$. This proves that $\chi$ is surjective.

To conclude, $\chi$ is a bijection, therefore $|\bar{K}|=|P_{n-1}(F)|=q^{n-1}+\cdots +q+1$. □