My professor said that a hyperplane in projective space can be assumed to be $V(x_0)$ by a linear change in coordinates. Im not quite sure what he means. A hyperplane in Hartshorne is defined to be the zero set of a linear homogeneous polynomial. So say $H = V(x_0 + x_1).$ Why can we assume $H = V(x_0)?$ Are the two isomorphic under some isomorphism of $\mathbb{P}^n?$ The only isomorphism I can think of is sending $x_0, x_1$ to $x_0$ is the pullback ring homomorphism of the respective coordinate rings but this is not an isomorphism.

  • $\begingroup$ Perhaps you could send $x_0$ to $(1:0:\dots:0)$, $x_1$ to $(0:1:0: \dots: 0)$, etc? A little linear algebra should prove that this is a linear automorphism of $\mathbb{P}^n$. $\endgroup$ – bounceback Oct 25 '18 at 22:30

The map \begin{align*} \mathbb{P}^n &\to \mathbb{P}^n\\ [x_0 : x_1 : \cdots : x_n] &\mapsto [x_0 + x_1 : x_1 : \cdots : x_n] \end{align*} is an isomorphism of $\mathbb{P}^n$ mapping $V(x_0+x_1)$ to $V(x_0)$.

This is really just an incarnation of changing bases. More generally, given vector space $V$ of dimension $n+1$ over a field $k$, a choice of basis $e_0, \ldots, e_n$ induces an isomorphism $\mathbb{P}(V) \to \mathbb{P}^n$ such that the following diagram commutes. The homogeneous coordinates on $\mathbb{P}(V)$ are basically just the linear maps $e_0^*, \ldots, e_n^*$ in the dual basis.


So, given a hyperplane $H = V(\ell)$ in $\mathbb{P}^n$ where $\ell = a_0 x_0 + \cdots + a_n x_n$ is some linear form, we can complete $v_0 = (a_0, \ldots, a_n)$ to a basis $\{v_0, v_1, \ldots, v_n\}$ for $k^{n+1}$. Letting $e_0, \ldots, e_n$ be the standard basis for $k^{n+1}$, then the change of basis map sending $e_i \to v_i$ induces a map $\mathbb{P}^n \to \mathbb{P}^n$ sending $V(\ell)$ to $V(x_0)$.


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