# Hyperplane is the zero set of a single variable by a linear change of coordinates?

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.

• 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 $\mathbb{P}^n$. – 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.
$$\hspace{3.75cm}$$
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 $$V(x_0)$$.