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. 
 A: 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 $V(x_0)$.
