Consider the projective space of dimension $n$ over $\mathbb{F}_q$, i.e., $\mathbb{P}^n({\mathbb{F}_q)}$. Take two distinct points $(a_0:\cdots:a_n)$ and $(b_0:\cdots:b_n)$ in $\mathbb{P}^n$. Now if $f=a_0X_0+\cdots+a_nX_n$ and $g=b_0X_0+\cdots+b_nX_n$ are two homogeneous linear polynomials then how can I show that the zero sets $Z(f)$ and $Z(g)$ are different in $\mathbb{P}^n$ ?
Help me. Thanks.
I will write $a=(a_0,\ldots,a_n)^T\in \Bbb F_q^{n+1}$, a column vector, and similarly for $b$.
Now, for a vector $v\in\Bbb F_q^{n+1}$ you have $f(v)=a^Tv$ and $g(v)=b^Tv$, where $a^T\in\Bbb F_q^{1\times(n+1)}$ is a matrix with just one row. We are looking for a vector $v$ which satisfies $a^Tv=0$ and $b^Tv\ne 0$ or the other way around. Assume for contradiction that $\ker(a^T)=\ker(b^T)$. Since $a$ is not the zero vector, we can operate on the "columns" (just single entries) of $a^T$ by an invertible matrix $g\in\operatorname{GL}_{n+1}(\Bbb F_q)$ so that $a^Tg = (1,0,\ldots,0)$. Now, $$ \ker(b^T g)=\ker(b^T)g^{-1}=\ker(a^T)g^{-1}=\ker(a^Tg)=\{ (0,\ast,\ldots,\ast) \}. $$ It follows that $b^T g = (\lambda,0,\ldots,0)$ for some $\lambda\in\Bbb F_q^\times$. Hence, $b^Tg=\lambda\cdot a^Tg$ and therefore $b^T = \lambda a^T,$ but this is just another way of saying that $[b]=[a]$, which is a contradiction to your assumption that the two projective points be distinct.