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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.

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    If you pass to $\Bbb A^n$, you can see that the two polynomials have zero sets which are hyperplanes with different normal lines. – 2017-02-03
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    I don't have much background on this. If you kindly elaborate it will help me a lot.@Arthur – 2017-02-03
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    Sorry, I meant $\Bbb A^{n+1}$. Anyways, $\Bbb P^n$ is a quotient space of $\Bbb A^{n+1}$, and in $\Bbb A^{n+1}$, the zero sets of $a_0X_0+\cdots+a_nX_n$ and $b_0X_0+\cdots+b_nX_n$ are hyperplanes that go through the origin. These hyperplanes have normal vectors $(a_0, \ldots,a_n)$ and $(b_0, \ldots,b_n)$ respectively. Note that the two given points being distinct in $\Bbb P^n$ means that the two normals are non-parallel. Therefore the two hyperplanes do not coincide (although they necessarily intersect). They still don't coincide when passing back into the quotient $\Bbb P^n$. – 2017-02-03
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    I changed $\mathbb{P^n}$ and $\mathbb{F_q}$ to $\mathbb{P}^n$ and $\mathbb{F}_q$ respectively. The former are orthographically incorrect. – 2017-02-03
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    Can it be shown a particular choice of point which is in one zero set but not in the other ? – 2017-02-03

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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.