[proofplan] We construct the incidence structure from the subspaces of the three-dimensional [vector space](/page/Vector%20Space) $V=\mathbb{F}_q^3$. The projective-plane axioms follow from elementary linear algebra: two distinct one-dimensional subspaces span a unique two-dimensional subspace, and two distinct two-dimensional subspaces in a three-dimensional space meet in a one-dimensional subspace. The order is computed by counting one-dimensional subspaces inside a two-dimensional vector space, and the nondegeneracy axiom is verified using the four points represented by $e_1,e_2,e_3,e_1+e_2+e_3$. [/proofplan]

[step:Define the incidence structure from subspaces of $\mathbb{F}_q^3$] Let $V := \mathbb{F}_q^3$, regarded as a three-dimensional vector space over $\mathbb{F}_q$. Define the point set \begin{align*} \mathcal{P} := \{P \subset V : P \text{ is a one-dimensional } \mathbb{F}_q\text{-linear subspace}\}, \end{align*} and define the line set \begin{align*} \mathcal{L} := \{\ell \subset V : \ell \text{ is a two-dimensional } \mathbb{F}_q\text{-linear subspace}\}. \end{align*} Define the incidence relation $I \subset \mathcal{P} \times \mathcal{L}$ by \begin{align*} (P,\ell) \in I \iff P \subset \ell. \end{align*} This is the incidence structure denoted $PG(2,q)$. [/step]

[step:Construct the unique projective line through two distinct projective points] Let $P,Q \in \mathcal{P}$ be distinct points. Choose nonzero vectors $p \in P$ and $q \in Q$. Since $P \ne Q$, the vectors $p$ and $q$ are linearly independent over $\mathbb{F}_q$; otherwise $q=\lambda p$ for some $\lambda \in \mathbb{F}_q^\times$, which would imply $Q=P$. Define \begin{align*} \ell := \operatorname{span}_{\mathbb{F}_q}\{p,q\} \subset V. \end{align*} Then $\ell$ is a two-dimensional $\mathbb{F}_q$-linear subspace of $V$, so $\ell \in \mathcal{L}$, and $P \subset \ell$, $Q \subset \ell$. If $m \in \mathcal{L}$ is another line with $P \subset m$ and $Q \subset m$, then $p,q \in m$. Since $m$ is a linear subspace, $\operatorname{span}_{\mathbb{F}_q}\{p,q\} \subset m$. Both $\ell$ and $m$ are two-dimensional subspaces, hence $\ell=m$. Therefore exactly one line is incident with both $P$ and $Q$. [/step]

[step:Find the unique projective point where two distinct projective lines meet] Let $\ell,m \in \mathcal{L}$ be distinct lines. Since both are two-dimensional subspaces of the three-dimensional vector space $V$, their intersection cannot be zero-dimensional. Indeed, choose a basis $(u_1,u_2)$ of $\ell$. If $\ell \cap m=\{0\}$, then the quotient map \begin{align*} \pi: V &\to V/m \end{align*} would restrict to an injective [linear map](/page/Linear%20Map) $\pi|_\ell:\ell \to V/m$. But $\dim_{\mathbb{F}_q}\ell=2$ and $\dim_{\mathbb{F}_q}(V/m)=1$, so no injective linear map from a two-dimensional vector space to a one-dimensional vector space exists. Thus $\ell \cap m$ contains a nonzero vector. Since $\ell \ne m$, the intersection $\ell \cap m$ is not two-dimensional; otherwise $\ell \cap m=\ell=m$. Hence $\ell \cap m$ is one-dimensional. Define \begin{align*} P := \ell \cap m. \end{align*} Then $P \in \mathcal{P}$, and $P \subset \ell$, $P \subset m$. If $Q \in \mathcal{P}$ is incident with both $\ell$ and $m$, then $Q \subset \ell \cap m=P$. Since $Q$ and $P$ are both one-dimensional subspaces, $Q=P$. Therefore exactly one point is incident with both $\ell$ and $m$. [/step]

[step:Count the points on each line and the lines through each point]Let $\ell \in \mathcal{L}$. Since $\ell$ is a two-dimensional vector space over $\mathbb{F}_q$, it has $q^2$ vectors, of which $q^2-1$ are nonzero. Each one-dimensional subspace of $\ell$ has exactly $q-1$ nonzero vectors, and two distinct one-dimensional subspaces have no nonzero vector in common. Therefore the number of points incident with $\ell$ is \begin{align*} \frac{q^2-1}{q-1}=q+1. \end{align*} Now let $P \in \mathcal{P}$. The lines through $P$ are precisely the two-dimensional subspaces $\ell \subset V$ with $P \subset \ell$. Such subspaces correspond bijectively to one-dimensional subspaces of the quotient vector space $V/P$. Since $\dim_{\mathbb{F}_q}(V/P)=2$, the same count gives exactly \begin{align*} \frac{q^2-1}{q-1}=q+1 \end{align*} lines through $P$.[/step]

[guided]We first count the points on a fixed projective line. Let $\ell \in \mathcal{L}$. By definition, $\ell$ is a two-dimensional vector space over the finite field $\mathbb{F}_q$, so it contains $q^2$ vectors. Exactly one of these vectors is the zero vector, hence $\ell$ contains $q^2-1$ nonzero vectors. A projective point incident with $\ell$ is a one-dimensional subspace $P \subset \ell$. Every such $P$ has the form \begin{align*} P=\{\lambda v:\lambda \in \mathbb{F}_q\} \end{align*} for some nonzero vector $v \in \ell$, and therefore $P$ has exactly $q-1$ nonzero vectors. Distinct one-dimensional subspaces share no nonzero vector: if $0 \ne w \in P \cap Q$, then both $P$ and $Q$ are the one-dimensional span of $w$, so $P=Q$. Thus the nonzero vectors of $\ell$ are partitioned into the nonzero vectors lying on each projective point contained in $\ell$. Dividing the number of nonzero vectors in $\ell$ by the number of nonzero vectors in each one-dimensional subspace gives \begin{align*} \#\{P \in \mathcal{P}:P \subset \ell\} = \frac{q^2-1}{q-1} = q+1. \end{align*} We next count the lines through a fixed projective point. Let $P \in \mathcal{P}$. The quotient vector space $V/P$ has dimension $2$ over $\mathbb{F}_q$. A line $\ell \in \mathcal{L}$ with $P \subset \ell$ determines the one-dimensional subspace $\ell/P \subset V/P$. Conversely, if $R \subset V/P$ is a one-dimensional subspace, then its inverse image under the quotient map \begin{align*} \pi: V &\to V/P \end{align*} is a two-dimensional subspace $\pi^{-1}(R) \subset V$ containing $P$. These two assignments are inverse to each other, so lines through $P$ are in bijection with one-dimensional subspaces of the two-dimensional vector space $V/P$. Applying the same nonzero-vector count in $V/P$ gives \begin{align*} \#\{\ell \in \mathcal{L}:P \subset \ell\} = \frac{q^2-1}{q-1} = q+1. \end{align*}[/guided]

[step:Exhibit four projective points with no three collinear] Let $e_1,e_2,e_3 \in V$ denote the standard basis vectors, and define \begin{align*} P_1 &:= \operatorname{span}_{\mathbb{F}_q}\{e_1\},& P_2 &:= \operatorname{span}_{\mathbb{F}_q}\{e_2\},& P_3 &:= \operatorname{span}_{\mathbb{F}_q}\{e_3\},& P_4 &:= \operatorname{span}_{\mathbb{F}_q}\{e_1+e_2+e_3\}. \end{align*} These four points are distinct because none of the four displayed nonzero vectors is a nonzero scalar multiple of another. We prove that no three of these points are collinear. Three projective points lie on a common projective line exactly when their representative nonzero vectors span a subspace of dimension at most $2$. Thus it suffices to check that any three of \begin{align*} e_1,\quad e_2,\quad e_3,\quad e_1+e_2+e_3 \end{align*} are linearly independent over $\mathbb{F}_q$. The triples $(e_1,e_2,e_3)$, $(e_1,e_2,e_1+e_2+e_3)$, $(e_1,e_3,e_1+e_2+e_3)$, and $(e_2,e_3,e_1+e_2+e_3)$ are linearly independent because, in each case, a relation \begin{align*} a u+b v+c w=0 \end{align*} forces the coefficient of at least one coordinate appearing only in $w$ among the chosen triple to give $c=0$, after which the independence of the relevant standard basis vectors gives $a=b=0$. Hence no three of $P_1,P_2,P_3,P_4$ are collinear. [/step]

[step:Conclude that the incidence structure is a projective plane of order $q$] The preceding steps verify the projective-plane axioms: any two distinct points are incident with a unique line, any two distinct lines are incident with a unique point, and there exist four points with no three collinear. The counting step shows that every line is incident with exactly $q+1$ points and every point is incident with exactly $q+1$ lines. Therefore the incidence structure $PG(2,q)$ is a projective plane of order $q$. [/step]