[proofplan] We verify the affine plane axioms directly from the field operations in $\mathbb{F}_q$. First, two distinct points determine exactly one line, with a separate treatment of the vertical and non-vertical cases. Next, for any point not lying on a given line, we construct the unique disjoint line through that point by preserving either the slope or the vertical coordinate. Finally, we exhibit three non-collinear points and compute that every line contains exactly $q$ points, so the order is $q$. [/proofplan] [step:Define the points, lines, and incidence relation] Let the point set be \begin{align*} P := \mathbb{F}_q^2. \end{align*} For each pair $a,b \in \mathbb{F}_q$, define the non-vertical line \begin{align*} L_{a,b} := \{(x,y) \in P : y = ax + b\}. \end{align*} For each $c \in \mathbb{F}_q$, define the vertical line \begin{align*} V_c := \{(x,y) \in P : x = c\}. \end{align*} Let \begin{align*} \mathcal{L} := \{L_{a,b} : a,b \in \mathbb{F}_q\} \cup \{V_c : c \in \mathbb{F}_q\} \end{align*} be the line set. A point $p \in P$ is incident with a line $\ell \in \mathcal{L}$ exactly when $p \in \ell$. [/step] [step:Show that two distinct points determine a unique line] Let $p_1 = (x_1,y_1) \in P$ and $p_2 = (x_2,y_2) \in P$ be distinct points. If $x_1 = x_2$, then $p_1,p_2 \in V_{x_1}$. No non-vertical line $L_{a,b}$ contains both points, because containment would imply $y_1 = ax_1 + b$ and $y_2 = ax_2 + b = ax_1 + b$, and hence $y_1 = y_2$, contradicting $p_1 \neq p_2$ when $x_1=x_2$. The only vertical line containing $p_1$ is $V_{x_1}$, so $V_{x_1}$ is the unique line through $p_1$ and $p_2$. If $x_1 \neq x_2$, then $x_2 - x_1 \neq 0$ in the field $\mathbb{F}_q$, so $(x_2-x_1)^{-1}$ exists. Define $a := (y_2-y_1)(x_2-x_1)^{-1}$ and $b := y_1 - ax_1$. Then $p_1 \in L_{a,b}$ by the definition of $b$, and \begin{align*} ax_2+b = a x_2 + y_1 - ax_1 = y_1 + a(x_2-x_1) = y_1 + (y_2-y_1)(x_2-x_1)^{-1}(x_2-x_1) = y_2, \end{align*} so $p_2 \in L_{a,b}$. No vertical line contains both points because $x_1 \neq x_2$. It remains to prove uniqueness among non-vertical lines. Suppose $L_{a',b'}$ contains both $p_1$ and $p_2$, where $a',b' \in \mathbb{F}_q$. Then $y_1 = a'x_1 + b'$ and $y_2 = a'x_2 + b'$. Subtracting the first equation from the second gives \begin{align*} y_2-y_1 = a'(x_2-x_1). \end{align*} Multiplying by $(x_2-x_1)^{-1}$ gives $a'=a$, and then $b'=y_1-a'x_1=b$. Hence $L_{a',b'}=L_{a,b}$. Therefore any two distinct points determine exactly one line. [guided] We must prove both existence and uniqueness of the line through two distinct points, and the proof splits according to whether the two points have the same first coordinate. Let $p_1=(x_1,y_1)$ and $p_2=(x_2,y_2)$ be distinct points of $P=\mathbb{F}_q^2$. First suppose $x_1=x_2$. Then both points lie on the vertical line \begin{align*} V_{x_1}=\{(x,y)\in P:x=x_1\}. \end{align*} Could there also be a non-vertical line through both points? If $L_{a,b}$ contained both points, then $y_1 = ax_1+b$ and $y_2 = ax_2+b$. Since $x_2=x_1$, the second equation becomes $y_2=ax_1+b$, so $y_1=y_2$. Together with $x_1=x_2$, this would give $p_1=p_2$, contradicting the assumption that the points are distinct. Therefore no non-vertical line contains both points. Also, a vertical line $V_c$ contains $p_1$ only if $c=x_1$, so the only vertical line through both points is $V_{x_1}$. Thus the unique line is $V_{x_1}$. Now suppose $x_1\neq x_2$. The field property is used exactly here: because $x_2-x_1\neq 0$, the element $x_2-x_1$ has a multiplicative inverse in $\mathbb{F}_q$. Define $a := (y_2-y_1)(x_2-x_1)^{-1}$ and $b := y_1-ax_1$. The choice of $b$ forces the first point to lie on $L_{a,b}$, since \begin{align*} ax_1+b = ax_1+y_1-ax_1 = y_1. \end{align*} The choice of $a$ forces the second point to lie on the same line: \begin{align*} ax_2+b = ax_2+y_1-ax_1 = y_1+a(x_2-x_1) = y_1+(y_2-y_1)(x_2-x_1)^{-1}(x_2-x_1) = y_2. \end{align*} So $L_{a,b}$ is a line through both points. No vertical line can contain both points, because a vertical line has a fixed first coordinate and $x_1\neq x_2$. For uniqueness, suppose another non-vertical line $L_{a',b'}$ contains both $p_1$ and $p_2$. Then $y_1 = a'x_1+b'$ and $y_2 = a'x_2+b'$. Subtracting gives \begin{align*} y_2-y_1=a'(x_2-x_1). \end{align*} Since $x_2-x_1$ is invertible, multiplication by $(x_2-x_1)^{-1}$ gives \begin{align*} a'=(y_2-y_1)(x_2-x_1)^{-1}=a. \end{align*} Substituting this into $y_1=a'x_1+b'$ gives \begin{align*} b'=y_1-a'x_1=y_1-ax_1=b. \end{align*} Thus $L_{a',b'}=L_{a,b}$. Hence in all cases there is exactly one line through two distinct points. [/guided] [/step] [step:Construct the unique parallel line through an external point] Let $\ell \in \mathcal{L}$ be a line and let $p_0=(x_0,y_0)\in P$ be a point with $p_0\notin \ell$. Here two lines are called parallel if they are disjoint. First suppose $\ell=L_{a,b}$ for some $a,b\in\mathbb{F}_q$. Since $p_0\notin L_{a,b}$, we have $y_0\neq ax_0+b$. Define \begin{align*} b_0:=y_0-ax_0. \end{align*} Then $p_0\in L_{a,b_0}$. Also $b_0\neq b$, because $b_0=b$ would imply $y_0=ax_0+b$. For any point $(x,y)\in P$, simultaneous containment in $L_{a,b}$ and $L_{a,b_0}$ would imply \begin{align*} y=ax+b=ax+b_0, \end{align*} so $b=b_0$, a contradiction. Hence $L_{a,b_0}$ is disjoint from $L_{a,b}$. If $m\in\mathcal{L}$ is any line through $p_0$ disjoint from $L_{a,b}$, then $m$ cannot be vertical unless $m=V_{x_0}$. But $V_{x_0}$ meets $L_{a,b}$ at the point $(x_0,ax_0+b)$, so $m$ is not vertical. Write $m=L_{a',b'}$. Since $m$ is disjoint from $L_{a,b}$, we must have $a'=a$; if $a'\neq a$, then $a-a'\neq 0$ and the point with \begin{align*} x=(b'-b)(a-a')^{-1}, \qquad y=ax+b \end{align*} lies on both lines. Since $m$ contains $p_0$, we have $y_0=ax_0+b'$, so $b'=y_0-ax_0=b_0$. Thus $m=L_{a,b_0}$. Now suppose $\ell=V_c$ for some $c\in\mathbb{F}_q$. Since $p_0\notin V_c$, we have $x_0\neq c$. The vertical line $V_{x_0}$ contains $p_0$ and is disjoint from $V_c$. If $m\in\mathcal{L}$ is any line through $p_0$ disjoint from $V_c$, then $m$ cannot be non-vertical, because every non-vertical line $L_{a,b}$ meets $V_c$ at $(c,ac+b)$. Hence $m$ is vertical, and since it contains $p_0$, it must be $V_{x_0}$. Therefore through $p_0$ there is exactly one line parallel to $\ell$. [/step] [step:Exhibit three non-collinear points] Consider the three points \begin{align*} p_1&:=(0,0), & p_2&:=(1,0), & p_3&:=(0,1) \end{align*} in $P=\mathbb{F}_q^2$. Since $\mathbb{F}_q$ is a field, $0\neq 1$, so these are three distinct points. The unique line through $p_1$ and $p_2$ is $L_{0,0}$, because both points satisfy $y=0$. The point $p_3=(0,1)$ does not lie on $L_{0,0}$, since $1\neq 0$. Therefore $p_1,p_2,p_3$ are not collinear. This verifies the non-degeneracy axiom for an affine plane. [/step] [step:Count the points on each line to identify the order] Let $a,b\in\mathbb{F}_q$. The map $\phi_{a,b}: \mathbb{F}_q \to L_{a,b}$ defined by $\phi_{a,b}(x) := (x, ax+b)$ is bijective. It is surjective by the definition of $L_{a,b}$, and it is injective because equality $(x,ax+b)=(x',ax'+b)$ implies $x=x'$. Hence $|L_{a,b}|=|\mathbb{F}_q|=q$. Similarly, for each $c\in\mathbb{F}_q$, the map $\psi_c: \mathbb{F}_q \to V_c$ defined by $\psi_c(y) := (c,y)$ is bijective, so $|V_c|=q$. Thus every line has exactly $q$ points. Combining this count with the two incidence axioms and the existence of three non-collinear points proved above, the incidence structure $AG(2,q)$ satisfies the definition of an [affine plane](/page/Affine%20Plane) of order $q$. [/step]