I am trying to understand the following corollary in James Oxley book:
A simple rank-r matroid M that is representable over $GF(q)$ has at most $\frac{q^r - 1}{q - 1}$ elements. Moreover, if $|E(M)| = \frac{q^r - 1}{q - 1},$ then $M$ is isomorphic to $PG(r-1, q).$
While reading this Corollary I was thinking about the representability of $U_{2,4},$ I know it is representable over $\mathbb F$ iff $|\mathbb F| \geq 3.$
My question is: Does this proposition tell us that $PG(1,q) \cong U_{2,q}$?
If this is true, what if I want that $AG(1,q) \cong U_{2,q},$ how can I show that?
If this is wrong, what should I add to the corollary (if it is actually needed ) to know that $AG(1,q) \cong U_{2,q}$?
Any clarification will be greatly appreciated.
Recall that $U_{m, q}$ has $q$ elements. Moreover, $PG(r-1, q)$ has $\frac{q^r-1}{q-1}$ elements and $AG(r-1, q)$ has $q^{r-1}$ elements. Now, if $r-1=1$, as it does in our case, this simplifies, after some algebra, to $PG(1, q)$ having $q+1$ elements, and $AG(1, q)$ having $q$ elements.
Now, this informs us that $PG(1, q)\not\cong U_{2,q}$, but perhaps $PG(1, q)\cong U_{2,q+1}$ and $AG(1, q)\cong U_{2,q}$. Both of these are in fact true. Note that it is sufficient to prove one of these two statements, since $AG(1,q)$ is obtained from $PG(1,q)$ by deleting a hyperplane, which in our case is a rank $1$ flat; a point, and the same is true about $U_{2,q}$ and $U_{2,q+1}$.
We prove $PG(1, q)\cong U_{2,q+1}$. Note that the rank of both matroids is two, and the number of both is $q+1$. Now, one way to define to define $PG(r-1, q)$ is the following: collect all vectors of size $r-1$ and entries in $GF(q)$ in a matrix $A$. Obtain $A'$ from $A$ by deleting the zero column, and all columns but one from every set of parallel columns (columns that are a constant multiple of each other). This in terms of a matroid corresponds to deleting all loops, and keeping one element from each parallel class. Then $M(A')$ is known as the simplification of $M(A)$, and we write $M(A') = \operatorname{si}(M(A))$. This matroid is $PG(r-1, q)$. In our case, $r-1=1$, and since our matroid is simple, $r(\{e, f\})=2$ for all $e$ and $f$ in $PG(1, q)$, i.e., all pairs of elements lie on a line. But this is exactly the definition of $U_{2,q+1}$, as desired.