Finite prime field representation of uniform matroid $U_{2,n}$ Suppose I have a uniform matroid $U_{2,n} = (E, I)$ (so $F \subset E$ has $F \in I \iff |F| \leq 2$) and want to represent it over $GF(p)$, i.e. I would like to construct a map $\phi : E \to GF(p)^2$ such that any subset $F = \{f_1, \dotsc, f_k \}$ of $E$ with $|F| \leq 2$ is independent ($F \in I$) $\iff \phi(f_1), \dotsc, \phi(f_k)$ are linearly independent in $GF(p)^2$.
Literature (books, Wikipedia etc.) tells me that $p = r + 1 = 3$ should be enough to construct such a function, but I don't see how we can construct $\phi$ such that any two elements from $E$ have linearly independent vectors in the image of $\phi$, but any three elements from $E$ have linearly dependent vectors from $\phi$.
I also want to show for every prime $p$, there exists $n \in \mathbb{N}$ such that $U_{4, n}$ has no prime representation but I have no idea where to start with this given my issues with the previous problem.
Any ideas or help appreciated, thanks.
 A: I am not certain that this is correct, but this is my solution for the $U_{2,n}$ problem. I welcome corrections and suggestions. 
The prime $p$ will depend on $n$, we need $p>n$, any will do. 
I define $\varphi\colon \{1,\ldots, n\}\to GF(p)^2$ by
\begin{align}
\varphi(1) &=(1,1),\\
\varphi(2) &=(1,2),\;\text{and so on},\\
\varphi(m) &=(1,m).
\end{align}
Any two of these will be independent: if $(x,y)=\lambda\cdot(x^\prime, y^\prime)$ for $\lambda\ne 0$, then $x=\lambda x^\prime$, which implies $\lambda=1$, since the "$x$-coordinate" is always $1$, then $y=y^\prime$...
Any three of these will not be independent: let the vectors be $(1,y),(1,y^\prime)$, and $(1,y^{\prime\prime})$, with $y<y^\prime<y^{\prime\prime}$ (we may assume they are distinct, else they are trivially dependent). Notice that for all $z\in\{0,\ldots,p-1\}$ there is a $t\in\Bbb{N}$ such that $t(y^\prime-y)=z$. In particular, there is a $t$ such that $t(y^\prime-y)=z=y^{\prime\prime}-y$. Now
\begin{align}
(1,y^\prime)-(1,y) &=(0,y^\prime-y)\\
t(0,y^\prime-y) &= (0,y^{\prime\prime}-y)\\
(0,y^{\prime\prime}-y) + (1,y) &= (1,y^{\prime\prime}),
\end{align}
i.e. $(1,y^{\prime\prime})$ is a linear combination of $(1,y^\prime)$ and $(1,y)$:
$$t\big((1,y^\prime)-(1,y)\big) + (1,y) = (1,y^{\prime\prime}).$$

As for $U_{4,n}$... A representation of this needs a function $\varphi\colon\{1,\ldots,n\}\to GF(p)^4$ such that $f_1,\ldots,f_k$ are independent iff $\varphi(f_1),\ldots,\varphi(f_k)$ are.
Let $n>p^4$, then since $|GF(p)^4|=p^4$, we must have $\varphi(x)=\varphi(x^\prime)$ for some $x\ne x^\prime$, which leads to a contradiction.
