Let $\operatorname{GF}(3)$ be the Galois field with $3$ elements so that $\operatorname{GF}(3)\cong\mathbb{Z}/3\mathbb{Z}$.
Consider the vector space $\operatorname{GF}(3)^3$. Obviously, this vector space has $3^3$ vectors.
Let $u := (1, 0, 0)\in\operatorname{GF}(3)^3$.
How to determine all linear subspaces of $\operatorname{GF}(3)^3$ that contain $u$?
(I don't want to name each of them but I want to characterize them)
Let $\Bbb{F}_3$ denote the field of three elements and $U$ the span of $u:=(1,0,0)$. The linear map $$q:\ \Bbb{F}_3^3\ \longrightarrow\ \Bbb{F}_3^2:\ (a,b,c)\ \longmapsto\ (b,c),$$ yields a bijection between linear subspaces of $\Bbb{F}_3^2$ and linear subspaces of $\Bbb{F}_3^3$ containing $\ker q=U$. Explicitly, the linear subspaces of $\Bbb{F}_3^2$ with their corresponding subspaces of $\Bbb{F}_3^3$ are: \begin{eqnarray*} 0\qquad&&\text{ corresponds with }\qquad q^{-1}(0)=U,\\ \langle(1,0)\rangle\qquad&&\text{ corresponds with } \qquad q^{-1}(\langle(1,0)\rangle)=\langle u,(0,1,0)\rangle,\\ \langle(1,1)\rangle\qquad&&\text{ corresponds with } \qquad q^{-1}(\langle(1,1)\rangle)=\langle u,(0,1,1)\rangle,\\ \langle(1,2)\rangle\qquad&&\text{ corresponds with } \qquad q^{-1}(\langle(1,2)\rangle)=\langle u,(0,1,2)\rangle,\\ \langle(0,1)\rangle\qquad&&\text{ corresponds with } \qquad q^{-1}(\langle(0,1)\rangle)=\langle u,(0,0,1)\rangle,\\ \Bbb{F}_3^2\qquad&&\text{ corresponds with } \qquad q^{-1}(\Bbb{F}_3^2)=\Bbb{F}_3^3. \end{eqnarray*}