Visualizing a decomposition of the Grassmannian $\operatorname{Gr}(2, F^3)$

I try to understand what the Grassmannian $\operatorname{Gr}(2, F^3)$ (over some field $F$) looks like geometrically (in as far this term makes sense when we do not specify the field), and in particular how two subsets (specified below) divide the whole thing among them. To my shame I do not even feel I can get the dimensions right.

Hopefully someone can help.

My intuition (based on the case that the field $F$ is $\mathbb{R}$) is that $\operatorname{Gr}(2, F^3)$ is isomorphic to $\mathbb{P}^2 = \operatorname{Gr}(1, F^3)$ because we can “identify” each 2-plane in 3-space with “the” line “orthogonal” to it. (I have no idea what this would mean over other fields than $\mathbb{R}$ but still suspect that the isomorphism survives in those cases.) That would settle the first question, and in particular make the Grassmannian 2-dimensional. But now about the subspaces.

Let the 3-dimensional space in which we classify the 2-dimensional subspaces be the space of degree 2 homogeneous polynomials in variables $x$ and $y$, so the space spanned by basis elements $x^2$, $xy$, $y^2$. Now let a two-dimensional plane $V$ in this space be of

• Type I if every two elements of $V$ have a common divisor of degree 1 in the ring $F[x, y]$
• Type II if no two linearly independent elements of $V$ have a common divisor of degree 1 in $F[x, y]$.

So $V_1 = \operatorname{span}(\{x^2, xy\})$ is of type I because all its elements are divisible by $x$ and $V_2 = \operatorname{span}(\{x^2, y^2\})$ is of type II, because if some $u = \alpha x^2 + \beta y^2$ and $v = \gamma x^2 + \delta y^2$ have some common divisor $w = \epsilon x + \zeta y$ (with Greek letters denoting scalars in $F$), while simultaneously $V_2 = \operatorname{span}(\{u, v\})$, we would have that every element of $V_2$ is divisible by $w$ including the original basis elements $x^2$ and $y^2$, which is impossible.

Similar reasoning shows that types I and II together make up all possible two-dimensional planes in this 3-dimensional space.

So my question is: what do the sets of $TI$ and $TII$ of Type I points and Type II points in $\operatorname{Gr}(2, 3)$ look like? What dimension do they have? Do they fall apart into smaller irreducible components? How do they relate to other known subsets of the Grassmannian like Schubert cells and would that provide a better notation for them?

W.r.t. the dimension question: it seems that for each homogeneous degree 1 polynomial $w$ there is exactly one Type I space for which $w$ is the common divisor of its elements. This would suggest that $TI$ is two-dimensional, as is the entirety of $\operatorname{Gr}(2, 3)$ (if the above is correct).

But does that mean that $TII$ is only 1- or 0-dimensional? Or that two 2-dimensional pieces can peacefully coexist? Both options sound implausible to me. Any help is welcome.