Consider the map $j: X\times Y \to\mathbb{G}(1,n), \ ([v],[w])\mapsto [v\wedge w]$ i.e. sending two points to their line where $X$ and $Y$ are two projective irreducible varieties. I would like to prove that this map is rational. When $X$ and $Y$ are disjoint, this map is obviously regular and thus also rational due to the nature of the Pluecker Coordinates.
If $X\neq Y$, then we can look at the open subset $X\times Y \setminus (X\cap Y)\times (X\cap Y)$. As $X\neq Y$, this is a non-empty subset and as $X$ and $Y$ are irreducible, it is also dense. This open set, however, was already covered above as you simply get varietes that are disjoint and thus you get a regular map. This means that $j$ is rational in this case.
Now, the case I am struggling on is $X=Y$. Why is $j: X\times X\to \mathbb{G}(1,n)$ rational? I have failed to come up with a suitable non-empty open subset unfortunately. Any help would be greatly appreciated.
Question: "Consider the map $j: X\times Y \to\mathbb{G}(1,n), \ ([v],[w])\mapsto [v\wedge w]$ i.e. sending two points to their line where $X$ and $Y$ are two projective irreducible varieties. I would like to prove that this map is rational. When $X$ and $Y$ are disjoint, this map is obviously regular and thus also rational due to the nature of the Pluecker Coordinates."
Answer: If $X \subseteq \mathbb{P}(V^*), Y \subseteq \mathbb{P}(W^*)$ and if $x:=[l] \in X, y:=[l']\in Y$ with $l \subseteq V, l' \subseteq W$ non-zero lines, it follows the "wedge product" $[l \wedge l']$ is not well defined. The lines $l,l'$ live in different vector spaces and cannot be "wedged together". Hence the map $j$ does not exist.
If $E$ is an $A$-module, you may form the module $\wedge^n E$ of $E$ with itself $n$-times. If $E$ is generated by one element $e$ it follows $E \wedge_A E=(0)$ is zero. Hence if $A$ is a field and $l\subseteq E$ is a line it follows $l \wedge l=(0)$ is zero.
Note: If $X:=\mathbb{G}(m,n)$ is the grassmannian of $m$-planes $W\subseteq V\cong \mathbb{C}^n$ in an $n$-dimensional complex vector space, $V$ it follows the exterior product $\wedge^m W \subseteq \wedge^m V$ is a line in $\wedge^m V$. This gives the Plucker embedding
$$p: X \rightarrow \mathbb{P}(\wedge^m V^*)$$
defined by $p([W]):=[\wedge^m W]$. Here you wedge a vector space $W$ with itself. You cannot wedge two vector spaces $W \wedge V$.