On the idealizer of the set of elementary wedge products of two vectors in $K^4$, for a field $K$ Let $K$ be a field of characteristic zero. Consider $V=K^4$ with standard basis vectors $e_1,e_2,e_3,e_4$. We can consider the second exterior product $\bigwedge^2 V $ of $V=K^4$ with  a basis given by $\{e_1\wedge e_2,e_1\wedge e_3,e_1\wedge e_4, e_2\wedge e_3, e_2\wedge e_4, e_3\wedge e_4\}$. For $\hat x=x_1e_1+x_2e_2+x_3e_3+x_4e_4 \in V$ and $\hat y=y_1e_1+y_2e_2+y_3e_3+y_4e_4 \in V$, we have the elementary wedge product 

$\hat x \wedge \hat y=\sum_{i&lt;j}(x_iy_j-x_jy_i) e_i \wedge e_j$.   

Now let $R=K[T_1,...,T_6]$ and   Consider  $$I :=\{f \in R : f(a_1b_2-a_2b_1,a_1b_3-a_3b_1,a_1b_4-a_4b_1,a_2b_3-a_3b_2,a_2b_4-a_4b_2,a_3b_4-a_4b_3) = 0, \forall (a_1,...,a_4); (b_1,...,b_4)\in K^4\}$$.

Then $I $ is an ideal of $R$. My question is :

Is $I$ always a principal, radical ideal ? 

(May assume $K$ is algebraically closed if need be)
First thing's first: $I$ is the ideal corresponding to the closed embedding $G_k(n)\hookrightarrow \Bbb P(\bigwedge^k K^n)$ of the Grassmanian of $k$-planes in $n$-space into projective space. This enables us to make many determinations about $I$ from knowing what's going on with these geometric objects.

As $n,k$ vary, $I$ is not always principal: if $I$ were principal, then $G_k(n)$ would be a subvariety of codimension at most one in $\Bbb P(\bigwedge^k K^n)$ by Krull's Height Theorem. But the dimension of $G_k(n)$ is $k(n-k)$ while the dimension of the projective space in question is $\binom{n}{k}-1$. It is not hard to see that $k(n-k)+1$ is not always greater than $\binom{n}{k}-1$: for instance, choosing $n=5,k=2$ gives that the former is $7$ while the latter is $9$. 

In your specific case of $k=2,n=4$, the ideal is principal and is generated by $p_{23}p_{14} - p_{13}p_{24} + p_{12}p_{34}$, which may be verified by examining the relations between the the coordinates you have written (we write $p_{ij}$ for the determinant of the minor composed of rows $i$ and $j$, or equivalently the coefficient of $e_i\wedge e_j$). This equation may be seen to generate a radical without too much trouble. In general, the equations which generate $I$ are called the Plucker relations, and it's known how to construct all of them (see wikipedia, for instance).

In general, as $n,k$ vary, $I$ is always radical, since the Grassmanian over a field is reduced. This is in fact a special case of a more general theorem: determinental ideals are radical (see Are the determinantal ideals prime?) for several links and sources.
Once you fix $x$ and $y$, your question is equivalent to determining the ideal vanishing at a point in $K^6$. Such an ideal is nothing, but the maximal ideal corresponding to the point.