0
$\begingroup$

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

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

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)

$\endgroup$
  • $\begingroup$ Why do you switch from $x_1e_1+\cdots+x_4e_4$ to $(a_1,\cdots,a_4)$ ? $\endgroup$ – mr_e_man Sep 11 '18 at 1:55
  • $\begingroup$ @mr_e_man: oh just random that is ... that doesn't affect the problem in any way ... $\endgroup$ – user521337 Sep 11 '18 at 2:23
0
$\begingroup$

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.

$\endgroup$
-1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I am very sorry ... I really wanted to ask about the idealizer of the set of all elementary wedge products $x \wedge y$ ... $\endgroup$ – user521337 Sep 11 '18 at 21:56
  • $\begingroup$ That's ok. I believe then you are talking about Grassmannians probably the embedding of $Gr(2,4)$ into $P^5$. Their defining equations are well known. However, I do not know much about this object. I hope you can google it, or some else will answer your questions. $\endgroup$ – Youngsu Sep 11 '18 at 22:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.