Let $X$ be a projective variety over an algebraically closed field, and $I$ be the homogenous ideal of $X$ and $J$ be an ideal with the same zero set. Suppose that I know $I=\langle f_1,...f_n \rangle$ with $f_i$ of degree $d$, and $J= \langle g_1,...g_m \rangle$ with $g_i$ also of degree $d$. Does it follow that $J$ is also a radical ideal (and hence equal to $I$)?

My motivation: I am trying to show that the quadratic relations (a postiori Plucker relations) that one gets by a certain tensor contraction, generate the ideal of the image of the grassmannian under the Plucker embedding. I know that elements of the coordinate ring of the image can't satisfy any relations of degree 1. So if the criterion above holds then these quadratic relations are actually Plucker relations.

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    $\begingroup$ I think I found a counterexample. The ideal of the inclusion of 1 dimensional subspaces into 3 dimensional subspaces which is projective ideal of degree 2, and is radical strictly contains the ideal of 1 dimensional subspaces into 2 dimensional subspaces+ the ideal of 2 dimensional subspaces into 3 dimensional subspaces, the sum also generated by polynomials of degree 2. They have the same zero set, so the ideal corresponding to $J$ in my question is not radical. $\endgroup$ Commented Apr 10, 2016 at 23:27

1 Answer 1


This criterion is not correct. Let $F_{123}$ be the variety $GL(4)/B$, $B$ a Borel subgroup, and $F_{12}=G/P_{12}$, $F_{23}=G/P_{23}$, where $P_{12}$ and $P_{23}$ are respective parabolic subgroups. These are all projective subvarieties of the product of grassmanians $G_1^4 \times\cdots\times G_4^4$.

Then if you give Plucker coordinates to this product of Grassmann varieties $$(\sum a_i e_i,\sum_{i<j} b_{ij} e_i \wedge e_j, c_1 e_2 \wedge e_3 \wedge e_4 + \cdots+ c_4 e_1 \wedge e_2 \wedge e_3, d e_1 \wedge e_2 \wedge e_2 \wedge e_4 ),$$ the relation $a_1c_3$, in $\mathscr{I}( F_{13})$ is not in the sum of ideals $\mathscr{I}( F_{13})+ \mathscr{I} (F_{23})$. This happens despite the fact that $ V\mathscr{I}(\mathscr{I}( F_{13})+ \mathscr{I} (F_{23}))=V\mathscr{I}(F_{123})$.

The geometric interpretation of this fact is that if a one dimensional subspace is included in a 2 dimensional subspace and that 2 dimensional subspace is included in a 3 dimensional subspace, this 1 dimensional subspace is included in this 3 dimensional subspace. The implication is that sum of quadratic projective ideals $\mathscr{I}( F_{13})+ \mathscr{I} (F_{23})$ is not radical. This is the minimal non-example of a determinantal variety where this happens because this is the minimal dimension in which there are nontrivial subspace inclusion relations.


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