# Is an ideal generated by multilinear, irreducible, homogeneous polynomials of different degrees always radical?

I asked this question on math.se and someone even put a bounty on it, yet there was no answer. Hence, I am asking here. Assume $\Bbbk$ to be a field of characteristic zero.

Definition. A polynomial $f\in\Bbbk[x_0,\ldots,x_n]$ is called multilinear if $\deg_{x_i}(f)=1$ for each $0\le i \le n$. In other words, $f$ is linear in each variable. If $f$ is homogeneous of degree $d$, then $f$ is a linear combination of monomials of the form $x_{i_1}\cdots x_{i_d}$ with $0\le i_1<i_2<\cdots<i_d\le n$.

Given an ideal $I=(f_1,\ldots,f_r)\subseteq\Bbbk[x_0,\ldots,x_n]$ with the property that the $f_i$ are irreducible, homogeneous, multilinear polynomials of (pairwise) different degrees, I am asking whether $I$ is radical.

I actually don't believe it holds in general - if this is the case, I would love to see a counterexample.

If it is true however, then I am sure that the assumption on the degree can not be dropped (see this example of an ideal generated by irreducible, homogeneous, multilinear polynomials which is not radical). I would also love to see a proof in this case, of course.

Thanks a lot in advance!

## 1 Answer

One general fact that comes to mind: If an ideal $I\subset \mathbb{k}[x_1,\dots,x_n]$ contains an element of the form $f = gx_1 + h$ where $g,h$ don't use $x_1$, and $g$ is a nonzerodivisor mod $I$, then the primary components of $I\cap \mathbb{k}[x_2,\dots,x_n]$ and $I$ are in bijection. This is birational projection and I learned it from Mike Stillman (see Proposition 23 in the appendix of http://arxiv.org/pdf/math/0301255.pdf).

Now here is almost a counter-example to your question:

$$I = \langle x_{1} x_{9}-x_{4}x_{8}, x_{4}x_{6}-x_{7}x_{9}, x_{2}x_{5}-x_{3}x_{9}, x_{2}x_{3}-x_{5}x_{6} \rangle \subset \mathbb{k}[x_1,\dots,x_9]$$

This ideal has 6 components, one of which is primary with minimal prime $\langle x_9, x_5, x_4, x_2 \rangle$.

If I read your hypotheses correctly, the only bit missing is the pairwise different degrees of the generators. I have an inkling that this may be a red herring. If I modify my example by adding some extra unrelated variables, then the embedded component over $\langle x_9, x_5, x_4, x_2 \rangle$ is essentially unchanged:

$$\langle x_{1}x_{9}-x_{4}x_{8}, x_{4}x_{6}y_{1}-x_{7}x_{9}y_{2}, x_{2}x_{5}y_{3}y_{4}-x_{3}x_{9}y_{5}y_{6}, x_{2}x_{3}y_{7}y_{8}y_{9}-x_{5}x_{6}y_{10}y_{11}y_{12} \rangle$$

The Binomials package in Macaulay2 quickly confirms that this ideal is not radical.

• Now let me get the math.se bounty :) – Thomas Kahle Sep 4 '13 at 13:12
• There is no Theorem 23 in that arxiv paper. – user26857 Dec 14 '15 at 17:35
• It's Proposition 23. I fixed the answer, thanks. – Thomas Kahle Dec 15 '15 at 5:45