I [asked this question on math.se](https://math.stackexchange.com/questions/475664/is-an-ideal-generated-by-multilinear-polynomials-of-different-degrees-always-rad) 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.



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> **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 $0\le i\_1<i\_2<\cdots<i\_d\le n$.
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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](https://math.stackexchange.com/questions/473266/ideal-defining-the-nilpotent-cone-of-mathfrakgl-nk/473527#473527) 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!



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


