Is an ideal generated by multilinear polynomials of different degrees always radical?

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 $0\le i_1<i_2<\cdots<i_d\le n$.

I tried to answer a question and ended up with 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. The question is now whether such an ideal is always radical.

If it is false, I would love to see a counterexample.

If it is true, 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 am not so sure if it matters for the polynomials to be irreducible. I would also love to see a proof, of course.

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@YACP: I added a definition. Also, I forgot to mention that the polynomials are homogeneous. –  Jesko Hüttenhain Aug 25 '13 at 10:15
Isn't $\left(x-y\right)^2$, but not $x-y$, in the ideal generated by $x+y$ and $xy$ ? (Unless $\mathbb k$ has characteristic $2$.) –  darij grinberg Aug 25 '13 at 11:47
@darijgrinberg: good one! Got one with irreducible polynomials? –  Jesko Hüttenhain Aug 25 '13 at 11:56
Oh, I missed that word (given it wasn't in the title). That sounds harder. –  darij grinberg Aug 25 '13 at 12:26
In characteristic $0$ one knows that an ideal is radical if the initial ideal $\operatorname{in}(I)$ is square free. Unfortunately I don't know if this is the case for your example, but maybe someone else can prove (or disprove) this. –  user26857 Aug 25 '13 at 15:30

This is a repost of my answer here.

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 Theorem 23 in this paper).

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.

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Thanks for the answer ! And especially because it confirms my first and last sentences in the bounty text :). It was "I would think the answer is no, but I am too lazy and not smart enough to find a counterexample." –  Cantlog Sep 4 '13 at 17:44