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.

Thanks a lot in advance.

  • $\begingroup$ 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 $2$.) $\endgroup$ – darij grinberg Aug 25 '13 at 11:47
  • $\begingroup$ @darijgrinberg: good one! Got one with irreducible polynomials? $\endgroup$ – Jesko Hüttenhain Aug 25 '13 at 11:56
  • $\begingroup$ Oh, I missed that word (given it wasn't in the title). That sounds harder. $\endgroup$ – darij grinberg Aug 25 '13 at 12:26
  • $\begingroup$ In characteristic $3$, the ideal $\left(x+y+z,xy+yz+zx\right)$ contains $\left(x-y\right)^3$ but not $x-y$. Remains to deal with characteristic $0$. $\endgroup$ – darij grinberg Aug 25 '13 at 12:35
  • 3
    $\begingroup$ 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. $\endgroup$ – 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.

  • $\begingroup$ 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." $\endgroup$ – Cantlog Sep 4 '13 at 17:44

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.