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Let $R=\Bbb K[x_1,\dots,x_n]/I$ and $Z=\mathsf Z(I)$ where $\Bbb K$ is any field with $char(\Bbb K)\neq 2$.

Is there a way to describe space of $f_1,f_2\in R$ that satisfies $$f_1+f_2\neq 0,\mbox{ }\forall z\in Z$$$$f_1f_2= 0,\mbox{ }\forall z\in Z$$ with space of $g_1,g_2\in R$ that satisfies $$g_1+g_2=1,\mbox{ }\forall z\in Z$$$$g_1g_2= 0,\mbox{ }\forall z\in Z?$$

I am interested in the case $I=(x_1^2-x_1,\dots,x_n^2-x_n)$. This is the space of multilinear polynomials.

Note that we could take $\mathsf Z(f_1)\cap\mathsf Z(I) = \mathsf Z(g_1)\cap\mathsf Z(I)$ and $\mathsf Z(f_2)\cap\mathsf Z(I) = \mathsf Z(g_2)\cap\mathsf Z(I)$.

We also have $\mathsf Z(f_1)\cap\mathsf Z(f_2)\cap\mathsf Z(I)=\mathsf Z(g_1)\cap\mathsf Z(g_2)\cap\mathsf Z(I)=\emptyset$.

I want to compare minimum of $deg(f_1)deg(f_2)$ with minimum of $deg(g_1)deg(g_2) = deg(g_1)^2 =$$ deg(g_2)^2$ where $deg$ refers to the total degree which is atmost $n$ for multilinear polynomials. If I know how much bigger the first space is dimensionally compared to the second space, I can probably do a monomial count to get degree comparison.

It is clear over $\Bbb F_2$, we have $deg(g_i)\leq deg(f_1)deg(f_2)$ since $g_i=f_i$ here.

Does the relation hold for all fields?

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  • $\begingroup$ Are you sure that what you describe are multilinear polynomials? Reading your posts, I was sure that multilinear were a subspace of the polynomial ring, which is not a ring. Does it really make much sense to multiply them? $\endgroup$
    – Alex Degtyarev
    Commented Dec 4, 2014 at 11:34
  • $\begingroup$ @AlexDegtyarev $(x_{1} x_{2}+ x_{2} x_{3})^2 = (x_{1} x_{2}+ x_{2} x_{3} + 2 x_{1} x_{2} x_{3})$ So both $(x_{1} x_{2}+ x_{2} x_{3})$ and $(x_{1} x_{2}+ x_{2} x_{3} + 2 x_{1} x_{2} x_{3})$ are multilinear? $\endgroup$
    – Turbo
    Commented Dec 4, 2014 at 11:36
  • $\begingroup$ You can define multiplication to produce something of degree one in each argument, e.g., like you did, or factoring out $x_i^2$, or something else. My question was whether it makes sense. E.g., why $x_i^2-x_i$ and not just $x_i^2$? Is there any background? $\endgroup$
    – Alex Degtyarev
    Commented Dec 4, 2014 at 11:39
  • $\begingroup$ @AlexDegtyarev Of course. Any ideal that produces a 'non-trivial' discrete set whose cardinality is exponential in $n$ would do. $\mathsf Z((x_1^2-x_1,\dots,x_n^2-x_n))$ is one such set and is of importance to computer science. This set describes the boolean cube. $\endgroup$
    – Turbo
    Commented Dec 4, 2014 at 11:42

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