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I do not work in algebraic geometry. So I am using Wikipedia definition of monomials, and take their real linear combinations. In particular, I am interested in such objects where each term is of degree 4 and number of variables is even (just a simple example to clarify what is in my mind; $f(x_1,x_2,x_3,x_4)=x_1^2x_2^2+x_3^2x_4^2-3x_1x_2x_3x_4$). For simplicity, I call such objects as monomial (though I guess form is a better word).

Let $f(x_1,\cdots,x_n)$ be any such monomial and $Z(f):=\{(z_1,\cdots,z_n):z_i\in\mathbb{C}, ~~f(z_1,\cdots,z_n)=0\}$. We know that if there are two objects $f$ and $g$, such that $Z(f)=Z(g)$, then $f=\lambda g$, for some scalar $\lambda$. Is there a way to check, given $f$, whether there is a $g$ with the same degree (4 for me) and same variables, $Z(f)\subseteq Z(g)$. Moreover if the above happens, whether $f=\lambda g$ for some scalar $\lambda$ or not.

In general also, whether there is any result regarding the above problem (for some general degree and without restricting the number of variables as odd only.

My motivation for the above problem: while working on the positive map between $C^*$ algebras, I noticed that whether a map is exposed or not can be reduced to some calculations which can be rewritten as above.

Since I do not work (and know algebraic geometry) i am not sure whether my writing is clear (or in worst case the question is meaningful). Also my use of terminologies may be confusing. I shall be more than happy to explain the details in comments and reedit the question, if necessary. Advanced thanks for any help.

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The objects you're talking about are homogeneous polynomials ( – Qiaochu Yuan Jun 10 '12 at 19:44
Your claim in the second paragraph is false without further hypotheses (take for example $f = x^2 y, g = xy^2$). A correct version of this claim is given by the Nullstellensatz ('s_Nullstellensatz). – Qiaochu Yuan Jun 10 '12 at 20:44
up vote 2 down vote accepted

Suppose $f$ and $g$ are any polynomials whatsoever in $\mathbb{C}[x_1, ... x_n]$ such that $Z(f) \subseteq Z(g)$. In other words, $g$ vanishes wherever $f$ vanishes. By the Nullstellensatz, it follows that $$g \in \sqrt{(f)}$$

where $(f)$ is the ideal generated by $f$ and $\sqrt{I}$ is the radical of an ideal. If $f$ is squarefree, then $\sqrt{(f)} = (f)$, and so it follows that there exists a polynomial $h$ such that $$g = hf.$$

But if $f, g$ are homogeneous of the same degree then this is only possible if $h$ is a scalar.

If $f$ is not squarefree, then $\sqrt{(f)}$ will be larger than $(f)$ and nontrivial possibilities for $g$ will exist. For example, as I mentioned in the comments, taking $f = x^2 y$ we have $\sqrt{(f)} = (xy)$ and so $g$ can be a multiple of $xy$ such as $x y^2$.

So the problem reduces to checking whether or not $f$ is squarefree. Unfortunately I don't know how to do this. If $f$ has a square factor $h^2$ then $h | \frac{\partial f}{\partial x_i}$ for all $i$ so one could try to compute the gcd of $f$ and $\frac{\partial f}{\partial x_i}$ for all $i$ using Gröbner basis techniques...? Just a guess.

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I suppose one could compute $\gcd(f, \frac{\partial f}{\partial x_i})$ in $\mathbb{C}(x_1, ... \widehat{x_i}, ..., x_n)[x_i]$ (where the hat denotes omission) for each $i$...? – Qiaochu Yuan Jun 10 '12 at 20:59

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