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The elementary symmetric polynomials appear when we expand a linear factorization of a monic polynomial: we have the identity $$ \prod _{j=1}^{n}(\lambda -X_{j})=\lambda ^{n}-e_{1}(X_{1},\ldots ,X_{n})\lambda ^{n-1}+e_{2}(X_{1},\ldots ,X_{n})\lambda ^{n-2}+\cdots +(-1)^{n}e_{n}(X_{1},\ldots ,X_{n}). $$

Let $\vec v\in \mathbb Z^n$, where the first element is always $0$, i.e. $e_1(...)=0$. How many solutions for $X_k\in\mathbb C$ do we get for the following: $$\vec v=\pmatrix{e_1\\e_2\\\vdots\\e_n}?$$

Some low dimensional numerical experiments point toward $n!$. Is this true?

draks ...
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1 Answers1

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No, this is not true. The condition on $e_1$ is irrelevant, as is the condition that the entries of $v$ are integers. You're asking how many ways there are to order the roots of some polynomial of degree $n$, and the answer depends on their multiplicities. If the roots each have multiplicity $1$ then the answer is $n!$ but, for example, if there's only one root with multiplicity $n$ then the answer is $1$. In general if the roots have multiplicities $m_i$ then the answer is

$$\frac{n!}{\prod m_i!}.$$

Qiaochu Yuan
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  • Is there a standard way to get this solutions? A certain kind of transform? – draks ... Dec 22 '15 at 07:01
  • @draks: you find all the roots of the polynomial, then permute them. Are you asking how to find all the roots or how to permute them? – Qiaochu Yuan Dec 22 '15 at 07:02
  • Is there no way for something like an exceptional kind of symmetry. I mean the following: Two different sets of $X_k$ lead to the same $\vec v$, which might have plenty of zeros. The question is motivated by graph-theoretical one... – draks ... Dec 22 '15 at 07:06
  • @draks: that depends on what you mean by "different." Like I said, if you write down the appropriate polynomial with coefficients given by the entries of $v$, then the $X_i$ must be, in some order, the roots of that polynomial (with multiplicity). – Qiaochu Yuan Dec 22 '15 at 07:11