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In Stewart and Tall's book on Algebraic Number Theory, they give a theorem of Newton:

Theorem 1.12. Let $R$ be a ring. Then every symmetric polynomial in $R[t_1, \ldots, t_n]$ is expressible as a polynomial with coefficients in $R$ in the elementary symmetric polynomials $s_1, \ldots, s_n$.

After proving this they give the following corollary.

Corollary 1.14. Suppose that $L$ is an extension of the field $K$, $p \in K[t]$, $\partial p = n$, and the zeros of $p$ are $\theta_1, \ldots, \theta_n \in L$. If $h(t_1,\ldots,t_n) \in K[t_1,\ldots,t_n]$ is symmetric, then $h(\theta_1, ..., \theta_n) \in K$.

Note: $\partial p$ denotes the degree of polynomial $p$.

My Question: I don't see how the corollary follows from the theorem in any straightforward way. Surely the way to apply the theorem is to first consider any symmetric polynomial $h(t_1,\ldots,t_n) \in K[t_1,\ldots,t_n]$ and then rewrite $h$ in terms of the elementary symmetric polynomials according to theorem 1.12. But I don't see why evaluating these elementary symmetric polynomials at roots $\theta_1,\ldots, \theta_n$ should always yield something in $K$.

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2 Answers 2

up vote 3 down vote accepted

Hint: Vieta's formulas hold in any field.

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Vieta's formulas are not mentioned directly or otherwise in this introductory text. Although Vieta certainly closes the gap I was looking for, I'm wondering if there is some simple explanation I'm overlooking. –  Fixee Mar 11 '12 at 2:50
    
@Fixee: The formulas are so ubiquitous they sometimes aren't even named, just taken for granted. The explanation is rather simple: expand $a_n(T-\theta_1)(T-\theta_2)\cdots(T-\theta_n)$ and you will find the coefficients of $p(T)$ as alternating elementary symmetric polynomials in the $\theta_i$'s. –  anon Mar 11 '12 at 2:53

Hint: The coefficient of $t^i$ in $p(t)$ is $(-1)^{n-i}a_ns_{n-i}(\theta_1,...,\theta_n)$ where $a_n$ is the coefficient of $t^n$ in $p(t)$. So we know that $s_i(\theta_1,...,\theta_n)\in K$.

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Far from a "hint", I think this completely answers the question. (And it was essentially given in a comment by @anon below his answer.) –  Fixee Mar 12 '12 at 0:20

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