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If $f(X)$ in $\mathbb{Q}[X]$ is an irreducible polynomial polynomial of degree $n \geq 2,$ with roots $\alpha_1, \alpha_2,\ldots,\alpha_n$ in $\mathbb{C},$ show that $\displaystyle \sum_{j=1}^{n}\frac{1}{\alpha_j^2}\in \mathbb{Q}.$

I have tried that

If $f(x) = b_nx^n+b_{n-1}x^{n-1}+\cdots+b_1x+b_0; ~ b_i \in \mathbb{Q} $ \begin{align} \sum_{j=1}^{n}\frac{1}{\alpha_j^2} &= \frac{1}{\alpha_1^2}+ \frac{1} {\alpha_2^2} + \cdots + \frac{1}{\alpha_n^2} \\ &=\frac{(\alpha_2\cdots \alpha_n)^2+(\alpha_1\alpha_3\cdots \alpha_n)^2+\cdots (\alpha_1\alpha_2\cdots \alpha_{n-1})^2} {(\alpha_1\alpha_2\cdots \alpha_{n})^2} \\ &= \frac{?}{(\alpha_1\alpha_2\cdots \alpha_{n})^2} \end{align} The denominator is $b_0^2$ and so rational. If I show that numerator is also rational then I am done. Is there any identity way which shows thats true. I am sure there must be but I don't exactly recall it. Also, is there any other way to prove the above result, specially using Galois theory. Thanks in advance.

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  • $\begingroup$ Do you know any theorems about symmetric polynomials (and perhaps rational functions)? $\endgroup$
    – David
    Jan 13 '16 at 6:30
  • $\begingroup$ Not exactly. I can take a look. Can you provide any reference? What result I need to look. $\endgroup$
    – Learner
    Jan 13 '16 at 6:50
  • $\begingroup$ The result is that any symmetric rational function of several variables is a rational function of the elementary symmetric polynomials. This theorem allows you to say in advance that a solution like Achille Hui's will be possible, but without writing down the equations. This theorem-for polynomials only-can be found in Artin's Algebra, as well as in most other algebra books at that level. A further argument is needed for symmetric rational functions, but in your case, it's okay because your rational function can easily be expressed as $f/g$ where $f$ and $g$ are symmetric polynomials. $\endgroup$
    – David
    Jan 13 '16 at 7:02
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Remember, if $\alpha_j$ are roots of $f(x)$, then $\frac{1}{\alpha_j}$ are roots of $x^n f\left(\frac{1}{x}\right)$.

A proof of $\sum_{j=1}^n \frac{1}{\alpha_j^2} \in \mathbb{Q}$ is not that different from a proof that $\sum_{j=1}^n \alpha_j^2 \in \mathbb{Q}$. Following the same spirit of the proof of the latter case, we have:

$$\sum_{j=1}^n \frac{1}{\alpha_j^2} = \left(\sum_{j=1}^n \frac{1}{\alpha_j}\right)^2 - 2\sum_{1\le i < j\le n} \frac{1}{\alpha_i\alpha_j} = \left(\frac{e_{n-1}}{e_n}\right)^2 - 2\frac{e_{n-2}}{e_n} $$ where $$e_k = \sum\limits_{1 \le j_1 < j_2 < \cdots j_k \le n} \prod\limits_{s=1}^k \alpha_{j_s} = (-1)^k\frac{b_{n-k}}{b_n}$$ is the $k^{th}$ elementary symmetric polynomial for the roots $\alpha_j$.

Since all $b_k \in \mathbb{Q}$, so does all $e_k$ and the expression you have.

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  • $\begingroup$ Thank you, this helps alot. $\endgroup$
    – Learner
    Jan 13 '16 at 6:51
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Using Galois theory, this result is trivial. Let $K = \mathbb{Q}(\alpha_1, \dots, \alpha_n)$. Then $K$ is the splitting field of $f$, hence is a Galois extension of $\mathbb{Q}$. Now if $c = \sum 1/\alpha_i^2$, then $c$ is fixed under any automorphism $\sigma$ of $K$, since $\sigma$ must permute the roots of $f$. Therefore $c$ belongs to the fixed field of the Galois group of $K$ over $\mathbb{Q}$. But that fixed field is $\mathbb{Q}$.

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  • $\begingroup$ @ David Thank you. .. $\endgroup$
    – Learner
    Jan 13 '16 at 6:52
  • $\begingroup$ Equivalently, the quantity in question is the trace of $1/\alpha_1^2$. $\endgroup$
    – Lubin
    Jan 14 '16 at 5:18

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