Proving that $r = \displaystyle\sum_{i=1}^k \frac{1}{\alpha_i} \in \mathbb{Q}$

Question

Let $f(x) \in \mathbb{Q}[x]$ be an irreducible polynomial of degree $n$ with roots $\alpha_1, \alpha_2, \dots, \alpha_k$ in some splitting field $K$, none of which are $0$. Prove that $r = \displaystyle\sum_{i=1}^k \frac{1}{\alpha_i} \in \mathbb{Q}$.

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This question is very similar in flavor to this post which is identical except the user there had $r=\displaystyle\sum_{i=1}^n \frac{1}{\alpha_i^2}$ with emphasis on the square. An answer to that post said:

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 $\mathbb{Q}$.

I was wondering why this answer is justified. I certainly agree that $\sigma \in \mathrm{Gal}(K/\mathbb{Q})$ permutes every root $\alpha_i$ $-$ but how does it follow that $\sum_i 1/\alpha_i$ is rational? We know that $\alpha_i$ resides in the splitting field $K$, but $\alpha_i$ may not necessarily be rational.

Answer 1

Let us first assume that $0$ is not a root of $f(x)$.

1. You can easily see that $c=\sum_{i=0}^k \frac{1}{\alpha_i}$ is invariant under any element of Gal$(K/\mathbb{Q})$. Thus $c$ belongs to the fixed field of Gal$(K/\mathbb{Q})$, but that field is precisely $\mathbb{Q}$.

2. If you prefer an alternate method, first, let $g(x) = \sum_{i=0}^k a_ix^{k-i}$, where $f(x)$ is defined $f(x):= \sum_{i=0}^k a_ix^i$. Next, let $h(x)$ be $g(x)$ divided by a scalar such that $h(x)$ is monic. Then on the one hand, $\alpha$ a root of $f(x)$ with multiplicity $\ell_i$ $\iff$ $\frac{1}{\alpha_i}$ a root of $h(x)$ with multiplicity $\ell_i$. On the other hand, $h(x)$ is in $\mathbb{Q}(x)$. Then $\sum_{i=1}^k \frac{1}{\alpha_i}$ is simply the coefficient of the $x^{k-1}$-term of $h(x)$.

3. Still another way: Writing $f$ as above with $f(x):= \sum_{i=0}^k a_ix^i$ [and $a_k=1$], note the equation $$a_1 = \sum_{i=1}^k \Big(\frac{1}{\alpha_i} \times (\alpha_1 \alpha_2 \cdots \alpha_k)\Big)$$ $$= \sum_{i=1}^k \Big(\frac{1}{\alpha_i} \times a_0\Big) = a_0\sum_{i=1}^k \frac{1}{\alpha_i},$$where recall $a_0,a_1$ are as in the equation $f(x)=\sum_{i=0}^ka_ix^i$. [I trust you are aware of how to express each of the coefficients of a monic polynomial $f$ as a symmetric functions of the roots of $f$.] But as $f(x)$ is in $\mathbb{Q}[x]$ and so $a_0,a_1$ are both rational, it follows that $\sum_{i=1}^k \frac{1}{\alpha_i}$ must also be rational.