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Been sitting with this question from for a whole bunch of hours now. I'm studying for an exam so I don't want to be stuck for too long, however much I dig endlessly pondering.

Let $f(t)$ be a separable polynomial in $\Bbb Q[t]$ with zeros $\alpha_i,i=1,\dots,4$. Let \begin{align*} \beta_1=\alpha_1\alpha_2+\alpha_3\alpha_4\\ \beta_2=\alpha_1\alpha_3+\alpha_2\alpha_4\\ \beta_3=\alpha_1\alpha_4+\alpha_2\alpha_3 \end{align*} and $g(t)=(t-\beta_1)(t-\beta_2)(t-\beta_3)$. Show that $g(t)\in \Bbb Q[t]$.

So, since the coefficients of $g$ are \begin{align*} \beta_1+\beta_2+\beta_3\\ \beta_1\beta_2+\beta_1\beta_3+\beta_2\beta_3\\ \beta_1\beta_2\beta_3 \end{align*} I just want to show that they're rational. A common approach is to try to express them in terms of \begin{align*} s_1&=\alpha_1+\alpha_2+\alpha_3+\alpha_4\\ s_2&=\alpha_1\alpha_2+\alpha_1\alpha_3+\alpha_1\alpha_4+\alpha_2\alpha_3+\alpha_2\alpha_4+\alpha_3\alpha_4\\ s_3&=\alpha_1\alpha_2\alpha_3+\alpha_1\alpha_2\alpha_4+\alpha_1\alpha_3\alpha_4+\alpha_2\alpha_3\alpha_4\\ s_4&=\alpha_1\alpha_2\alpha_3\alpha_4 \end{align*} which I know are rational 'cause they're the coefficients of $f$. But this is where I don't get any further. I know that if the coefficients of $g(t)$ are symmetric polynomials in the roots, then they can be expressed in terms of $s_1,s_2,s_3,s_4$ but f.ex. $\beta_1+\beta_2+\beta_3$ for sure ain't symmetric, since with $\begin{pmatrix}1 & 2 & 4 & 3\end{pmatrix}$ we get the term $\alpha_2\alpha_4$ which isn't a term of $\beta_1+\beta_2+\beta_3$.

What do you think? Am I not thinking enough?

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    $\begingroup$ Are you sure that your definition of $\beta_2$ is correct? I think it should be $\alpha_1\alpha_3+\alpha_2\alpha_4$. $\endgroup$
    – Alex Fok
    Dec 21, 2015 at 9:44
  • $\begingroup$ Haha, I was sure beginning to think there was a typo. I did the corresponding change of $\beta_1$ but that didn't yield much. I'll try yours too and report back. $\endgroup$ Dec 21, 2015 at 10:41
  • $\begingroup$ @AlexFok It worked, is there any other way of fixing the typo up? $\endgroup$ Dec 21, 2015 at 12:01
  • $\begingroup$ I made the correction for you. $\endgroup$
    – Lubin
    Dec 22, 2015 at 0:25
  • $\begingroup$ You don’t need to show that the $\beta_j$ are individually invariant under any permutation of the $\alpha_i$, but just that $\beta_1+\beta_2+\beta_3$ is, for instance. And the others you quote. $\endgroup$
    – Lubin
    Dec 22, 2015 at 0:27

1 Answer 1

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You only want to show that the triple $\{\beta_1,\beta_2,\beta_3\}$ is invariant under all permutations of the roots $\{\alpha_i\}$, since the fixed field of $\Bbb Q(\alpha_1,\cdots,\alpha_4)$ under the permutation group $S_4$ is the field of the elementary symmetric functions in the $\{\alpha_i\}$, which are the coefficients of the polynomial, assumed to be in $\Bbb Q$.

So it only remains to show that every simple exchange of two of the $\alpha_i$ leaves the set $\{\beta_1,\beta_2,\beta_3\}$ unchanged. For instance, $\alpha_1\leftrightarrow\alpha_2$ leaves $\beta_1$ fixed, but interchanges $\beta_2$ and $\beta_3$. You’ll find the same result for any other simple exchange. So all is well.

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