In the Problems from the Book by Titu Andreescu, there is a proof of Example 9 on page 494 with the following:

Example 9. Let $f$ be a monic polynomial with integer coefficients and let $p$ be a prime number. If $f$ is irreducible in $\mathbb{Z}[x]$ and $\sqrt[p]{(-1)^{\deg f}f(0)}$ is irrational, then $f(X^p)$ is also irreducible in $\mathbb{Z}[x]$.

Solution. Consider $\alpha$ a complex zero of $f$ and let $n=\deg f$ and $g(X)=X^p$ and $h=g-\alpha$. Using previous results, it suffices to prove that $h$ is irreducible in $\mathbb{Q}[\alpha][X]$. Because $\mathbb{Q}[\alpha]$ is a subfield of $\mathbb{C}$, it suffices to prove that $\alpha$ is not the $p-th$ power of an element of $\mathbb{Q}[\alpha]$. Suppose there is $u\in \mathbb{Q}[x]$ of degree at most $n-1$ such that $\alpha=u^p(\alpha)$ . Let $\alpha_1, \alpha_2, \dots, \alpha_n$ be the zeroes of $f$. Because $f$ is irreducible and $\alpha$ is one of its zeroes, $f$ is the minimal polynomial of $\alpha$, so $f$ must divide $u^p(X)-X$. Therefore $\alpha_1\cdot\alpha_2\cdots\alpha_n=(u(\alpha_1)\cdot u(\alpha_2)\cdots u(\alpha_n))^p.$ Finally, using the fundamental theorem of symmetric polynomials, $u(\alpha_1)\cdot u(\alpha_2)\cdots u(\alpha_n)$ is rational. But $\alpha_1\cdot \alpha_2\cdots \alpha_n=(-1)^nf(0)$, implies $\sqrt[p]{(-1)^nf(0)} \in \mathbb{Q}$, a contradiction.

My question is how does the fundamental theorem of symmetric polynomials imply that $u(\alpha_1)\cdot u(\alpha_2)\cdots u(\alpha_n)$ is rational?

The theorem says that any symmetrical polynomial can be uniquely expressed as a polynomial in elementary symmetric polynomials (one formulation is here The Fundamental Theorem of Symmetric Polynomials), but how does that apply in this case and leads to rationality?

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    $\begingroup$ @markvs No, $f \ne u$. $\endgroup$ Nov 24, 2021 at 22:23
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    $\begingroup$ I guess $f$ is irriducible of $\mathbb Q$, right? $\endgroup$ Nov 24, 2021 at 22:24
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    $\begingroup$ If $f$ has rational coefficients, then by Vieta theorem, all the elementary symmetric functions applied to roots of $f$ give rational numbers. Then by the fund. theorem, all symmetric polynomial functions with rational coeffs applied to roots give rational numbers. The function $x_1x_2...x_n$ is symmetric. $\endgroup$
    – markvs
    Nov 24, 2021 at 22:30
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    $\begingroup$ Isn't the solution a bit misleading? To get that $u(\alpha_1)\cdots u(\alpha_n) \in \mathbb Q$ you only need to know that $\alpha_1,\dots,\alpha_n$ are the roots of a polynomial of degree $n$ with coefficients in $\mathbb Q$, right? $\endgroup$ Nov 24, 2021 at 22:43
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    $\begingroup$ To show that $u(\alpha_1)\cdots u(\alpha_n) \in \mathbb Q$ you don't need symmetric polynomials at all. Simply take a field homomorphism and apply it to $u(\alpha_1)\cdots u(\alpha_n)$ and realize that it is fixed because it only permutes the $\alpha_i$. $\endgroup$ Nov 24, 2021 at 22:48


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