# How does the fundamental theorem of symmetric polynomials imply that this number is rational?

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?

• @markvs No, $f \ne u$. Nov 24, 2021 at 22:23
• I guess $f$ is irriducible of $\mathbb Q$, right? Nov 24, 2021 at 22:24
• 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. Nov 24, 2021 at 22:30
• 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? Nov 24, 2021 at 22:43
• 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 $\alpha_i$. Nov 24, 2021 at 22:48