# Is there a polynomial whose cube is a symmetric polynomial?

The discriminant of variables $$x_1, \dots, x_n$$, $$d=d(x_1, \dots, x_n)=\prod_{i is a classical example of a polynomial whose square $$d^2$$ is symmetric in its entries. This turns out to be very useful for understanding what do some field extensions look like. So the question is:

Does there exist a polynomial $$p(x_1, \dots,x_n)$$ which is not symmetric but $$p^3$$ is? More generally, for some $$k>1$$, is there a polynomial such that $$p, p^2, \dots, p^{k-1}$$ are not symmetric and $$p^k$$ is?

Nice question! The answer is no, there are no such polynomials.

Recall that if $$F$$ is a field, $$x_1, \dots x_n$$ are indeterminates and $$e_1, \dots e_n$$ are the elementary symmetric polynomials in the $$x_i$$, then $$F(x_1, \dots x_n)$$ is a Galois extension of $$F(e_1, \dots e_n)$$ with Galois group $$S_n$$ acting by permutation on the $$x_i$$ (in fact it is the splitting field of the polynomial $$\prod (t - x_i)$$ whose coefficients are the $$e_i$$). Below we'll need $$\text{char}(F) \neq 2$$ because we'll be dealing with a lot of minus signs.

In this language the existence of the discriminant can be explained as follows. Suppose there is a polynomial $$p \in F(x_1, \dots x_n)$$ which is not symmetric but such that $$p^2$$ is symmetric. Then $$p$$ would be a root of the quadratic polynomial $$t^2 - p^2$$, which is an irreducible polynomial over $$F(e_1, \dots e_n)$$ splitting over $$F(x_1, \dots x_n)$$. This means that the Galois group $$S_n$$ acts transitively on its roots, so for any $$\sigma \in S_n$$ we would have that either $$\sigma p = p$$ or $$\sigma p = -p$$, and the latter case must occur at least once. This produces a nonzero homomorphism $$S_n \to \{ \pm 1 \}$$, and it's a classic result that there is exactly one such homomorphism, the sign homomorphism. So $$p$$ must be a polynomial which satisfies $$\sigma p = \text{sgn}(\sigma) p$$, generating the unique quadratic extension of $$F(e_1, \dots e_n)$$ contained in $$F(x_1, \dots x_n)$$, which by the fundamental theorem of Galois theory must be the fixed field $$F(x_1, \dots x_n)^{A_n}$$ of the unique subgroup of index $$2$$ in $$S_n$$, namely the alternating group $$A_n$$, the kernel of the sign homomorphism.

(More explicitly, because $$S_n$$ is generated by transpositions and every transposition is conjugate, the above condition is equivalent to the condition that $$p$$ is sent to $$-p$$ by any transposition of two of the variables $$x_i$$. This actually implies that $$p$$ must vanish if any two of the variables $$x_i$$ are set equal, so as a polynomial it must be divisible by $$x_i - x_j$$ for all $$i \neq j$$!)

Now suppose $$p$$ is a polynomial such that $$p^k$$ is symmetric, $$k \ge 3$$. Arguing as above, $$p$$ is a root of the polynomial $$t^k - p^k$$, and so again $$S_n$$ acts on the roots of this polynomial (but not necessarily transitively). So if $$\sigma \in S_n$$ is a permutation then $$\sigma p$$ is another root of this polynomial, meaning $$(\sigma p)^k = p^k$$, which gives that $$\frac{\sigma p}{p}$$ is a $$k^{th}$$ root of unity. This produces a homomorphism

$$S_n \ni \sigma \mapsto \frac{\sigma p}{p} \in \mu_k(F) = \{ \zeta \in F : \zeta^k = 1 \}$$

from $$S_n$$ to the group of $$k^{th}$$ roots of unity in $$F$$. This group is abelian, so this homomorphism factors through the abelianization of $$S_n$$, but this abelianization is just $$C_2$$ (and the abelianization map $$S_n \to C_2$$ is the sign homomorphism). This gives that we must in fact have $$\sigma p = \pm p$$ for all $$\sigma \in S_n$$ exactly as above, meaning that $$(t - p)(t + p) = t^2 - p^2$$ must already have coefficients in $$F(e_1, \dots e_n)$$, so $$p^2$$ must already be symmetric.

Let $$A$$ be a domain in which the polynomial $$x^k-1$$ is split ( always OK if $$k\le 2$$). Say we have two elements in $$A$$ such that $$a^k = b^k$$. Then $$a=\epsilon \cdot b$$, with $$\epsilon^k = 1$$. Indeed, we have the factoring $$a^k - b^k = \prod_{\omega} (a - \epsilon b)$$

Back to our problem: $$P\in F[x_1, \ldots, x_n]$$, $$P^k$$ symmetric, means $$\phi(P) =\epsilon(\phi)\cdot P$$ for every $$\phi \in S_n$$, where $$\epsilon\colon S_n \to \mu_k$$, is a group morphism, $$\mu_k$$ the $$k$$-th roots of $$1$$, and proceed as elsewhere.