Factors of a symmetric polynomial

Question

Is any factor of a symmetric polynomial still symmetric ?? The field is set over $\mathbb Q$. While over $\mathbb C$, I've known the counterexample $x^2 + y^2 = (x+iy)(x-iy)$.

Answer 1

You can produce counterexamples over any field: given a non-symmetric polynomial $P(x,y)$, the polynomial $Q(x,y)=P(x,y)\cdot P(y,x)$ is symmetric but has non-symmetric factors.

In fact this is pretty much the only way of producing counterexamples : if a symmetric polynomial $Q(x,y)$ is a product of irreducible factors $P_1(x,y)\cdots P_m(x,y)$, then for any $1\leq j\leq m$ $P(y,x)$ must divide $Q(y,x)=Q(x,y)$, hence $P_j(y,x)=P_k(x,y)$ for some $k$. Thus any symmetric polynomial can be written as a product of symmetric polynomials and polynomials of the form $P(x,y)\cdot P(y,x)$.