# When is a linear combination of the elementary symmetric polynomials reducible?

Let $n\ge 2$ and consider the polynomial ring $\mathbb F [X_1,...,X_n]$, where $\mathbb F$ is a field. Let $e_j:=e_j(X_1,...,X_n)$ be the elementary symmetric polynomial of degree $j$ in $X_1,...,X_n$ (https://en.wikipedia.org/wiki/Elementary_symmetric_polynomial).

Now is there a way to characterize for which $(c_0,...,c_n)\in \mathbb F^{n+1} \setminus \{0\}$, is the polynomial $f(X_1,...,X_n)=\sum_{j=0}^n c_je_j \in \mathbb F[X_1,...,X_n]$ reducible in $\mathbb F [X_1,...,X_n]$ ?

For example, if $n$-many among $c_0,...,c_n$ are zero, i.e. if we have $f=e_k$ for some $k$, then we must have $k=n$, because $e_1,...,e_{n-1}$ are all irreducible as seen here Is an elementary symmetric polynomial an irreducible element in the polynomial ring? . Apart from this, I don't know ...

Suppose that $f=p_1 p_2\cdots p_k$, where each $p_i$ is irreducible. Since $f$ is a symmetric function, for every $w\in S_n$ and every $i$ we must have $w\cdot p_i = cp_j$ for some $j$, where $c$ is a nonzero constant (depending on $w$ and $i$). In each factor $p_i$ pick out the term of highest degree $d$ that is first in lex order, e.g., for $d=3$ the lex order is $$X_1^3<X_1^2X_2<X_1^2 X_3<\cdots<X_1^2 X_n<X_1 X_2^2<\cdots<X_n^3.$$ Unless $p_i$ has the form $a+bX_k$ and no other factor is of this form (for the same $k$), the highest term (in lex order) of $f$ will have an exponent greater then one, so $f$ is not a linear combination of $e_j$'s. Thus we must have $$f=c(a+X_1)(a+X_2)\cdots (a+X_n).$$
• Thanks for your answer ... I have some trouble understanding all the "lex order" argument of yours ... can you please explicitly say how the lex order between two monomials is defined ? And then what do you mean by " pick out the term of highest degree $d$ that is first in lex order " ? – user111524 May 26 '18 at 4:43
• You do not even need lex order, just look at this product as a polynomial in, say, $x_1$. It must be linear, thus one factor is linear and other factors are constant. So, each factor $p_i$ is linear with respect to the variables from some set $A_i$, and these $A_i$ must be disjoint. But by symmetry, if $|A_j|=k$, all subsets of size $k$ must appear between $A_i$'s. This all is possible only if $A_i$'s have size 1. – Fedor Petrov May 26 '18 at 6:26
• @FedorPetrov: the $A_i$ s must be disjoint because in the expression for $f$, no $X_i$ has exponent more than $1$ ... am I right ? Although I do not understand why if some $|A_i|=k$, then all subsets of size $k$ must occur ... could you please explain this point ? – user111524 May 26 '18 at 7:14
• They are disjoint, since $x_1$ belongs to exactly one if them. The same holds for any other variable. The family of sets $A_i$'s must be invariant under permutations, this follows from the uniqueness of the factorization into irreducible factors. – Fedor Petrov May 26 '18 at 10:03