Let $S=\mathbb{C}[x_1,x_2,\dots,x_n]$ be a polynomial ring. Let $e_a$ denotes the elementary symmetric polynomials of degree $a$ in $S$.

For $n=2$:


For $n=3$:


In general for any $n$ and $a$, one has $$ e_a(x_1,x_2,\dots,x_n):=\sum_{1 \leq i_{1} < i_{2} < \cdots < i_a \leq n} x_{i_1}x_{i_2}\cdots x_{i_a} $$

Question: Let $n \in \mathbb{N}$ with $n \geq 3$. Is it true that $e_a$ is an irreducible element in $\mathbb{C}[x_1,x_2,\dots,x_n]$ for $a=2,3,\dots,{n-1}$.

For $n=1$, $e_1$ is an irreducible element. For $n=2$, $e_1$ is an irreducible element. For $n=3$, $e_1$ and $e_2$ are irreducible element.

Fact: $e_1$ is be definition, an irreducible element. And, $e_n$ is trivially reducible. My Question is therefore, to know, whether $e_2,e_3,\dots,e_{n-1}$ are irreducible elements in $\mathbb{C}[x_1,x_2,\dots,x_n]$ for $n \geq 4$.

Similar results: Power sum symmetric polynomials and complete homogeneous symmetric polynomials are irreducible elements in $\mathbb{C}[x_1,x_2,\dots,x_n]$ for $n \geq 3$. For complete symmetric polynomial, see Is complete homogeneous symmetric polynomials, an irreducibile element in Polynomial ring?.

Therefore it is natural to ask for the elementary symmetric polynomials.



2 Answers 2


For $\alpha\neq n$, the symmetric polynomial is of the form $f\cdot x_n + g$ where $f,g$ are non-zero elements of $A={\mathbb C}[x_1,...,x_{n-1}]$ with no common factor.

Thus $${\mathbb C}[x_1,...,x_n]/(e_\alpha)=A[x_n]/(f x_n+g)=A[g/f]\subset K$$

where $K$ is the quotient field of $A$. It follows that ${\mathbb C}[x_1,...,x_n]/(e_\alpha)$ is a domain, so $e_\alpha$ is irreducible.

  • 3
    $\begingroup$ Basically one is already done with the first sentence. If $R$ is an integral domain, then elements $fx+g$ are irreducible $R[x]$ iff $f,g$ are coprime (only need to worry about constant factors, etc.). $\endgroup$ May 31, 2012 at 15:39
  • $\begingroup$ I think the solution given by Steven Landsburg is a complete proof. I do agree with the Martin suggestion as well. Thanks. $\endgroup$
    – Neeraj
    May 31, 2012 at 16:38
  • $\begingroup$ Late to the party, but why exactly is $A\left[x_n\right] / \left(fx_n+g\right) = A\left[g/f\right]$ (an isomorphism of $A$-algebras, I presume)? I see that the canonical $A$-algebra homomorphism $A\left[x_n\right] / \left(fx_n+g\right) \to A\left[g/f\right]$ is surjective, but its injectivity is not clear to me (nor do I see how to construct its inverse). I agree with Martin Brandenburg's argument, though. $\endgroup$ Feb 15, 2018 at 1:36
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    $\begingroup$ @darijgrinberg Suppose $p=\sum a_ix^i$ maps to $0$. Then $\sum a_ig^i/f^i=0$. Multiplying through by the right power of $f$ gives $\sum a_ig^if^{n-i}=0$. Then because $A$ is a $UFD$ and $g,f$ have no factors in common, we must have $f|a_n$. Write $a_n=fb$, and now modify the expression for $p$ by eliminating $a_n$ and replacing $a_{n-1}$ with $a_{n-1}+gb$. This reduces the minimum degree of an expression for an element in the kernel, so we are done by induction. Have I missed something? $\endgroup$ Feb 15, 2018 at 2:43
  • $\begingroup$ @StevenLandsburg: Oh, thanks. I abstracted too much and didn't think the UFDness of $A$ would be useful. $\endgroup$ Feb 15, 2018 at 2:52

Doesn't this follow quite quickly by setting one variable equal to 0?

Edit: I was thinking this way. Factors of homogeneous polynomials are homogeneous. Setting the final variable $x_n$ to 0 therefore deals easily in an inductive proof except for the case of $e_a$ with a = n-1. There you have to divide the variables with index up to n-1 into two subsets, and consider the product to two factors of the type "monomial + $x_n$ times something". Because the square of $x_n$ can't actually occur in the product, one of the factors is a monomial; and this is going to be a contradiction except if it is a constant.

  • $\begingroup$ In all fairness, this is rather a comment than an answer, isn't it? $\endgroup$ May 31, 2012 at 13:49
  • $\begingroup$ A hint? I thought there was an elementary inductive proof. $\endgroup$ May 31, 2012 at 14:21
  • $\begingroup$ I did not state the contrary :-) $\endgroup$ May 31, 2012 at 14:45

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