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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.


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up vote 8 down vote accepted

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.

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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.). – Martin Brandenburg May 31 '12 at 15:39
I think the solution given by Steven Landsburg is a complete proof. I do agree with the Martin suggestion as well. Thanks. – Neeraj May 31 '12 at 16:38

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.

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In all fairness, this is rather a comment than an answer, isn't it? – Vladimir Dotsenko May 31 '12 at 13:49
A hint? I thought there was an elementary inductive proof. – Charles Matthews May 31 '12 at 14:21
I did not state the contrary :-) – Vladimir Dotsenko May 31 '12 at 14:45

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