9
$\begingroup$

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 $n=2$:

$e_1=x_1+x_2$;
$e_2=x_1x_2$.

For $n=3$:

$e_1=x_1+x_2+x_3$;
$e_2=x_1x_2+x_1x_3+x_2x_3$,
$e_3=x_1x_2x_3.$

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.

Thanks.

Improve this question
$\endgroup$
1

2 Answers 2

Reset to default
12
$\begingroup$

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.

Improve this answer
$\endgroup$
5
  • 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.). $\endgroup$
    – Martin Brandenburg
    Commented 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
    Commented 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. $\endgroup$
    – darij grinberg
    Commented Feb 15, 2018 at 1:36
  • 1
    $\begingroup$ @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? $\endgroup$
    – Steven Landsburg
    Commented 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. $\endgroup$
    – darij grinberg
    Commented Feb 15, 2018 at 2:52
4
$\begingroup$

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.

Improve this answer
$\endgroup$
3
  • $\begingroup$ In all fairness, this is rather a comment than an answer, isn't it? $\endgroup$
    – Vladimir Dotsenko
    Commented May 31, 2012 at 13:49
  • $\begingroup$ A hint? I thought there was an elementary inductive proof. $\endgroup$
    – Charles Matthews
    Commented May 31, 2012 at 14:21
  • $\begingroup$ I did not state the contrary :-) $\endgroup$
    – Vladimir Dotsenko
    Commented May 31, 2012 at 14:45

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .