Is complete homogeneous symmetric polynomials, an irreducibile element in Polynomial ring? - MathOverflow

Is complete homogeneous symmetric polynomials, an irreducibile element in Polynomial ring?

2012-05-26T14:31:58Z

Let $S=\mathbb{C}[x_1,x_2,\dots,x_n]$ be a polynomial ring. Let $n \geq 3$. Let $h_a$ denotes the complete homogeneous symmetric polynomial of degree $a$. $$ h_a=\text{ sum of all monomials of degree } a.$$ For example: for $n=3$ and $a=2$, one has: $$h_2=x_1^2+x_2^2+x_3^2+x_1x_2+x_1x_3+x_2x_3.$$ Question: Is it true that $h_a$ is an irreducible element in $\mathbb{C}[x_1,x_2,\dots,x_n]$.

The $h_a$ was introduced by Sir Issac Newton in seventeenth centuary along with many other symmetric polynomials such as Power sum symmetric polynomials and elementary symmetric polynomials.

It is known that $p_a=x_1^a+\cdots+x_n^a$ is an irreducible element in $\mathbb{C}[x_1,x_2,\dots,x_n]$ for $n \geq 3$. I am interested to know similar result for the complete homogeneous symmetric polynomial.

Thank you
Neeraj Kumar.

Answer by Patricia Hersh for Is complete homogeneous symmetric polynomials, an irreducibile element in Polynomial ring?

2012-05-26T15:19:43Z

You were probably mainly interested in the case of $a\le n$. Here's a quick counterexample for $a>n$, letting $n=1$ and $a=2$. Notice then that $h_2 = (h_1)^2$.

If there were a factorization with $a\le n$, it would need to involve nonsymmetric polynomials. But since $S$ is a unique factorization domain, this would mean for any nonsymmetric factor, we would also need all the elements of its orbit as factors. While I was editing this, Will Sawin came up with an elegant proof, whereas I was just grinding through cases, so I will omit those cases.

Since Will mentions still needing the base case of $n=a=3$, notice that we would need a nonsymmetric factor, which means a nonsymmetric factor with either (a) terms of the form $x_ix_j^2$, with nontrivial orbit, yielding too high a degree, or (b) it has terms of the form $x_i$ with some included and some omitted, contradicting the product including all $x_i^3$ terms, or (c) it has some terms $x_ix_j$ with $i\ne j$ and omits others, again necessitating other factors in the orbit, giving too high a degree for the product, or (d) it has some terms $x_i^2$ and not others, giving the same contradiction.

Answer by Denis Serre for Is complete homogeneous symmetric polynomials, an irreducibile element in Polynomial ring?

2012-05-26T16:15:48Z

To go further in Patricia's direction, take $n=2$ and $a=3$. Then $h_a=(x+y)(x^2+y^2)$.

Answer by Igor Rivin for Is complete homogeneous symmetric polynomials, an irreducibile element in Polynomial ring?

2012-05-26T16:43:28Z

This is most relevant to Patricia's answer: There is a paper by Schinzel where he addresses the question of whether you have to factorize into symmetric factors (the answer is yes, under some conditions). I have no access to the paper, but the math review is useful

Answer by Will Sawin for Is complete homogeneous symmetric polynomials, an irreducibile element in Polynomial ring?

2012-05-26T16:55:51Z

We can extend Patricia's argument somewhat. For $n=3$ and $a=4$, symmetrically it is $s_1^4-3s_1^2s_2+s_2^2+2s_1s_3$ which is irreducible (view it as a polynomial in s_2, complete the square, constant term is not a perfect square) so there are no symmetric factors, so it must have 3 asymmetric factors, which must be linear, then another linear symmetric factor. But there are no linear symmetric factors!

Answer by François Brunault for Is complete homogeneous symmetric polynomials, an irreducibile element in Polynomial ring?

2012-05-26T17:41:50Z

EDIT : prompted by Will Sawin's comment, the argument now works for every $n \geq 3$. Thanks !

The polynomial $h_a(x_1,\ldots,x_n)$ is irreducible for every $a \geq 1$ and $n \geq 3$.

Recall that if $h_a = FG$ with $F$ and $G$ non constant then $F$ and $G$ have to be homogenous. By Bézout's theorem, the hypersurfaces $F=0$ and $G=0$ intersect in the projective space $\mathbf{P}^{n-1}(\mathbf{C})$ since $n \geq 3$. This gives a singular point on the hypersurface $h_a=0$. So it suffices to prove that $h_a,\frac{\partial h_a}{\partial x_1},\ldots,\frac{\partial h_a}{\partial x_n}$ have no common zero in $\mathbf{C}^n \backslash \{0\}$ . This fact is true for every $a \geq 1$ and $n \geq 2$, and we prove this by induction.

For $a=1$ it is easy. For $n=2$ it amounts to the fact that the polynomial $T^a+\cdots+T+1 = (T^{a+1}-1)/(T-1)$ has distinct roots.

In general, we have $$h_a = \sum_{a_1+\cdots+a_n=a} x_1^{a_1} \cdots x_n^{a_n}$$ so that $$\frac{\partial h_a}{\partial x_i} = \sum_{a_1+\cdots+a_n=a-1} (a_i+1) x_1^{a_1} \cdots x_n^{a_n}.$$ Note that $\sum_{i=1}^n \frac{\partial h_a}{\partial x_i} = (a+n-1) h_{a-1}$. Moreover $h_a=x_i h_{a-1}+R$ for some polynomial $R$ not depending on $x_i$, so that $$\frac{\partial h_a}{\partial x_i}=h_{a-1}+x_i \frac{\partial h_{a-1}}{\partial x_i}.$$ If $x=(x_1,\ldots,x_n)$ is a common zero of $h_a$ and all its partial derivatives then $h_{a-1}(x)=0$ and $x_i \frac{\partial h_{a-1}}{\partial x_i}(x)=0$ for all $i$. By induction, we must have $x_i=0$ for some $i$. Assume for example $x_n=0$. Then $(x_1,\ldots,x_{n-1}) \in \mathbf{C}^{n-1}$ provides in fact a common zero of $h_a(x_1,\ldots,x_{n-1})$ and all its partial derivatives, so applying the induction hypothesis for $n-1$ we get $x=0$.