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

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.

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You can reduce to the case $n=3$ using an induction argument, but I can't immediately see how to solve $n=3$. –  Will Sawin May 26 '12 at 15:51
You should have a look to McDonald's book about symmetric and Hall polynomials –  Denis Serre May 26 '12 at 16:14
View the polynomial as a monic polynomial in $x_n$. The leading term is $1$, the constant term is the polynomial with $a$ the same and $n$ one less. By induction on $n$, this is irreducible, so each factor must have constant term $1$ or that polynomial. But since the polynomial is homogeneous, the factors are homogeneous, thus degree $0$ or degree $a$, thus no nontrivial factors. –  Will Sawin May 26 '12 at 16:30
For the case $n=3$ it would suffice to prove that the projective curve $h_a(x,y,z)=0$ is smooth, in other words that $h_a$ has no common zero with all its partial derivatives in $\mathbf{P}^2(\mathbf{C})$. –  François Brunault May 26 '12 at 16:51
@Will Sawin: Thank you for your explanation! And I also agree with Patricia that answers should not be hidden as comments. –  Martin Brandenburg May 26 '12 at 20:08

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 $x=0$.

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There is no reason this argument should fail for $n>3$. –  Will Sawin May 26 '12 at 19:26
@Will Sawin : Thanks for pointing this out ! The same argument should prove that the hypersurface $h_a(x_1,\ldots,x_n)=0$ is smooth for any $n \geq 3$, which implies that $h_a$ is irreductible. Alternatively, your argument reduces the general case to the case $n=3$. –  François Brunault May 26 '12 at 19:35
Thanks Brunault and Sawin –  Neeraj May 26 '12 at 22:26

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.

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It is well known in symmetric functions that $h_1,…,h_n$ are algebraically independent in ${\bf Q}[x_1,…,x_n]$ and that the symmetric functions $h_{\lambda }=h_{\lambda_1}⋯h_{\lambda_k}$ for $\lambda = (\lambda_1,\dots ,\lambda_k)$ with $\lambda_1\ge \lambda_2\ge \cdots \ge\lambda_k$ and $\sum \lambda_i = d$ form a vector space basis for the degree $d$ homogeneous piece of the ring of symmetric functions in $n$ variables. Hence my claim about needing nonsymmetric factors if there were going to be a nontrivial factorization. –  Patricia Hersh May 26 '12 at 17:39

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

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A related paper of Schinzel seems to be *Reducibility of a special symmetric form, Acta Mathematica Universitatis Ostraviensis, vol. 14 (2006), issue 1, pp. 71-74 (dml.cz/handle/10338.dmlcz/137486) –  Chandan Singh Dalawat May 26 '12 at 16:58
The main theorem of that paper requires, in our notation, $n>a+1$, the case Patricia Hersh already handled. –  Will Sawin May 26 '12 at 17:00
This makes me curious for $n\le a+1$ if there is any more general result (i.e. not just for homogeneous symmetric functions) about when irreducibility in the ring of symmetric functions implies irreducibility in ${\bf Q}[x_1,\dots ,x_n]$. But then I guess there are examples like the square of the Vandermonde determinant that limit how much one can hope for. –  Patricia Hersh May 26 '12 at 19:52

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

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For generally, for $a>2$ odd, we have that $h_a = \sum_{i=0}^{a} x^i y^{a-i}$ is reducible since it is normed in $x$ and has $-y$ as a zero. For $a$ even I think it will be irreducible ... –  Martin Brandenburg May 26 '12 at 16:27
One might want to go a little bit further, as Neeraj specified $n\geq 3$. –  Will Sawin May 26 '12 at 16:28
Oh, right. For $n \geq 3$ these trivial examples vanish. –  Martin Brandenburg May 26 '12 at 16:31
On $a$ even: We are working over $\mathbb C$. Every polynomial in two variables, homogenous in degree at least two, is reducible. –  Will Sawin May 26 '12 at 16:38

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!

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