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.
Best Answer
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 $x=0$.