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.



Is a complete homogeneous symmetric polynomial irreducible?
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 $x=0$.


