Is a complete homogeneous symmetric polynomial irreducible? 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.
 A: 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.
A: To go further in Patricia's direction, take $n=2$ and $a=3$. Then $h_a=(x+y)(x^2+y^2)$.
A: 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
A: 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 $x=0$.
A: 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!
