Let $\sigma\_{m, r}$ be the degree-$r$ elementary symmetric polynomial in $m$ variables. Let $X\_{m, r}$ be the zero set of $\sigma\_{m, r}$ and $S\_{m, r}$ its singular locus. I.e.,
$S\_{m,r}$ is the set of common zeroes of $\sigma\_{m, r}$ and its partials $\frac{\partial}{\partial x\_1} \sigma\_{m,r}, \cdots, \frac{\partial}{\partial x\_m} \sigma\_{m,r}$.


Is it the case that for $r > 2$, $S\_{m,r}$ is the set of points with $m + 2 - r$ coordinates equal to 0? I've verified this in Mathematica for small $m, r$.



Singular locus of zero set of elementary symmetric polynomial
Notice that the vanishing of $\frac{\partial}{\partial x\_1} \sigma\_{m,r}, \cdots, \frac{\partial}{\partial x\_m} \sigma\_{m,r}$ implies the vanishing of $\sigma\_{m,r}$ since 
$$\sigma\_{m,r}=\frac{1}{r}\left( x\_1\frac{\partial}{\partial x\_1} \sigma\_{m,r}+ \cdots + x\_m\frac{\partial}{\partial x\_m} \sigma\_{m,r}\right),$$
so we can just focus on the system of equations $\frac{\partial}{\partial x\_i}\sigma\_{m,r}=0$ as the defining equations for $S\_{m,r}$. We can give a proof by induction. Start with $r=2$, where the defining equations become 
$$\frac{\partial}{\partial x\_i}\sigma\_{m,r}=\left(\sum\_{k=1}^m x\_k\right)-x\_i=0$$
therefore $x\_i=0$ for all $i$. For larger $r$ we proceed as follows. First we notice that
$$\frac{\partial}{\partial x\_i}\sigma\_{m,r}=\sigma\_{m,r-1}-x\_i(\frac{\partial}{\partial x\_i}\sigma\_{m,r-1}).$$
So the system of equations is equivalent to
$$x\_i\frac{\partial}{\partial x\_i}\sigma\_{m,r-1}=0.$$
Suppose that $q$ of the variables vanish, and the remaining $m-q$ are nonzero. If $m-q\geq r-1$ then the elementary symmetric function are nontrivial, and by induction (use the statement for $m'=m-q, r'=r-1$) we have that $x\_i\frac{\partial}{\partial x\_i}\sigma\_{m,r-1}$ cannot all be zero. Now we can conclude that $m-q\le r-2$ or in other words $q\geq m+2-r$. Of course we can check that this is in fact enough, because if $m+2-r$ variables vanish then every monomial in our equations vanishes.


