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 $m, r$.


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.

  • $\begingroup$ Sorry, how do you get from \sigma_{m,r-1} - x_i d/dx_i \sigma_{m,r-1} = 0 to just x_i d/dx_i \sigma_{m,r-1} = 0? $\endgroup$ – Izaak Meckler Mar 9 '17 at 22:56
  • $\begingroup$ @IzaakMeckler if you add all of them up, you get $\sigma_{m,r-1}=0$. $\endgroup$ – Gjergji Zaimi Mar 9 '17 at 22:57
  • $\begingroup$ Got it! Do you know if this result holds in characteristic > 0 as well? This proof breaks down I guess since r might not be invertible. $\endgroup$ – Izaak Meckler Mar 9 '17 at 23:57
  • $\begingroup$ It fails already in characteristic 2. The singular set of $\sigma_{5,2}$ in $\mathbb F_2$ contains $(1,1,1,1,1)$. $\endgroup$ – Gjergji Zaimi Mar 10 '17 at 0:01

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