Elementary Symmetric Polynomials and Determinant

Question

I am trying to show that \begin{equation} \sqrt{\Delta} = \prod\limits_{1 \leq i < j \leq n} \left( x_i - x_j \right) = \det \begin{pmatrix} \dfrac{\partial \sigma_{n,1}}{\partial x_1} & \cdots & \dfrac{\partial \sigma_{n,n}}{\partial x_1} \\ \vdots & \ddots & \vdots \\ \dfrac{\partial \sigma_{n,1}}{\partial x_n} & \cdots & \dfrac{\partial \sigma_{n,n}}{\partial x_n} \end{pmatrix} \end{equation} where $\sigma_{n,j}$ are the elementary symmetric polynomials of $n$ variables $x_1, x_2, · · · , x_n$ Does anyone know of another method to do so other than induction since I am not sure how the inductive step works out~

Answer 1

We work over the ring $R=\mathbb{Z}[x_1,\cdots,x_n]$.

Let $$P=\prod\limits_{1 \leq i < j \leq n} \left( x_i - x_j \right)\in R,\qquad\qquad D= \begin{pmatrix} \dfrac{\partial \sigma_{n,1}}{\partial x_1} & \cdots & \dfrac{\partial \sigma_{n,n}}{\partial x_1} \\ \vdots & \ddots & \vdots \\ \dfrac{\partial \sigma_{n,1}}{\partial x_n} & \cdots & \dfrac{\partial \sigma_{n,n}}{\partial x_n} \end{pmatrix}\in M_n(R)$$

For $i\neq j$, the $i$'th row of $D$ may be obtained from the $j$'th row, by replacing every occurence of $x_i$ with $x_j$. Thus if we subtract the $j$'th row from the $i$'th row, we obtain a row where every entry is divisible by $x_i-x_j$. As the determinant of the matrix is invariant under such operations, we see that ${\rm Det}(D)$ is divisible by $x_i-x_j$ for all $i\neq j$.

As $R$ is a UFD, we conclude that $P$ divides ${\rm Det}(D)$. That is we have some $C\in R$ with $PC={\rm Det}(D)$.

Write $C=C_0+\cdots+C_k$, where each $C_i$ is a homogeneous element of $R$, of degree $i$. Note that both $P$ and ${\rm Det}(D)$ are homogeneous elements of $R$ of degree $n(n-1)/2$. As the product of two homogeneous elements of $R$, of degrees $i,j$ is a non-zero homogeneous element of $R$ of degree $i+j$, we see that $C=C_0\in \mathbb{Z}$.

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Edit I originally showed $C_0=1$ through direct calculation of a coefficient. This calculation remains intact below this edit. However a slicker method by induction was pointed out by @user26857 and explained by @dxiv (see their comments below). Their method:

Let $R_n=R, P_n=P, c_n=C_0, D_n=D$. We argue by induction. Suppose we have shown that $c_{n-1}=1$, so that $P_{n-1}={\rm Det}(D_{n-1})$. We will show that $c_n=1$.

Let $\phi\colon R_n \to R_{n-1}$ denote the map sending $x_n\mapsto 0$ and restricting to the identity on $R_{n-1}$.

Then $\phi(P_n)=x_1\cdots x_{n-1} P_{n-1}$ and $\phi(D_n)$ is the matrix $D_{n-1}$ with an extra row and column attached at the end, where the only non-zero entry in the final column is the diagonal entry $x_1\cdots x_{n-1}$ at the bottom.

We have: \begin{eqnarray}\phi(P_n)&=&c_n{\,\,\rm Det}(\phi(D_n))\\&=&c_nx_1\cdots x_{n-1}{\,\,\rm Det}(D_{n-1})\\&=&c_n x_1\cdots x_{n-1}P_{n-1}\\&=&c_n\phi(P_n),\end{eqnarray} so $c_n=1$ as required.

For the base case note that $$(x_1-x_2)={\,\rm Det}\left(\begin{array}{cc} 1&x_2\\1&x_1\end{array}\right).$$

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Original proof that $C_0=1$:

The coefficient in $P$ on $x_1^{n-1}x_2^{n-2}\cdots x_{n-1}$ is $1$. We must show that the corresponding coefficient on ${\rm Det}(D)$ is also $1$.

We have $D_{ij}= \dfrac{\partial \sigma_{n,j}}{\partial x_i}$, which is the sum of all products of $j-1$ distinct $x_r$, with each $r\neq i$. We now consider the expansion of ${\rm Det}(D)$, as a sum over permutations $\sigma\in S_n$ of products $\epsilon_\sigma D_{1\sigma_1}\cdots D_{n\sigma_n}$. (Here $\epsilon_\sigma$ denotes the sign of the permutation $\sigma$).

How can the monomial $x_1^{n-1}x_2^{n-2}\cdots x_{n-1}$ arise in this expansion? That is, how can we obtain this monomial, by multiplying together one monomial from $D_{i\sigma_i}$, for $i=1,\cdots,n$?

Firstly, for the exponent on $x_1$ to be $n-1$, we must have $\sigma_1=1$, so that we get a factor of $x_1$ from each column other than the first, which cannot contribute such a factor.

In particular, the monomial picked up from the second column must be $x_1$, so for the exponent on $x_2$ to be $n-2$ we must have $\sigma_2=2$, and there must be a factor of $x_2$ in each monomial picked up from later columns.

Again this means that the monomial picked up from the third column is $x_1x_2$, so $\sigma_3=3$ and there is a factor of $x_3$ in the monomials picked up from all subsequent columns.

Repeating this argument for each column, we conclude that the only occurrence of $x_1^{n-1}x_2^{n-2}\cdots x_{n-1}$, in the expansion of ${\rm Det}(D)$ occurs in the $\sigma=$Id term, as the product of monomials $$1x_1(x_1x_2)(x_1x_2x_3)\cdots (x_1x_2x_3\cdots x_{n-1}).$$

Thus the coefficient on $x_1^{n-1}x_2^{n-2}\cdots x_{n-1}$ in ${\rm Det}(D)$ is $1$ as required.