1 Zarhin’s result

First we recall the main result of [1]. Let f be a monic complex polynomial of degree n with distinct roots \(\alpha _1,\alpha _2,\ldots ,\alpha _n\). Define

$$\begin{aligned} M(f)=\bigl (f'(\alpha _1),f'(\alpha _2),\ldots ,f'(\alpha _n)\bigr ), \end{aligned}$$

then the derivative \(dM_f\) of M for each such f has rank \(n-1\).

Since the map \({\mathbf \alpha } = (\alpha _1,\alpha _2,\ldots ,\alpha _n) \mapsto f(x)=(x-\alpha _1) (x-\alpha _2)\cdots (x-\alpha _n)\) is a regular (n!-sheeted) covering at points \({\mathbf \alpha }\) where \(\alpha _i\) are distinct, the study of the rank of the derivative of M can be done equivalently at f or \({\mathbf \alpha }\). We do so at \({\mathbf \alpha }\).

The matrix \(T=T(f)\) of the derivative map is given by

$$\begin{aligned} T_{ij}&=\frac{\partial }{\partial \alpha _j}\,f'(\alpha _i) =\frac{\partial }{\partial \alpha _j}\,(\alpha _i-\alpha _1)(\alpha _i-\alpha _2)\cdots \widehat{(\alpha _i-\alpha _i)}\cdots (\alpha _i-\alpha _n)\\&=-(\alpha _i-\alpha _1)(\alpha _i-\alpha _2)\cdots \widehat{(\alpha _i-\alpha _j)}\cdots \widehat{(\alpha _i-\alpha _i)}\cdots (\alpha _i-\alpha _n),\qquad i\ne j, \\ T_{ii}&=\sum _j(\alpha _i-\alpha _1)(\alpha _i-\alpha _2)\cdots \widehat{(\alpha _i-\alpha _j)}\cdots \widehat{(\alpha _i-\alpha _i)}\cdots (\alpha _i-\alpha _n). \end{aligned}$$

Our proof of Zarhin’s result shows that the matrix T has some remarkable properties and so it might be of independent interest.

We will simplify the notation by writing

Then, for \(i \ne j,\) \(T_{ij}=f_{ij}(\alpha _i)\).

For \(X \subset \{1,2,\ldots n\} \) with m elements, we let T[X] denote the submatrix of T obtained by omitting the \(j^\mathrm{th}\) row and column for each \(j \in X\) and let \(D[X]= \det T[X]\); so the principal minor of T is D[n]. We also let \(\mathrm{\Delta }(g)\) denote the discriminant of a polynomial g.

We note that the sum of the columns of T is zero and so \(\mathrm{rank}\,T <n\). Since the discriminant of a polynomial with distinct roots is non-zero, the proof will be completed by

Proposition

For each k, \(1 \le k \le n \),

$$\begin{aligned} D[k]=(-1)^{n-1 \atopwithdelims ()2} (n-1)!\, \mathrm{\Delta }(f_k)= (-1)^{n-1 \atopwithdelims ()2} (n-1)!\!\!\prod _{\begin{array}{c} 1 \le i <j\le n\\ i,j \ne n \end{array}}\!\!\!(\alpha _i-\alpha _j)^2. \end{aligned}$$

In particular,

$$\begin{aligned} D[n]=(-1)^{n-1 \atopwithdelims ()2} (n-1)!\,\mathrm{\Delta }(f_n)=(-1)^{n-1 \atopwithdelims ()2} (n-1)!\!\!\prod _{1 \le i <j<n}\!\!(\alpha _i-\alpha _j)^2. \end{aligned}$$

Proof

We prove the result for \(k=n\) (that is, we are considering the principal minor of T) and the proof is, apart from notation, the same for other values of k. Interchanging both the ith and jth rows and the ith and jth columns of T for \(1 \le i <j <n\) interchanges i and j but does not change the determinant D[n] of the principal minor. So D[n] is a symmetric polynomial in \(\alpha _1,\alpha _2, \ldots , \alpha _{n-1}\). If we set \(\alpha _i = \alpha _j\) then the \(i^\mathrm{th}\) and \(j^\mathrm{th}\) rows of T are equal, so if \(1 \le i <j <n\) and \(\alpha _i = \alpha _j\) then \(D[n]=0.\) Now we recall the well known result:

If \(P(x_1,x_2,\dots , x_r)\) is a symmetric polynomial which vanishes when any pair of the x’s are equal then p is a multiple of \(\prod _{1 \le i <j\le r}(x_i-x_j)^2\).

So D[n] is a multiple of \(\mathrm{\Delta }(f) =\prod _{1 \le i <j<n}(\alpha _i-\alpha _j)^2\), which clearly has total degree but the total degree of each \(T_{ij}\) is \(n-2\) so D[n] also has total degree . So for some constant c.

To determine the value of c,  we consider each \(T_{ij}\) as an element of the polynomial ring where \(R= { {\mathbb {C}}}[\alpha _2,\alpha _3, \ldots , \alpha _n]\). We use induction on n; it starts trivially at \(n=2\).

The degrees of the various \(T_{ij}\) as polynomials in \(\alpha _1\) are given by

figure a

Moreover, the coefficient of \(\alpha _1^{n-2}\) in \(T_{1j}\) is \(-1\) for \(j>1\) and, since the sum of the columns of T is zero, the coefficient of \(\alpha _1^{n-2}\) in \(T_{11}\) is \(n-1\). There are no occurrences of \(\alpha _1\) in the first column (except for \(T_{11}\)) and so when calculating the determinant by using the first row, the terms that contribute \(\alpha _1^{2(n-2)}\) to D[n] all come from the product \(T_{11} D[1,n]\). (Note that the highest degree term involving \(\alpha _1\) in \(T_{11} \) is \(\alpha _1^{n-2}\) and that in each entry of T[1, n] it is \(\alpha _1\).) The coefficient of \(\alpha _1\) in the entries of T[1, n] (since these entries are all linear in \(\alpha _1\)) are given by

$$\begin{aligned} \frac{\partial }{\partial \alpha _1}\,T_{ij} = \frac{\partial }{\partial \alpha _1}\,f_{ij}(\alpha _i) =-f_{1ij}(\alpha _i),\qquad i\ne j,\quad i, j >1. \end{aligned}$$

But these are precisely the negatives of the off-diagonal entries of the matrix \(T(f_1)\) that one obtains from the polynomial \(f_1(x)=(x-\alpha _2)(x-\alpha _3)\cdots (x-\alpha _n)\). The entries of the first column of T do not involve \(\alpha _1\) and since the sum of the columns of T is 0, \(\alpha _1\) does not appear in the sum of the columns of T[n]. But the sum of the columns of \(T(f_1)\) is also 0 and so we conclude that the matrix \(T[1,n] - \alpha _1T(f_1)\) is independent of \(\alpha _1\). Hence, by applying the induction hypothesis to \(f_1\), the term in D[n] involving \(\alpha _1^{2{n-2}} \) is

$$\begin{aligned} (n-1)\alpha _1^{n-2}\mathrm{det}(-I_{n-2})\alpha _1^{n-2\atopwithdelims ()2}&(n-2)!\,\mathrm{\Delta }(f_{n1})\\ {}&= (-1)^{n-1 \atopwithdelims ()2} (n-1)!\,\mathrm{\Delta }(f_{1n})\alpha _1^{2(n-2)} \end{aligned}$$

proving the proposition. \(\square \)

2 Multiple roots

Now we consider a monic polynomial \(f(x) \in {{\mathbb {C}}}[x]\) of degree n with multiple roots. Let \(R(f)=\{ \alpha _1,\ldots , \alpha _r \}\) denote the set of all its roots, \(\#\, R(f)=r\), \(r<n\), and \(R_k(f)\) the set of roots of f that have multiplicity exactly k. We order the roots so that their multiplicities are in decreasing order and suppose that \(\#\,R_1(f)=s\); clearly \(s<n\). The first \(r-s\) rows of the matrix M are zero, so \(\mathrm{rank}\,M \le s\). Somewhat tentatively, we make the following conjecture and sketch some of the calculations that support it.

Conjecture

The rank of M is s.

Consider an submatrix N of M formed from a set of s columns and the last s rows of M. We find that if all the determinants \(\mathrm{det}\,N\) are zero then a pair of roots of the polynomial are equal. In particular, calculations that we have carried out suggest that \(\mathrm{det}\,N\) is always of the form

$$\begin{aligned} \pm c \prod (a-b)^t g \end{aligned}$$

where the product is over a nonempty set of pairs of distinct roots a, b of f and g is a polynomial in \(\{ \alpha _1,\ldots , \alpha _r \}\). In various cases, we describe the powers t, the constant c and the polynomials g:

  • Let f(x) have only one root, say, \(\alpha _r\) of multiplicity 1, then the principal minor N is and it is easy to calculate that

    $$\begin{aligned} N=-k_1\frac{f_r(\alpha _r)}{(\alpha _r-\alpha _1)} \end{aligned}$$

    where \(k_1\) is the multiplicity of the root \(\alpha _1\) and (as in Sect. 1) \(f_r(x)\) is f(x) with the factor \(x-\alpha _r\) omitted.

  • Let f(x) have the root \(\alpha _1\) with multiplicity \(k>1\) and the other roots be \(\alpha _2, \ldots ,\alpha _r\) all of multiplicity 1 then, the principal minor, \(\mathrm{det}\,N\) has factors \(\alpha _1-\alpha _{\ell }\) with index \(t=k-1\) and the factors \(\alpha _m-\alpha _{\ell }\), \(\ell ,m >1\), with index \(t=2\) and c is .

  • Let f(x) have the root \(\alpha _1\) with multiplicity \(k>1\), the root \(\alpha _2\) with multiplicity \(\ell >1\) and the other roots of multiplicity 1. Then, when \(r=5\), the determinant of the minors has the form indicated but with a non-trivial factor g. The principal minor has \(g=\alpha _1+2\alpha _2 -3\alpha _3\) and one of the other minors has \(g=\alpha _1+2\alpha _2 -3\alpha _4\). If both these g vanish then we have that \(\alpha _3 = \alpha _4\) and if some other factor of the determinant vanishes then, again two of the \(\alpha \)’s are equal which contradicts our hypothesis.

This final calculation seems to indicate that it may be difficult to verify the conjecture by a direct calculation.