An explicit root-separation bound for squarefree integer polynomials

If $P(z)=0$ and $\mathrm{\mid }z\mathrm{\mid }>1+H$, then $\mathrm{\mid }{a}_d\mathrm{\mid }\mathrm{\mid }z{\mathrm{\mid }}^d\le H{\sum }_{j=0}^{d-1}\mathrm{\mid }z{\mathrm{\mid }}^j<H\mathrm{\mid }z{\mathrm{\mid }}^d\mathrm{/}(\mathrm{\mid }z\mathrm{\mid }-1)<\mathrm{\mid }z{\mathrm{\mid }}^d\le \mathrm{\mid }{a}_d\mathrm{\mid }\mathrm{\mid }z{\mathrm{\mid }}^d$, because $\mathrm{\mid }{a}_d\mathrm{\mid }\ge 1$ and $\mathrm{\mid }z\mathrm{\mid }-1>H$. This contradiction proves $\mathrm{\mid }z\mathrm{\mid }\le R$ for every root.

Let ${\alpha }_1,\dots ,{\alpha }_d$ be the distinct roots from [F1] and set $D=a_d^{2d-2}{\prod }_{i<j}({\alpha }_i-{\alpha }_j{)}^2$. Form the coefficient map $\mathrm{\Phi }(u,v)=uP+v{P}^{\mathrm{\prime }}$ from pairs with $\mathrm{deg}u\le d-2$, $\mathrm{deg}v\le d-1$ to polynomials of degree at most $2d-2$. Its square coefficient matrix $S$ has integer entries, so $\det S$ is an integer-coefficient polynomial in $a_d,\dots ,a_0$.

Divide each output polynomial by $P$ and record its quotient and remainder. In coefficient order from highest degree down, the map $(q,r)\mapsto qP+r$ has triangular diagonal consisting of $d-1$ copies of $a_d$ and $d$ copies of $1$, so the inverse coordinate change has determinant $\pm a_d^{-(d-1)}$. Under this change, $\mathrm{\Phi }$ has block form $\left(\begin{array}{cc}I& \ast \\ 0& M\end{array}\right)$, where $M$ is multiplication by $P^{\mathrm{\prime }}$ on $\mathbb{C}[x]\mathrm{/}(P)$. Hence $\det S=\pm a_d^{d-1}\det M$.

For each $i$ define $L_i(x)={\prod }_{j\mathrm{\ne }i}(x-{\alpha }_j)\mathrm{/}({\alpha }_i-{\alpha }_j)$. These degree-$(d-1)$ polynomials satisfy $L_i({\alpha }_j)=1$ when $i=j$ and $0$ otherwise. For every degree-$<d$ representative $q$, the difference $q-{\sum }_{i}q({\alpha }_i)L_i$ has degree $<d$ and vanishes at all $d$ distinct roots; [F7] and [F9] force this difference to be zero. Thus evaluation is an isomorphism $\mathbb{C}[x]\mathrm{/}(P)\to {\mathbb{C}}^d$. It conjugates $M$ to the diagonal matrix with entries $P^{\mathrm{\prime }}({\alpha }_i)$, so [F4] gives $\det M={\prod }_i{P}^{\mathrm{\prime }}({\alpha }_i)$. By [F2], $P^{\mathrm{\prime }}({\alpha }_i)=a_d{\prod }_{j\mathrm{\ne }i}({\alpha }_i-{\alpha }_j)$; pairing the two factors for each $i<j$ yields $\det S=\pm a_d^{2d-1}{\prod }_{i<j}({\alpha }_i-{\alpha }_j{)}^2=\pm a_{d}D$.

Set the variable $a_d$ to zero in the integer-coefficient polynomial $\det S$. Then $\mathrm{deg}P\le d-1$ and $\mathrm{deg}{P}^{\mathrm{\prime }}\le d-2$, so every $uP+v{P}^{\mathrm{\prime }}$ has degree at most $2d-3$; the coefficient map misses the $x^{2d-2}$ coordinate and its determinant is zero. Therefore the constant term in $a_d$ of $\det S$ is zero, so $a_d$ divides $\det S$ as an integer-coefficient polynomial. At the given integer coefficients, $\det S\mathrm{/}{a}_d$ is an integer, and [step 3.1] shows that $D$ is an integer up to sign.

Squarefreeness makes every root distinct, so every factor in $D$ is nonzero and $\mathrm{\mid }D\mathrm{\mid }\ge 1$.

Let $\delta ={\min }_{i<j}\mathrm{\mid }{\alpha }_i-{\alpha }_j\mathrm{\mid }$ and choose a pair attaining it. Every other pairwise distance is at most $2R$ by [step 1.1] and the triangle inequality [F5]. Thus $1\le \mathrm{\mid }D\mathrm{\mid }\le H^{2d-2}{\delta }^2(2R{)}^{d(d-1)-2}$. Rearranging gives $\delta \ge H^{-(d-1)}(2R{)}^{-(d(d-1)-2)\mathrm{/}2}=\sigma (P)$.