Proof. Let $\eta = \kappa /(n-1)$ . Let $a_2,\ldots ,a_{n-1}$ be integers with $|a_i|<H^i$ for $i=2,\ldots ,n-1$ . The discriminant of $x^n + a_2x^{n-2} + \cdots + a_n$ is a polynomial $F(a_n)$ in $a_n$ of degree $n-1$ with leading term $C_n a_n^{n-1}$ , where $C_n$ is a nonzero constant. Let $s_1,\ldots ,s_{n-1}\in {\mathbb C}$ be the $n-1$ roots of $F(x)$ . Then

$$ \begin{align*}F(a_n) = C_n(a_n-s_1)\cdots(a_n-s_{n-1}).\end{align*} $$

Since $|F(a_n)| \leq H^{n(n-1)-\kappa }$ , it follows that $|(a_n-s_1)\cdots (a_n-s_{n-1})|\ll _n H^{n(n-1)-\kappa }$ . Hence,

(2.1) $$ \begin{align} |a_n - s_i| \ll_n H^{n-\eta} \end{align} $$

for some $i\in \{1,\ldots ,n-1\}$ . The number of integers $a_n$ satisfying (2.1) for some i is $O(H^{n-\eta }+1)$ . Multiplying this by the number of choices for $a_2,\ldots ,a_{n-1}$ then gives the desired bound.

Proof. Let $v_p$ denote the p-adic valuation. Suppose $f(x)\in {\mathcal W}_{p^k}$ , and let $\ell = v_p(\Delta (f))\geq 2k$ . Let $r_1,\ldots ,r_n$ denote the n roots of $f(x)$ in $\overline {{\mathbb Q}_p}$ . Suppose the p-adic valuation $\theta = v_p(r_1-r_2)$ is the largest among all differences of roots of $f(x).$ Then $p^{\ell -2\theta }\mid \Delta '(f)$ . If $\theta \leq (\ell -k)/2$ , then $f\in {\mathcal W}_{p^k}^{(1)}$ .

Now suppose that $\theta> (\ell -k)/2\geq k/2$ . Then $v_p(r_i-r_j)<\theta $ for all distinct pairs $i,j$ with $\{i,j\}\neq \{1,2\}$ ; indeed, if $v_p(r_i-r_j)=\theta $ , then $(r_1-r_2)^2(r_i-r_j)^2\mid \Delta (f)$ implying $\ell \geq 4\theta>2\ell -2k$ , a contradiction. Thus, either $r_1,r_2$ are defined over ${\mathbb Q}_p$ or are conjugate over some quadratic extension of ${\mathbb Q}_p$ . That is, we have $q(x) = (x-r_1)(x-r_2) = x^2 - bx + c$ for some $b,c\in {\mathbb Z}_p$ . We claim that $b/2\in {\mathbb Z}_p$ . This is clear for $p\neq 2$ , while if $p=2$ , then $b^2 - 4c = (r_1-r_2)^2$ is divisible by $2^{2\theta }$ and, hence, is divisible by $2^2$ since $\theta>k/2\geq 1/2$ .

Now $q(b/2) = - \frac 14(r_1-r_2)^2$ and $q'(b/2) = 0$ . Since $f(x) = q(x)h(x)$ for some $h(x)\in {\mathbb Z}_p[x]$ , we have that $\frac 14(r_1-r_2)^2$ divides $f(b/2)$ and $f'(b/2)$ . Hence, $f\in {\mathcal W}_{p^e}^{(2)}$ , where $e= \lceil \theta \rceil - v_p(2)$ .

Proof. The first containment is proved in the proof of [Reference Bhargava, Shankar and Wang7, Theorem 4.4]. To obtain the second containment, fix $f\in {\mathcal W}_m$ for some $m>M$ . Let $\prod _{i=1}^n p_i^{k_i}$ be the prime factorisation of m, and let

$$ \begin{align*}m_1 = \prod_{p_i\colon f\in {\mathcal W}_{p_i^{k_i}}^{(1)}} p_i^{k_i},\qquad m_2 = \prod_{p_i\colon f\in {\mathcal W}_{p_i^{\lceil k_i/2\rceil-v_{p_i}(2)}}^{(2)}} p_i^{\lceil k_i/2\rceil-v_{p_i}(2)}.\end{align*} $$

Then $f\in {\mathcal W}_{m_1}^{(1)}\cup {\mathcal W}_{m_2}^{(2)}$ and $m_1m_2^2\geq m/2> M/2$ . Hence, either $m_1>q_1$ or $m_2>q_2$ , and, therefore, $f\in \bigcup _{m>q_1}{\mathcal W}_{m}^{(1)}\cup \bigcup _{m>q_2}{\mathcal W}_{m}^{(2)}$ , as desired.

Proof. Let $B\in W$ be the generic element. Fix $P\in M_n({\mathbb Q}[\sqrt {-1}])$ , such that $PA_0P^t$ is the identity matrix $I_n$ . Let $B' = PBP^t$ . Then the entries of $B'$ lie in ${\mathbb Z}[\sqrt {-1}][b_{11},\ldots ,b_{nn}]$ . Let $c_1,\ldots ,c_n \in \overline {{\mathbb Q}(b_{11},\ldots ,b_{nn})}$ be the eigenvalues of $B'$ , and let $h(c_1,\ldots ,c_n)\in \text {O}_n({\mathbb Q}[\sqrt {-1}](b_{11},\ldots ,b_{nn},c_1,\ldots ,c_n))$ be a change-of-basis matrix, such that $h(c_1,\ldots ,c_n)B'h(c_1,\ldots ,c_n)^t = B'' = \text {diag}(c_1,\ldots ,c_n)$ . Note that we have fixed an order of the eigenvalues here. For any other order, we simply multiply $h(c_1,\ldots ,c_n)$ by the corresponding permutation matrix.

Following [Reference Wang18, §2], we now have the following explicit construction of the $2^{2g}$ common isotropic g-planes with respect to the quadratic forms defined by $I_n$ and $B''$ . Let $D_1,\ldots ,D_{n} \in {\mathbb Q}(c_1,\ldots ,c_n)$ be a nonzero solution to the following system of linear equations:

$$ \begin{align*} D_1 + D_2 + \cdots + D_n &= 0\\ D_1c_1 + D_2c_2 + \cdots + D_nc_n &= 0\\ &\vdots \\ D_1c_1^{2g-1} + D_2c_2^{2g-1} + \cdots + D_nc_n^{2g-1} &=0. \end{align*} $$

Note that we may take

(4.1) $$ \begin{align} D_i = \pm \prod_{j\neq i}(c_j - c_i)^{-1}. \end{align} $$

This is obtained by noting that the kernel of an incomplete Vandermonde matrix is spanned by the last row of the inverse of the completed Vandermonde matrix. None of the $D_i$ is equal to $0$ , and so for each choice of $d_i\in {\mathbb Q}(\sqrt {D_i})$ with $d_i^2 = D_i$ , we have a g-plane

(4.2) $$ \begin{align} Y = \mathrm{Span}\{(d_1,\ldots,d_n), (d_1c_1,\ldots,d_nc_n),\ldots, (d_1c_1^{g-1},\ldots,d_nc_n^{g-1})\}, \end{align} $$

which is isotropic with respect to the quadratic forms defined by $I_n$ and $B''$ . Negating all of the $d_i$ ’s gives the same Y, and so we have $2^{2g}$ distinct g-planes. Now

$$ \begin{align*}P^th(c_1,\ldots,c_n)^tY=\mathrm{Span}\{(\ell_{11},\ldots,\ell_{1n}),\ldots,(\ell_{g1},\ldots,\ell_{gn})\}\end{align*} $$

gives a g-plane that is isotropic with respect to the quadratic forms defined $A_0$ and B, where each $\ell _{ij}$ is a linear form in $d_1,\ldots ,d_n$ with coefficients in ${\mathbb Q}[\sqrt {-1}](b_{11},\ldots ,b_{nn},c_1,\ldots ,c_n)$ .

Next we apply the Plücker embedding to send each $P^th(c_1,\ldots ,c_n)^tY$ to a point

$$ \begin{align*}[G_0(d_1,\ldots,d_n)\colon\cdots\colon G_N(d_1,\ldots,d_n)]\end{align*} $$

in projective space, where each $G_i$ is a homogeneous polynomial of degree g with coefficients in ${\mathbb Q}[\sqrt {-1}](b_{11},\ldots ,b_{nn},c_1,\ldots ,c_n).$ As $d_1,\ldots ,d_{n}$ vary, we get $2^{2g}$ points $P_1,\ldots ,P_{2^{2g}}$ this way. Now let $L_1(x_0,\ldots ,x_N) = \alpha _0x_0 + \cdots + \alpha _Nx_N$ and $L_2(x_0,\ldots ,x_N) = \beta _0x_0 + \cdots + \beta _Nx_N$ be two linear forms with integer coefficients $\alpha _i,\beta _i$ to be chosen later. For $i = 1,\ldots ,2^{2g}$ , let

$$ \begin{align*}Q_i = [L_1(P_i)\colon L_2(P_i)]\in{\mathbb P}^1({\mathbb Q}[\sqrt{-1}](b_{11},\ldots,b_{nn},c_1,\ldots,c_n)(d_1,\ldots,d_n)).\end{align*} $$

For now, we only require that none of the $Q_i$ equals $[1:0]$ . We then see that there is a binary $2^{2g}$ -ic form $J(c_1,\ldots ,c_n)(x,y)$ defined over ${\mathbb Q}[\sqrt {-1}](b_{11},\ldots ,b_{nn},c_1,\ldots ,c_n)$ vanishing on $Q_1,\ldots ,Q_{2^{2g}}$ . By scaling, we may assume that

$$ \begin{align*}J(c_1,\ldots,c_n)(x,y)\in{\mathbb Z}[\sqrt{-1}][b_{11},\ldots,b_{nn},c_1,\ldots,c_n][x,y].\end{align*} $$

Let $J_1(c_1,\ldots ,c_n)(x) = J(c_1,\ldots ,c_n)(x,1)$ . Note that if $B_0\in W({\mathbb Q})$ is distinguished, then one of the $P_i$ is defined over ${\mathbb Q}$ , in which case, $J_1(c_1,\ldots ,c_n)(x)$ has a root over ${\mathbb Q}$ .

Finally, we note that the homogeneous polynomials $G_0,\ldots ,G_N$ are independent of the ordering of the eigenvalues $c_1,\ldots ,c_n$ , since permuting the $c_i$ ’s permutes the coordinates of Y, which is then cancelled by the extra permutation matrix in $h(c_1,\ldots ,c_n)$ . Therefore, the coefficients of $J_1(c_1,\ldots ,c_n)(x)$ are symmetric in $c_1,\ldots ,c_n$ and so are polynomials in the coefficients of the characteristic polynomial of B. We let $F(B,x)$ denote this polynomial.

It remains to prove that for some choice of coefficients $\alpha _i,\beta _i$ for the linear forms $L_1,L_2$ , the polynomial $F(B,x)\in {\mathbb Z}[\sqrt {-1}][b_{11},\ldots ,b_{nn}][x]$ is irreducible. It suffices to exhibit some $B_0\in W({\mathbb Q}[\sqrt {-1}])$ , such that $G_{{\mathbb Q}[\sqrt {-1}]}$ , the absolute Galois group of ${\mathbb Q}[\sqrt {-1}]$ , acts transitively on the $2^{2g}$ distinct roots of $F(B_0,x)$ . For any $c_1,\ldots ,c_n$ , let $E_i=(c_1-c_i)\cdots (c_{i-1}-c_i)(c_{i+1}-c_i)\cdots (c_n-c_i) = \pm D_i^{-1}$ . Let M be a large integer so that there are at least $n-1$ distinct primes $q_1,\ldots ,q_{n-1}$ lying inside $(M - \sqrt {M}, M)$ . Let $c_n = M$ , and let $c_i = M - q_i$ for $i = 1,\ldots ,n-1$ . Then for any $i,j=1,\ldots ,n-1$ with $j\neq i$ , we have $q_i\mid E_i$ and $q_i\nmid E_j$ . Note also that $E_n = (-1)^{n-1}q_1\cdots q_{n-1}$ . For each $i=1,\ldots ,n-1$ , let $\sigma _i$ denote an element in the absolute Galois group $G_{{\mathbb Q}[\sqrt {-1}]}$ that negates $\sqrt {q_j}$ for all $j=1,\ldots ,n-1$ for which $j\neq i$ and fixes all other square roots that appear (including $\sqrt {q_i}$ ). Since n is odd, we see that $\sigma _i(d_j)=-d_j$ for all $j\neq i$ and $\sigma _i(d_i) = d_i$ . Hence,

$$ \begin{align*}\sigma_i([d_1\colon\cdots\colon d_n]) =[d_1\colon\cdots\colon d_{i-1}\colon -d_i\colon d_{i+1}\colon\cdots\colon d_n].\end{align*} $$

That is, the absolute Galois group $G_{{\mathbb Q}[\sqrt {-1}]}$ acts transitively on the set $\{[d_1\colon \cdots \colon d_n]\mid d_i^2=D_i\}$ . Hence, it also acts transitively on the $2^{2g} Y$ ’s defined in (4.2) as the $d_i$ ’s vary and so also acts transitively on $P_1,\ldots ,P_{2^{2g}}$ . We may simply choose the integers $\alpha _i,\beta _i$ so that $Q_1,\ldots ,Q_{2^{2g}}$ are distinct. Let $B'=\text {diag}(c_1,\ldots ,c_n)$ , and let $B_0 = P^{-1}B'(P^{-1})^t\in W({\mathbb Q}[\sqrt {-1}])$ . Then $G_{{\mathbb Q}[\sqrt {-1}]}$ acts transitively on the $2^{2g}$ distinct roots of $F(B_0,x)$ .

Proof. Given $(B,c)$ , we proceed as before to orthogonally diagonalise $B'=PBP^t$ into $B'' = \text {diag}(c_1,\ldots ,c_{n-1},c)$ . Then we take $B''' = B'' - cI_n = \text {diag}(c_1-c,\ldots ,c_{n-1}-c,0)$ and use the construction above in the odd degree case with $c_1,\ldots ,c_n$ replaced by $c_1-c,\ldots ,c_{n-1}-c$ to obtain a polynomial whose coefficients are polynomials that are symmetric in $c_1 - c, \ldots , c_{n-1}-c$ . Note that $(x-c_1)\cdots (x-c_{n-1}) = \det (xA_0-B)/(x-c)\in R[x]$ . Hence, $(x-(c_1-c))\cdots (x-(c_{n-1}-c))\in R[x]$ , and so any polynomial that is symmetric in $c_1 - c, \ldots , c_{n-1}-c$ belongs to R.

Proof. Define $F_1(B,x)$ to be $N_{{\mathbb Q}[\sqrt {-1}]/{\mathbb Q}}F(B,x)$ in the odd case and to be $N_{K/{\mathbb Q}(b_{11},\ldots ,b_{nn})}F(B,c,x)$ in the even case. Theorems 4.1 and 4.2 imply that if $B_0\in W({\mathbb Q})$ is distinguished, then the Galois group of $F_1(B_0,x)$ has index at least $2^{2g}$ in the generic Galois group of $F_1(B,x)$ . More precisely, they imply that every irreducible factor of $F_1(B,x)$ has degree at least $2^{2g}$ , and that if $B_0$ is distinguished, then $F_1(B_0,x)$ has a rational root.

We wish to upper bound the number of $B_0\in W({\mathbb Z})$ , with $b_{ij}<Y_{ij}$ , such that $f(B_0,x)$ has a rational root in x for some irreducible factor $f(B,x)$ of $F_1(B,x)$ . When all the $Y_{ij}$ ’s are the same, say Y, the required bound follows immediately from [Reference Castillo and Dietmann9, Theorem 1]. Indeed, when applied to $f(B,x)$ with generic Galois group $G_f$ , [Reference Castillo and Dietmann9, Theorem 1] states that the number of $B_0\in W({\mathbb Z})$ with each $|b_{ij}|<Y$ , such that the Galois group of $f(B_0,x)$ is $K\subset G_f$ , is $O(Y^{\mathrm {dim}(W)-1+|G_f/K|^{-1}+\epsilon })$ . Since $B_0$ being distinguished implies that $f(B_0,x)$ has a rational root for some irreducible factor $f(B,x)$ of $F(B,x)$ , and the degree of $f(B,x)$ is at least $2^{2g}$ , we have $|G_f/K|\geq 2^{2g}$ and the result follows.

The proof of [Reference Castillo and Dietmann9, Theorem 1] in the case where we require $f(B_0,x)$ to have a rational root (as opposed to a general Galois subgroup $K\subset G_f$ ) is much simpler than the general case (as the construction of a polynomial, associated to $f(B_0,x)$ , having a rational root may be skipped in this case). We now describe how this proof also carries through without any change for general $Y_{ij}$ . First, $f(B,x)$ can be assumed to be monic in x by replacing $f(B,x)=g_0(B)x^m+g_1(B)x^{m-1}+\cdots + g_m(x)$ by $g_0(B)^{m-1}f(B,x/g_0(B))$ . In order to make this reduction, it is necessary to provide an upper bound for the number of $B_0\in W({\mathbb Z})$ with $|b_{ij}|<Y_{ij}$ , such that $g_0(B_0)=0$ . This number is clearly bounded by the right-hand side of (4.3), as a fibring argument readily shows. Specifically, we fibre over all but one of the coefficients, denoted by b. Fixing values for each $b_{ij}\neq b$ yields a polynomial $g(b)=g_0(B)$ in one variable. If $g(b)$ is not identically zero, then $g(b)=0$ has $O(1)$ different solutions, yielding a sufficient saving. Meanwhile, the condition of $g(b)$ being identically zero imposes one or more polynomial vanishing conditions on the coefficients $b_{ij}\neq b$ , and the number of such values of $b_{ij}$ also satisfies the required bound by induction.

The result for monic polynomials is proved in [Reference Castillo and Dietmann9, Lemma 7] by fibring over all but one of the coefficients and then using induction. This proof (for the case when all the $Y_{ij}$ ’s are the same) carries over without change for general $Y_{ij}$ , when the fibring is done over the variables $b_{ij}$ for which the $Y_{ij}$ are the smallest.

Proof. We begin by noting that the proofs of [Reference Bhargava4, Theorem 3.5 and Lemma 3.6] imply the bound

(5.2) $$ \begin{align} \#\bigcup_{\substack{m>M }}|\mu(m)|\{f\in {\mathcal W}_{m}^{(1)}:H(f)<H\}= O_\epsilon\bigg(\frac{H^{n(n+1)/2+\epsilon}}{M^{1-\epsilon}}\bigg)+O(H^{n(n+1)/2-1}). \end{align} $$

Briefly, the proof is as follows: first, there exists a polynomial $P\in {\mathbb Z}[V_n]$ , belonging to the algebra generated by $\Delta $ and $\Delta '$ , which does not involve the constant coefficient $a_n$ . Hence, we may consider P as a polynomial in $a_1,\ldots , a_{n-1}$ . Second, a bound of size $O(H^{n(n+1)/2-n-1})$ is easily obtained on the number of possible values of $a=(a_1,\ldots ,a_{n-1})$ , of bounded height, for which $P(a)=0$ . Third, we fibre over the $O(H^{n(n+1)/2-n})$ values of such a for which $P(a)\neq 0$ . For each such a, it is clear that $P(a)$ has at most $O(H^\epsilon )$ different divisors. Since $P(f)=P(a_1,\ldots ,a_{n-1})\equiv 0\pmod {m}$ for $f(x)=x^n+\sum _{i=1}^n a_{i}X^{n-i}\in {\mathcal W}_m$ , fixing a constrains the value of $m>M$ to be one of these $O(H^\epsilon )$ divisors of $P(a)$ . Fourth, and finally, we consider $\Delta $ to be a polynomial $\Delta _a(a_n)$ in $a_n$ . We then note that the condition $m\mid \Delta (f)=\Delta _a(a_n)$ implies that there are at most $O((H^n/m^{1-\epsilon })+1)=O((H^n/M^{1-\epsilon })+1)$ choices for $a_n$ , concluding the proof of (5.2).

It is precisely this last step which breaks down when m is not required to be squarefree. We note that $\Delta _a(a_n)$ is a polynomial of degree $n-1$ in $a_n$ , whose leading coefficient $(-1)^{n(n-1)/2} n^{n-1}$ does not depend on a. Suppose $p^k\parallel m$ for some prime p and some fixed $m\mid P(a)$ . Let $\ell \ll _n 1$ be a nonnegative integer, such that $g(a_n) = \Delta _a(a_n)/p^\ell \in {\mathbb Z}[x]$ and at least one of its coefficients is not divisible by p. The condition $m^2\mid \Delta (f)$ now becomes $p^{2k-\ell }\mid g(a_n)$ for every prime divisor p of m. Let $\delta = \lceil \frac {2k-\ell }{n-1}\rceil $ , and let $g(x) = f_1(x)\cdots f_j(x)$ be a factorisation in $({\mathbb Z}/p^\delta {\mathbb Z})[x]$ , where j is maximal. Since the reduction $\bar {g}(x)\in {\mathbb F}_p[x]$ of $g(x)$ modulo p is a nonzero polynomial of degree at most $n-1$ , we have $\bar {g} = \bar {f_1}\cdots \bar {f_j}$ , and so $j\leq n-1.$ In order for $p^{2k-\ell }\mid g(a_n)$ , we then must have $p^\delta \mid f_i(a_n)$ for some $i = 1,\ldots ,j$ . This implies that $(x-a_n)\mid f_i(x)$ in $({\mathbb Z}/p^\delta {\mathbb Z})[x]$ , and so by maximality of j, we see that $f_i(x)$ is linear. Hence, the density of integers $a_n$ , such that $p^{2k-\ell }\mid g(a_n)$ is at most $(n-1)/p^\delta .$ Multiplying over all prime divisors p of m gives that the density of integers $a_n$ , such that $m^2\mid \Delta _a(a_n)$ is $O(1/m^{2/(n-1)-\epsilon })$ . Combining with the proof of [Reference Bhargava4, Theorem 3.5 and Lemma 3.6] recalled above gives (5.1).

Proof of Theorem 1.5

To bound the number of elements in $\cup _{m>M} {\mathcal W}_{m}^{(2)}$ , we recall the setup of [Reference Bhargava, Shankar and Wang7]. The proofs of [Reference Bhargava, Shankar and Wang7, Theorems 2.3 and 3.2] imply that we have a map $\sigma _m:{\mathcal W}_{m}^{(2)}\to \frac 14 W({\mathbb Z})$ , injecting into the set of distinguished elements, such that the resolvent of $\sigma _m(f)$ is f and $Q(\sigma _m(f))=m$ . Here, Q is an invariant defined on the set of distinguished elements of $W({\mathbb Z})$ , given explicitly in [Reference Bhargava, Shankar and Wang7, §§2.1 and 3.1]. To bound the number of elements in $\cup _{m>M} {\mathcal W}_{m}^{(2)}$ having height less than H, it thus suffices to bound the number of $G({\mathbb Z})$ -orbits on distinguished elements of $W({\mathbb Z})$ having height less than H and Q-invariant larger than M. This is precisely what is carried out via geometry-of-numbers arguments in [Reference Bhargava, Shankar and Wang7, §2 and 3]. Moreover, using Corollary 4.3 instead of the Selberg sieve in the proofs of [Reference Bhargava, Shankar and Wang7, Propositions 2.6 and 3.5] improves the error terms there to $O_\epsilon (H^{\mathrm {dim}(W)-1+1/2^{2g}+\epsilon })$ . We thus obtain the following bound:

(5.3) $$ \begin{align} \#\bigcup_{\substack{m>M }}\{f\in {\mathcal W}_{m}^{(2)}:H(f)<H\}= O_\epsilon(H^{n(n+1)/2+\epsilon}/M)+O_\epsilon(H^{n(n+1)/2-1+1/2^{2g}+\epsilon}). \end{align} $$

We note that (5.3) is a strengthening of [Reference Bhargava, Shankar and Wang7, Theorem 1.5(b)]. Optimising by taking $\alpha = (n-1)/(n+3)$ and $\beta = 2/(n+3)$ in Proposition 3.2 gives Theorem 1.5.

Proof of Theorem 1.6

Applying an inclusion-exclusion sieve, we obtain

(5.4) $$ \begin{align} \#\big\{f\in V_n({\mathbb Z}): H(f)<H \mbox{ and } \Delta(f) \mbox{ squarefree}\big\}= \sum_{m\geq 1}\mu(m)\#\{f\in {\mathcal W}_m:H(f)<H\}. \end{align} $$

We break up the sum over m into three ranges, namely, the large range consisting of $m\geq H^{n/2}$ , the middle range consisting of $H\leq m< H^{n/2}$ and the small range consisting of $m<H$ . We will obtain precise estimates for the sum of m over the small range and prove that the sum over m in the middle and large ranges are negligible, where we say that a number is negligible if it is $O_\epsilon (H^{n(n+1)/2-1+1/2^{2g}+\epsilon })$ .

First, note that Theorem 1.5 implies the bound

(5.5) $$ \begin{align} \sum_{m\geq H^{n/2}}|\mu(m)|\#\{f\in {\mathcal W}_m:H(f)<H\}= O_\epsilon(H^{n(n+1)/2-1+1/2^{2g}+\epsilon}). \end{align} $$

Therefore, the sum over the large range is negligible.

Next we consider the middle range. That is, we sum the right-hand side of (5.4) over m in the range $H<m\leq H^{n/2}$ and prove that the sum is $O_\epsilon (H^{n(n+1)/2-1+\epsilon })$ . We fibre over integer tuples $a=(a_1,\ldots ,a_{n-1})$ . For any integer tuple a, let $\Delta _a(a_n)$ denote as above the discriminant of $x^n + a_1x^{n-1} + \cdots + a_n$ and let $\theta _a(m)$ denote the density of integers $a_n$ , such that $m^2\mid \Delta _a(a_n).$ Let B denote the set of integer tuples $a=(a_1,\ldots ,a_{n-1})$ , such that $|a_i|<H^i$ for $i=1,\ldots ,n-1$ . Then we have

(5.6) $$ \begin{align} \sum_{H<m\leq H^{n/2}}|\mu(m)|\#\{f\in {\mathcal W}_m:H(f)<H\}= \sum_{a\in B}\sum_{H<m\leq H^{n/2}}|\mu(m)|(\theta_a(m)\cdot 2H^n + O(1)). \end{align} $$

Since $\#B = O(H^{n(n+1)/2 - n})$ , we see that the sum of the $O(1)$ term is negligible.

Now the discriminant $\Delta (\Delta _a)$ of $\Delta _a(a_n)$ is a polynomial in $a_1,\ldots ,a_{n-1}$ . We claim that $\Delta (\Delta _a)$ has a term involving only $a_{n-1}$ . Indeed, when $a_1=\cdots =a_{n-2}=0$ , we have

$$ \begin{align*}\Delta_a(a_n) = \Delta(x^n + a_{n-1}x + a_n) = (-1)^{n(n-1)/2}n^na_n^{n-1}-(-1)^{n(n+1)/2}(n-1)^{n-1}a_{n-1}^n,\end{align*} $$

and so

$$ \begin{align*}\Delta(\Delta_a) = C_na_{n-1}^{n(n-2)},\end{align*} $$

for some nonzero constant $C_n$ depending only on n. As a consequence, given any values for $a_1,\ldots ,a_{n-2}$ , $\Delta (\Delta _a)$ will be a nonzero polynomial in $a_{n-1}$ . Hence, we have

$$ \begin{align*}\#\{a\in B\mid \Delta(\Delta_a) = 0\} = O(H^{n(n+1)/2 - n - (n-1)}).\end{align*} $$

Hence, the contribution to the right-hand side of (5.6) over $a\in B$ with $\Delta (\Delta _a) = 0$ is negligible.

Suppose now $a\in B$ with $\Delta (\Delta _a)\neq 0$ . Take any squarefree m with $H<m\leq H^{n/2}$ . Let $d = \gcd (\Delta (\Delta _a),m)$ , and let $m_1 = m/d$ . For any prime $p\mid d$ and $p\nmid n$ , the polynomial $\Delta _a(a_n)$ mod p is a nonzero polynomial (since its leading coefficient is nonzero) with a repeated factor, in which case, $\theta _a(p) = O(1/p).$ For any prime $p\mid m_1$ and $p\nmid n$ , the polynomial $\Delta _a(a_n)$ mod p is a nonzero polynomial without a repeated factor, in which case, $\theta _a(p) = O(1/p^2).$ Hence, we have $\theta _a(m) = O(1/(dm_1^{2-\epsilon })),$ where we absorb any common divisors of d and n, or of $m_1$ and n, into the implied constant. Denoting by $\sideset {}{'}\sum $ a sum over squarefree numbers, we have

$$ \begin{align*} \sum_{\substack{a\in B\\ \Delta(\Delta_a)\neq 0}} \sideset{}{'}\sum_{H<m\leq H^{n/2}} \theta_a(m) &\ll_\epsilon \sum_{\substack{a\in B\\ \Delta(\Delta_a)\neq 0}}\bigg( \sideset{}{'}\sum_{\substack{1\leq d\leq H\\ d\mid\Delta(\Delta_a)}} \sideset{}{'}\sum_{\frac{H}{d}<m_1\leq \frac{H^{n/2}}{d}}\frac{1}{dm_1^{2-\epsilon}}+\sideset{}{'}\sum_{\substack{H< d\leq H^{n/2}\\ d\mid\Delta(\Delta_a)}} \sideset{}{'}\sum_{1<m_1\leq \frac{H^{n/2}}{d}}\frac{1}{dm_1^{2-\epsilon}}\bigg)\\ &\ll_\epsilon \sum_{\substack{a\in B\\ \Delta(\Delta_a)\neq 0}} \sideset{}{'}\sum_{\substack{1\leq d\leq H^{n/2}\\ d\mid\Delta(\Delta_a)}} \frac{1}{H^{1-\epsilon}}\\ &\ll_\epsilon H^{n(n+1)/2-n -1+\epsilon}, \end{align*} $$

where the last bound follows because the number of divisors of the nonzero integer $\Delta (\Delta _a)$ with each $a_i$ bounded by some fixed power of H is $O_\epsilon (H^\epsilon )$ . Therefore, we have proved that

(5.7) $$ \begin{align} \sum_{H<m\leq H^{n/2}}|\mu(m)|\#\{f\in {\mathcal W}_m:H(f)<H\}=O_\epsilon(H^{n(n+1)/2-1+\epsilon}). \end{align} $$

It remains to consider the small range $1\leq m \leq H$ . For this range, note that $m^2\leq H^2$ is less than the range of $a_2$ , and so we fibre over $a_1$ only. Denote the density of ${\mathcal W}_m$ in $V_n({\mathbb Z})$ by $\theta (m)$ . When $m=p$ is a prime, we have $\theta (p) = O(1/p^2)$ . When m is squarefree in general, we have $\theta (m) = O_\epsilon (1/m^{2-\epsilon }).$ For any integer $a_1$ , let $V_n(a_1,{\mathbb Z})$ denote the set of monic polynomials of degree n whose $x^{n-1}$ -coefficient is $a_1$ and let $\theta (a_1,m)$ denote the density of ${\mathcal W}_m\cap V_n(a_1,{\mathbb Z})$ in $V_n(a_1,{\mathbb Z})$ . When m is coprime to n, we simply have $\theta (a_1,m) = \theta (m).$

We fibre over $a_1$ and break the regions for $a_2,\ldots ,a_n$ into intervals of length $m^2$ to obtain

$$ \begin{align*} &\sum_{1\leq m\leq H}\mu(m)\#\{f\in {\mathcal W}_m:H(f)<H\}\\ &=\displaystyle\sum_{1\leq m\leq H}\mu(m)\sum_{|a_1|<H}(\theta(a_1,m)2^{n-1}H^{n(n+1)/2-1}+O(H^{n(n+1)/2-3})) \\ &=2^{n-1}H^{n(n+1)/2-1}\sum_{d\mid n}\sum_{\substack{1\leq m_1\leq H/d\\ \gcd(m_1,n) = 1}} \mu(d)\mu(m_1) \sum_{|a_1| < H} \theta(a_1,d)\theta(a_1,m_1) + O(H^{n(n+1)/2-1})\\ &= 2^{n-1}H^{n(n+1)/2-1}\sum_{\substack{1\leq m_1\leq H/d\\ \gcd(m_1,n) = 1}}\mu(m_1)\theta(m_1)\sum_{d\mid n}\mu(d)\sum_{|a_1| < H}\theta(a_1,d)+ O(H^{n(n+1)/2-1}). \end{align*} $$

For any squarefree $d\mid n$ , we also have

$$ \begin{align*}\sum_{|a_1|<H}(\theta(a_1,d)2^{n-1}H^{n(n+1)/2-1} + O(H^{n(n+1)/2-3})) = \theta(d)2^nH^{n(n+1)/2} + O(H^{n(n+1)/2-1}),\end{align*} $$

as both sides count monic polynomials of degree n with height bounded by H having discriminant divisible by $d^2$ . Hence,

$$ \begin{align*}\sum_{|a_1| < H}\theta(a_1,d) = \theta(d)\cdot 2H + O(1),\end{align*} $$

and, thus,

$$ \begin{align*}\sum_{d\mid n}\mu(d)\sum_{|a_1| < H}\theta(a_1,d) = 2H\sum_{d\mid n}\mu(d)\theta(d) + O(1),\end{align*} $$

since the sum of $O(1)$ over $d\mid n$ is independent of H. Finally, combined with

$$ \begin{align*}\sum_{\substack{1\leq m_1<H/d\\ \gcd(m_1,n) = 1}}\mu(m_1)\theta(m_1) \ll_\epsilon \sum_{m_1\geq1}\frac{1}{m_1^{2-\epsilon}} = O(1)\end{align*} $$

and

$$ \begin{align*}\sum_{1\leq m\leq H}\mu(m)\theta(m) = \sum_{m\geq1}\mu(m)\theta(m) - \sum_{m>H}\mu(m)\theta(m) = \lambda_n - O_\epsilon(\sum_{m>H}\frac{1}{m^{2-\epsilon}}) = \lambda_n - O_\epsilon(\frac{1}{H^{1-\epsilon}}),\end{align*} $$

we have

(5.8) $$ \begin{align} \sum_{1\leq m\leq H}\mu(m)\#\{f\in {\mathcal W}_m:H(f)<H\} = \lambda_n 2^n H^{n(n+1)/2} + O_\epsilon(H^{n(n+1)/2-1+\epsilon}). \end{align} $$

The first estimate of Theorem 1.6 now follows from (5.4), (5.5), (5.7) and (5.8). The second estimate follows similarly.

Proof. For an integer k, the transformation $f(x)\mapsto f(x+k)$ does not change membership in ${\mathcal W}_m$ and changes the height of f by at most $O(|k|)$ . Hence, the set of elements in ${\mathcal W}_m$ having vanishing subleading coefficient and height $<H$ are each equivalent (under some transformation $f(x)\mapsto f(x+k)$ ) to $\gg H$ elements in ${\mathcal W}_m$ having height $\ll H$ . Therefore,

$$ \begin{align*} \#\bigcup_{\substack{m>M }}\{f\in {\mathcal W}^{\circ}_{m}:H(f)<H\}\ll_n \frac{1}{H}\#\bigcup_{\substack{m>M }}\{f\in {\mathcal W}_{m}:H(f)<H\}. \end{align*} $$

The result now follows from Theorem 1.5.

Proof of Theorem 1.2

Let $N_n^{\mathrm {prim}}(X)$ denote the number of primitive number fields of degree n having absolute discriminant less than X. As explained in the Introduction, the set of primitive number fields with absolute discriminant less than X injects into the set of integer monic polynomials of degree n with vanishing subleading coefficient and height $\ll X^{1/(2n-2)}$ . Let $S_X$ denote the image of this injection. Choose $\kappa =n-1$ . By Proposition 2.1, it follows that away from a set $S_X'$ of size $O(X^{\frac {n+2}{4}-\frac {1}{2n-2}})$ , every element in S has absolute discriminant $\gg X^{\frac {n-1}{2}}$ . Since the absolute discriminant of the field corresponding to an element in $S_X$ is less than X by definition, the absolute discriminant of any element in $S_X\backslash S_X'$ is divisible by $m^2$ for some $m\gg X^{\frac {n-3}{4}}$ . By Theorem 5.2, we thus deduce that

$$ \begin{align*} \begin{array}{rcl} N_n^{\mathrm{prim}}(X)&\leq& \#S_X\;\;=\;\;\#S_X'+\#(S_X\backslash S_X') \\[.2in]&\ll_\epsilon&\displaystyle X^{\frac{n+2}{4}-\frac{1}{2n-2}}+ \frac{X^{\frac{n+2}{4}+\epsilon}}{X^{\frac{n-3}{2(n+3)}}}+X^{\frac{n+2}{4}-\frac{1}{2n-2}+\frac{1}{2^{2g}(2n-2)}+\epsilon}. \end{array} \end{align*} $$

Since $\frac {n-3}{2(n+3)}\geq \frac {1}{2n-2}$ for $n\geq 6$ , we have proved the version of Theorem 1.2 where $N_n(X)$ is replaced by $N_n^{\mathrm {prim}}(X)$ .

Finally, we note that the bound [Reference Schmidt14, Equation (1.2)] with $L={\mathbb Q}$ implies that the number of imprimitive number fields of degree n with absolute discriminant less than X is at most $O(X^{\frac {n}{8}+\frac {1}{2}})$ . This completes the proof of Theorem 1.2.