Proof. Let e be the ramification index of $K/F$ . For $s \in \frac 1 e \mathbb Z$ , we denote by $B_s$ the closed ball of K of radius $q^{-s}$ and centre $0$ . If $\pi $ denotes a uniformiser of K, the set $B_s$ can be alternatively defined by $B_s = \pi ^{es} {\mathcal O}_K$ . We deduce from the latter equality that $\lambda _K(B_s) = \Vert \pi ^e \Vert ^{sr} = q^{-sr}$ .

We consider an element $a \in U$ such that $f(a) = 0$ . From our assumption, we know that $f'(a) \neq 0$ . Therefore, applying [Reference Caruso, Roe and Vaccon6, Lemma 3.4], we get the existence of a positive integer $S_a$ having the following property: for any $s \in \frac 1 e \mathbb Z$ , $s \geq S_a$ , the function f induces a bijection from $a + B_s$ to $f'(a) B_s$ . Up to enlarging $S_a$ , we can further assume that $\Vert f'(x) \Vert = \Vert f'(a) \Vert $ for all $x \in a + B_{S_a}$ . We deduce for these two facts that

(2.2)

for all $s \geq S^{\prime }_a = S_a - \log _q \Vert f'(a) \Vert $ .

From the previous discussion, we also derive that a is the unique zero of f in $B_{S_a}$ . In other words, the set of zeros of f is discrete. By compacity, it follows that f has only finitely many zeros in U. Let us call them $a_1, \ldots , a_m$ . Set $S = \max (S_{a_1}, \ldots , S_{a_m})$ and $S' = \max (S^{\prime }_{a_1}, \ldots , S^{\prime }_{a_m})$ and for $i \in \{1, \ldots , m\}$ , write $U_i = a_i + B_{S_{a_i}}$ . Up to again enlarging $S_{a_i}$ s, we can assume that the $U_i$ s are pairwise disjoint. Summing the equalities in equation (2.2), we find

(2.3)

provided that $s \geq S'$ . Let V be the complement in U of $U_1 \sqcup \cdots \sqcup U_m$ . It is compact, and the function f does not vanish on it. Hence, if s is large enough, we have $\Vert f(x) \Vert> q^{-s}$ for all $x \in V$ . For those s, we thus get

(2.4)

Combining equations (2.3) and (2.4), we find that the equality

holds true when s is sufficiently large. Passing to the limit, we get the theorem.

Proof. For simplicity, write

Let Z be the minimal monic polynomial of x over F. By our assumptions, Z has degree r and coefficients in ${\mathcal O}_F$ . This implies that the map $\Omega _{n-r} \times \Omega _{r-1} \to \Omega _n$ taking $(Q, R)$ to $QZ + R$ preserves the measure. Performing the corresponding change of variables, we end up with the equality

We consider the evaluation morphism $\alpha _x: F \otimes _{{\mathcal O}_F} \Omega _{r-1} \to K$ taking a polynomial R to $R(x)$ . It is F-linear and bijective since the domain of $\alpha _x$ is restricted to polynomials of degree strictly less than r. Its inverse $\alpha _x^{-1}$ is F-linear, so it is continuous. Thus, there exists a positive constant $\gamma $ such that $\Vert R(x) \Vert \geq \gamma \cdot \Vert R \Vert $ for all $R \in \Omega _{r-1}$ .

Let us assume for a moment that we are given a polynomial $R \in \Omega _{r-1}$ such that $\Vert R(x) \Vert \leq q^{-s}$ . Based on the preceding, we find that $\Vert R \Vert \leq \gamma ^{-1} q^{-s}$ , from which we further deduce that $\Vert R'(x) \Vert \leq \gamma ^{-1} q^{-s}$ . Since $Z'(x)$ does not vanish, we conclude that $Z'(x)$ must divide $R'(x)$ provided that s is large enough. One can then perform the change of variables $Q \mapsto Q - \frac {R'(x)}{Z'(x)}$ in the inner integral and get

$$ \begin{align*} \int_{\Omega_{n-r}} \hspace{-1em} \Vert Q(x)Z'(x)+R'(x) \Vert^r \, dQ & = \int_{\Omega_{n-r}} \hspace{-1em} \Vert Q(x)Z'(x) \Vert^r \, dQ \\ & = \Vert Z'(x) \Vert^r \cdot \int_{\Omega_{n-r}} \hspace{-1em}\Vert Q(x) \Vert^r \, dQ. \end{align*} $$

We are then left with

(2.6)

In order to estimate the last factor, we come back to the evaluation morphism $\alpha _x$ . Since it is a F-linear isomorphism, it must act on the measures by multiplication by some scalar (namely, its determinant). In other words, there exists a positive constant $\delta $ such that $\lambda _K(\alpha _x(H)) = \delta \cdot \mu _n(H)$ for all measurable subsets H of $\Omega _{r-1}$ . Taking $H = \Omega _{r-1}$ , we find

$$ \begin{align*}\delta = \lambda_K\big(\alpha_x(\Omega_{r-1})\big) = \lambda_K\big({\mathcal O}_F[x]\big) = \frac 1{\#\big({\mathcal O}_K/{\mathcal O}_F[x]\big)}.\end{align*} $$

As in the proof of Theorem 2.1, we let $B_s$ be the closed ball of K of radius $q^{-s}$ centred at $0$ . By definition, $\alpha _x^{-1}(B_s)$ consists of polynomials R such that $\Vert R(x) \Vert \leq q^{-s}$ . Moreover, if s is sufficiently large, $\alpha _x^{-1}(B_s)$ sits inside $\Omega _{r-1}$ , and so

Plugging this input in equation (2.6), we end up with

(2.7) $$ \begin{align} I_s = \Vert Z'(x) \Vert^r \cdot \#\big({\mathcal O}_K/{\mathcal O}_F[x]\big) \cdot \int_{\Omega_{n-r}} \hspace{-1em} \Vert Q(x) \Vert^r \, dQ \end{align} $$

when s is sufficiently large. The sequence $(I_s)_{s \geq 0}$ is then eventually constant and converges to the limit given by the above formula. In order to conclude the proof of the proposition, it remains to relate the norm of $Z'(x)$ with that of the discriminant of K. We endow K with the symmetric F-bilinear form $b : K \times K \to F$ , $(u,v) \mapsto \mathrm {Tr}_{K/F}(uv)$ . Also let $\mathcal B_x = (1, x, \ldots , x^{r-1})$ be the canonical basis of ${\mathcal O}_F[x]$ over ${\mathcal O}_F $ , and set $\mathcal B^{\prime }_x = \big (\frac {x^{r-1}}{Z'(x)}, \frac {x^{r-2}}{Z'(x)}, \ldots , \frac 1 {Z'(x)}\big )$ . Both $\mathcal B_x$ and $\mathcal B^{\prime }_x$ are F-basis of K, and it follows from [Reference Serre19, §III.6, Lemma 2] that the matrix of b in the basis $\mathcal B_x$ and $\mathcal B^{\prime }_x$ is lower-triangular with all diagonal entries equal to $1$ . Its determinant is then $1$ as well. Performing a change of basis, we find that

$$ \begin{align*} \det \mathrm{Mat}_{\mathcal B_x}(b) = \pm\:N_{K/F}\big(Z'(x)\big),\\[-15pt] \end{align*} $$

where, by definition, $\mathrm {Mat}_{\mathcal B_x}(b)$ is the matrix of b in $\mathcal B_x$ and $N_{K/F}$ is the norm map of K over F. We now consider a basis $\mathcal B$ of ${\mathcal O}_K$ over ${\mathcal O}_F$ . By definition, the discriminant of K is the determinant of $\mathrm {Mat}_{\mathcal B}(b)$ . Therefore, if P denotes the transition matrix from $\mathcal B$ to $\mathcal B_x$ , we derive from the change-of-basis formula that $\det \mathrm {Mat}_{\mathcal B_x}(b) = (\det P)^2 \cdot D_K$ and so

$$ \begin{align*} N_{K/F}\big(Z'(x)\big) = \pm\: (\det P)^2 \cdot D_K.\\[-15pt] \end{align*} $$

On the other hand, noticing that P is also the matrix of $\alpha _x$ in the canonical basis, we deduce that

$$ \begin{align*} \Vert\!\det P \Vert = \lambda_K\big({\mathcal O}_F[x]\big) = \frac 1 {\#\big({\mathcal O}_K/{\mathcal O}_F[x]\big)}.\\[-15pt] \end{align*} $$

Combining the two previous equalities, we end up with

$$ \begin{align*}\Vert Z'(x) \Vert^r = \Vert N_{K/F}\big(Z'(x)\big) \Vert = \frac 1 {\#\big({\mathcal O}_K/{\mathcal O}_F[x]\big)^2} \cdot \Vert D_K \Vert.\\[-15pt]\end{align*} $$

Plugging this relation into equation (2.7), we obtain the proposition.

Proof. As usual, let r be the degree of the extension $K/F$ . Let $K^{\mathrm {new}}$ be the subset of K consisting of elements x for which $F[x] = K$ . Since K contains only finitely many subextensions and each subextension is a closed subspace of K, we deduce that $K^{\mathrm {new}}$ is open in K. We set ${\mathcal O}_K^{\mathrm {new}} = {\mathcal O}_K \cap K^{\mathrm {new}}$ .

To start with, we assume that U is compact and included in ${\mathcal O}_K^{\mathrm {new}}$ . We then have $Z^{\mathrm {new}}_{U,n} = Z_{U,n}$ . Since a random polynomial has almost surely only simple roots, we deduce from the p-adic Kac-Rice formula (compare Theorem 2.1) that

It is clear that $I_s(x,P)$ is nonnegative everywhere. As in the proof of Theorem 2.1, let $B_s$ be the closed ball of K of radius $q^{-s}$ and centre $0$ . From [Reference Evans8, Proposition 2.3], we deduce that

$$ \begin{align*}\int_{P^{-1}(B_s)} \hspace{-0.2em} \Vert P'(x) \Vert^r \, dx \,=\, \int_{B_s} \# P^{-1}(y) \, dy \,\leq\, n \lambda_K(B_s) \,=\, n q^{-rs}\end{align*} $$

for any polynomial $P \in \Omega _n$ . Hence

$$ \begin{align*}\int_U I_s(x, P) \, dx = q^{sr} \int_{U \cap P^{-1}(B_s)} \Vert P'(x) \Vert^r \, dx \,\leq\, n.\end{align*} $$

We can then apply Lebesgue’s dominated convergence theorem and get

$$ \begin{align*}\mathbb E[Z_{U,n}] = \lim_{s \to \infty} \, \int_{\Omega_n} \int_U I_s(x, P) \, dx\, dP = \lim_{s \to \infty} \, \int_U \int_{\Omega_n} I_s(x, P) \, dP\, dx,\end{align*} $$

where the second equality comes from Fubini’s theorem. We now want to use a similar argument to swap the limit on s and the integral over U. In order to proceed, we fix an element $x \in U$ and denote by $\alpha _x : F \otimes _{{\mathcal O}_F} \Omega _n \to K$ the evaluation morphism at x already considered in the proof of Proposition 2.2. Noticing that $\Vert P'(x) \Vert ^r \leq 1$ for all $P \in \Omega _n$ , we find

$$ \begin{align*} \int_{\Omega_n} I_s(x, P) \, dP & \leq q^{sr} \cdot \mu_n\big(\Omega_n \cap \alpha_x^{-1}(B_s)\big) \\ & \leq q^{sr} \cdot \mu_n\big(\alpha_x^{-1}(B_s)\big) = \#\big({\mathcal O}_K/{\mathcal O}_F[x]\big). \end{align*} $$

We saw in Section 2.3.1 that the function $x \mapsto \#\big ({\mathcal O}_K/{\mathcal O}_F[x]\big )$ is continuous; hence it is integrable on the compact set U. The dominated convergence theorem then again applies and gives

$$ \begin{align*}\mathbb E[Z_{U,n}] = \int_U \, \lim_{s \to \infty} \int_{\Omega_n} I_s(x, P) \, dP\, dx = \int_U \rho_{K,n}(x) \, dx.\end{align*} $$

Theorem 2.6 is then proved when U is compact and included in ${\mathcal O}_K^{\mathrm {new}}$ .

One can extend the result to any open subset of ${\mathcal O}_K$ using a standard limit argument. Precisely, if U is open in ${\mathcal O}_K$ , one can construct an increasing sequence $(U_m)_{m \geq 0}$ of compact open subsets of ${\mathcal O}_K^{\mathrm {new}}$ such that $\bigcup _{m \geq 0} U_m = U \cap {\mathcal O}_K^{\mathrm {new}}$ . Applying what we have done before with $U_m$ , we find

(2.11) $$ \begin{align} \mathbb E[Z_{U_m,n}] = \int_{U_m} \rho_{K,n}(x) \: dx. \end{align} $$

Moreover, the sequence $Z_{U_m,n}$ is nondecreasing and simply converges to $Z^{\mathrm {new}}_{U,n}$ . By the monotone convergence theorem, we find that $\mathbb E[Z_{U_m,n}]$ converges to $\mathbb E[Z^{\mathrm {new}}_{U,n}]$ . Passing to the limit in equation (2.11), we obtain the theorem for U.

For a general U, we write $U = U_0 \sqcup U_\infty $ with $U_0 = U \cap {\mathcal O}_K$ and $U_\infty = U \backslash U_0$ . Since both sides of equation (2.10) are additive with respect to U, it is enough to prove the theorem for $U_\infty $ . For this, as in Section 2.3.2, we consider the measure-preserving map $\tau : \Omega _n \to \Omega _n$ defined by

$$ \begin{align*}\tau\big(a_n X^n + \cdots + a_1 X + a_0\big) = a_0 X^n + \cdots + a_{n-1} X + a_n.\end{align*} $$

An element $x \in K$ is a root of a polynomial P if and only if $x^{-1}$ is a root of $\tau (P)$ . It is also obvious that $x \in K^{\mathrm {new}}$ if and only if $x^{-1} \in K^{\mathrm {new}}$ . Thus we get $\mathbb E\big [Z^{\mathrm {new}}_{U_\infty ,n}\big ] = \mathbb E\big [Z^{\mathrm {new}}_{V_\infty ,n}\big ]$ , where $V_\infty $ is the image of $U_\infty $ under the map $x \mapsto x^{-1}$ . In addition, $V_\infty \subset {\mathcal O}_K$ ; we can then apply the theorem with $V_\infty $ and conclude that

$$ \begin{align*}\mathbb E\big[Z^{\mathrm{new}}_{U_\infty,n}\big] = \int_{V_\infty} \rho_{K,n}(x) \: dx = \int_{U_\infty} \rho_{K,n}(x^{-1}) \cdot \Vert x \Vert^{-2r} \: dx = \int_{U_\infty} \rho_{K,n}(x) \: dx,\end{align*} $$

which finally proves the theorem in all cases.

Proof. Since $K/F$ is unramified, its discriminant has norm $1$ and thus does not contribute. We fix an element $\zeta \in {\mathcal O}_K$ with norm $1$ such that ${\mathcal O}_K = {\mathcal O}_F[\zeta ]$ . The family $\mathcal B = (1, \zeta )$ is a basis of ${\mathcal O}_K$ over ${\mathcal O}_F$ . Let $x \in {\mathcal O}_K$ , $x \not \in {\mathcal O}_F$ , and write $x = a + b \zeta $ with $a, b \in {\mathcal O}_F$ , $b \neq 0$ . We have ${\mathcal O}_F[x] = {\mathcal O}_F[b \zeta ]$ , which shows that a basis of ${\mathcal O}_F[x]$ over ${\mathcal O}_F$ is simply $\mathcal B_x = (1, b\zeta )$ . Comparing $\mathcal B$ and $\mathcal B_x$ , we find $\#\big ({\mathcal O}_K/{\mathcal O}_F[x]\big ) = \#\big ({\mathcal O}_F/b{\mathcal O}_F\big ) = \Vert b \Vert ^{-1}$ . Observing in addition that a is the closest element of x in F, we can reinterpret the norm of b as the distance of x to F. Putting all ingredients together, we obtain the announced formula for $\rho _{K,2}(x)$ . This formula has been established when $x \not \in {\mathcal O}_F$ ; however, it obviously also holds true when $x \in {\mathcal O}_F$ since $\rho _{K,2}(x)$ vanishes in this case by definition.

We now move to the computation of $\rho _{K,n}$ for $n \geq 3$ . We continue to consider an element $x \in {\mathcal O}_K$ , $x \not \in {\mathcal O}_F$ and write $x = a + b \zeta $ with $a, b \in {\mathcal O}_F$ , $b \neq 0$ . Performing a change of variables or, more simply, coming back to Definition 2.3, we have

$$ \begin{align*}\rho_{K,n}(x) = \frac 1 {\#\big({\mathcal O}_K/{\mathcal O}_F[x]\big)} \cdot \int_{{\mathcal O}_F^2} \Vert u + vx \Vert^2 \: du \: dv = \Vert b \Vert \cdot \int_{{\mathcal O}_F^2} \Vert u + vx \Vert^2 \: du \: dv.\end{align*} $$

Replacing t by $t - ua$ , this reduces to

$$ \begin{align*}\rho_{K,n}(x) = \Vert b \Vert \cdot \int_{{\mathcal O}_F^2} \Vert u + vb\zeta \Vert^2 \: du \: dv = \Vert b \Vert \cdot \int_{{\mathcal O}_F^2} \max\big(\Vert tu\Vert^2, \Vert vb \Vert^2 \big) \: du \: dv.\end{align*} $$

As in Section 2.3.3, we decompose ${\mathcal O}_F$ as the disjoint union ${\mathcal O}_F = \{0\} \sqcup \bigcup _{s \geq 0} U_s$ , where $U_s$ consists of elements of norm $q^{-s}$ . Decomposing the integral accordingly and writing $\Vert b \Vert = q^{-\delta }$ , we find

$$ \begin{align*}\rho_{K,n}(x) = \left( 1 - \frac 1 q\right)^2 q^{-\delta} \cdot \sum_{s=0}^\infty \sum_{t=0}^\infty q^{-s-t-2\min(s, t+\delta)}.\end{align*} $$

Computing the latter double sum is painful but straightforward. We split the domain of summation into three regions, namely

$$ \begin{align*} D_1 & = \big\{ \, (s,t) \in \mathbb Z_{\geq 0}^2 \quad \text{such that} \quad s < \delta \,\big\}, \\ D_2 & = \big\{ \, (s,t) \in \mathbb Z_{\geq 0}^2 \quad \text{such that} \quad \delta \leq s \leq t + \delta \,\big\}, \\ D_3 & = \big\{ \, (s,t) \in \mathbb Z_{\geq 0}^2 \quad \text{such that} \quad s> t + \delta \,\big\}. \end{align*} $$

We then compute the double sum separately on each domain:

• ○ over $D_1$ : $\displaystyle \sum _{s=0}^{\delta -1} \sum _{t=0}^\infty q^{-3s-t} = \frac {q^4}{(q-1)(q^3-1)}\cdot (1 - q^{-3\delta })$ ,

• ○ over $D_2$ : $\displaystyle \sum _{s=\delta }^\infty \sum _{t=s-\delta }^\infty q^{-3s-t} = q^{-3\delta } \sum _{s=0}^\infty \sum _{t=s}^\infty q^{-3s-t} = \frac {q^5}{(q-1)(q^4-1)} \cdot q^{-3\delta }$ ,

• ○ over $D_3$ : $\displaystyle \sum _{t=0}^\infty \sum _{s=t+\delta +1}^\infty q^{-s-3t-2\delta } = q^{-3\delta -1} \sum _{t=0}^\infty \sum _{s=t}^\infty q^{-s-3t} = \frac {q^4}{(q-1)(q^4-1)} \cdot q^{-3\delta }$ .

Summing the contributions and noticing $q^{-\delta } = \Vert b \Vert = \mathrm {dist}(x,F)$ , we finally find the expression given in the statement of the proposition. As previously, we notice that this formula continues to hold when $x \in {\mathcal O}_F$ , given that $\rho _{K,n}(x)$ vanishes by definition in this case.

Proof. The proof is similar to that of Proposition 3.2, so we only sketch the argument. We fix a uniformiser $\pi $ of K. The family $(1,\pi )$ forms a basis of ${\mathcal O}_K$ over ${\mathcal O}_F$ . Let $x \in {\mathcal O}_K$ , $x \not \in {\mathcal O}_F$ , and write $x = a + b\pi $ with $a, b \in {\mathcal O}_F$ , $b \neq 0$ . From the fact that $(1, b\pi )$ is a ${\mathcal O}_F$ -basis of ${\mathcal O}_F[x]$ , we deduce that the cardinality of ${\mathcal O}_K/{\mathcal O}_F[x]$ is $\Vert b \Vert ^{-1}$ . Noticing that $\mathrm {dist}(x, F) = \Vert b \pi \Vert = \sqrt {q} \cdot \Vert b \Vert $ , we get the announced formula for $\rho _{K,2}(x)$ .

When $n \geq 3$ , we start with the formula

$$ \begin{align*}\rho_{K,n}(x) = \frac {\Vert D_K \Vert}{\#\big({\mathcal O}_K/{\mathcal O}_F[x]\big)} \cdot \int_{{\mathcal O}_F^2} \Vert u + vx \Vert^2 \: du \: dv.\end{align*} $$

Decomposing the integral into slices where $\Vert u \Vert $ and $\Vert v \Vert $ are constant, we obtain

$$ \begin{align*}\frac{\rho_{K,n}(x)}{\Vert D_K \Vert} = \left( 1 - \frac 1 q\right)^2 q^{-\delta} \cdot \sum_{s=0}^\infty \sum_{t=0}^\infty q^{-s-t-\min(2s, 2t+2\delta+1)},\end{align*} $$

where $\delta $ is defined by $\Vert b \Vert = q^{-\delta }$ . Finally, splitting the previous double sum into three parts exactly as we did in the unramified case, we end up after some calculations with the formula displayed in the statement of the proposition.

Proof. We first assume that $K/F$ is unramified. Writing ${\mathcal O}_K = {\mathcal O}_F[\zeta ]$ as in the proof of Proposition 3.2, we find

$$ \begin{align*}\int_{{\mathcal O}_K} \mathrm{dist}(x,F)^d \: dx \,=\, \int_{{\mathcal O}_F} \Vert b \Vert^d \: db = q^d \alpha_d,\end{align*} $$

the last equality being nothing but equation (2.8) established in Section 2.3.3. Similarly noticing that $x = a + b \zeta $ lies in $\mathfrak m_K$ if and only if both a and b falls in the maximal ideal $\mathfrak m_F$ of ${\mathcal O}_F$ , we obtain

$$ \begin{align*}\int_{\mathfrak m_K} \mathrm{dist}(x,F)^d \: dx \,=\, \lambda(\mathfrak m_F) \cdot \int_{\mathfrak m_F} \Vert b \Vert^d \: db \,=\, q^{-1} \int_{\mathfrak m_F} \Vert b \Vert^d \: db.\end{align*} $$

Fixing a uniformiser $\pi _F$ of F and performing the change of variables $b = \pi _F t$ , we finally get

$$ \begin{align*}\int_{\mathfrak m_K} \mathrm{dist}(x,F)^d \: dx \,=\, q^{-d-2} \int_{{\mathcal O}_F} \Vert t \Vert^d \: dt = q^{-2} \alpha_d.\end{align*} $$

The argument in the totally ramified case is similar. We pick a uniformiser $\pi $ of K and, writing ${\mathcal O}_K = {\mathcal O}_F[\pi ]$ , find

$$ \begin{align*}\int_{{\mathcal O}_K} \mathrm{dist}(x,F)^d \: dx \,=\, \int_{{\mathcal O}_F} \Vert b \pi \Vert^d \: db \,=\, \Vert \pi \Vert^d \cdot q^d \alpha_d \,=\, q^{d/2} \alpha_d.\end{align*} $$

For the integral over $\mathfrak m_K$ , we observe that $a + b\pi \in {\mathcal O}_K$ if and only if $a \in \mathfrak m_F$ . Therefore

$$ \begin{align*}\int_{\mathfrak m_K} \mathrm{dist}(x,F)^d \: dx \,=\, \lambda(\mathfrak m_F) \cdot \int_{{\mathcal O}_F} \Vert b \pi \Vert^d \: db \,=\, q^{d/2\,-\,1} \alpha_d,\end{align*} $$

which concludes the proof.

Proof. We assume first that $K/F$ is unramified. Let $x \in {\mathcal O}_K$ , $x \not \in {\mathcal O}_F$ . By compacity, there exists $a \in {\mathcal O}_F$ such that $\Vert x - a \Vert = \mathrm {dist}(x, F)$ . We pick such an element a and write $x - a = \pi ^\delta \zeta $ , where $\pi $ is a fixed uniformiser of F and $\zeta \in {\mathcal O}_K$ has norm $1$ . Let $k_F$ and $k_K$ denote the residue fields of F and K, respectively, and let $\bar \zeta $ be the image of $\zeta $ in $k_K$ . We claim that $\bar \zeta \not \in k_F$ ; otherwise, there would exist $b \in {\mathcal O}_F$ with $b \equiv \zeta \pmod \pi $ and so $\Vert x - (a + \pi ^\delta b) \Vert \leq q^{-\delta -1} < q^{-\delta } = \Vert x - a \Vert $ , contradicting the minimality property of a. Given that the extension $k_K/k_F$ has prime degree, we deduce that $k_K = k_F[\bar \zeta ]$ and, consequently, that ${\mathcal O}_K = {\mathcal O}_F[\zeta ]$ . The family $(1, \zeta , \ldots , \zeta ^{r-1})$ is then a basis of ${\mathcal O}_K$ over ${\mathcal O}_F$ , while $(1, \pi ^\delta \zeta , \ldots , (\pi ^\delta \zeta )^{r-1})$ is a basis of ${\mathcal O}_F[x]$ over ${\mathcal O}_F$ . This shows that

$$ \begin{align*}\#\big({\mathcal O}_K/{\mathcal O}_F[x]\big) = q^{\delta + 2\delta + \cdots + (r-1)\delta} = q^{\delta r(r-1)/2} = \mathrm{dist}(x,F)^{-r(r-1)/2},\end{align*} $$

from which we deduce the claimed formula for $\rho _{r,K}(x)$ .

We now move to the totally ramified case. Let $\pi _F$ (respectively, $\pi _K$ ) denote a fixed uniformiser of F (respectively, K). Then ${\mathcal O}_K = {\mathcal O}_F[\pi _K]$ and the family $\mathcal B = (1, \pi _K, \ldots , \pi _K^{r-1})$ is a ${\mathcal O}_F$ -basis of ${\mathcal O}_K$ . Let $x \in {\mathcal O}_K$ . As before, if $x \in {\mathcal O}_F$ , both sides of the equality we want to prove vanish, and there is nothing more to do. We then assume $x \not \in {\mathcal O}_F$ and consider $a \in {\mathcal O}_F$ such that $\Vert x - a \Vert = \mathrm {dist}(x,F)$ . Write $x - a = \pi _K^\delta u$ with $\delta \in \mathbb Z_{\geq 0}$ and $u \in {\mathcal O}_K^\times $ . The minimality condition in the definition of a implies that $\delta \not \equiv 0 \pmod r$ . In addition, the family $\mathcal B_x = (1, \pi _K^\delta u, \ldots , (\pi _K^\delta u)^{r-1})$ is a ${\mathcal O}_F$ -basis of ${\mathcal O}_F[x]$ . For $i \in \{0, \ldots , r{-}1\}$ , we decompose $(\pi _K^\delta u)^i$ in the basis $\mathcal B$ : that is, we write

(3.1) $$ \begin{align} (\pi_K^\delta u)^i = \sum_{j=0}^{r-1} a_{ij} \pi_K^j \end{align} $$

with $a_{ij} \in {\mathcal O}_F$ . Set $c = q^{1/r}$ . Taking norms in equation (3.1), we get $c^{-\delta i} = \max _{1 \leq j < r} \big (\Vert a_{ij} \Vert {\cdot } c^{-j}\big )$ . In other words, $\Vert a_{ij} \Vert \leq c^{j-\delta i}$ for all j, and the equality holds for at least one index j. On the other hand, the inequality is certainly strict as soon as r does not divide $j-\delta i$ because $\Vert a_{ij} \Vert $ is a power of $q = c^r$ . Therefore, if $j_i$ denotes the remainder in the division of $\delta i$ by r, we conclude that $\Vert a_{ij} \Vert \leq c^{j-\delta i}$ for all j with equality if and only if $j = j_i$ . Let A be the change-of-basis matrix from $\mathcal B$ to $\mathcal B_x$ . By definition, its entries are exactly the $a_{ij}$ s. Let P be the permutation matrix associated to $i \mapsto j_i$ , and define

$$ \begin{align*}B = \left(\begin{matrix} \pi_F^{-\delta_0} \\ & \ddots \\ & & \pi_F^{-\delta_{r-1}} \end{matrix} \right) \cdot P^{-1} \cdot A \qquad \text{with} \quad \delta_i = \frac{\delta i - j_i}r.\end{align*} $$

The estimations on $\Vert a_{ij} \Vert $ we have obtained previously show that B has entries in ${\mathcal O}_F$ and that $B \text { mod } \pi _F$ is lower-triangular with nonzero diagonal entries. Hence B is invertible over ${\mathcal O}_F$ , which gives $\Vert \!\det B \Vert = 1$ . Since P is clearly invertible as well, we conclude that

$$ \begin{align*}\#\big({\mathcal O}_K/{\mathcal O}_F[x]\big) = \Vert\!\det A \Vert^{-1} = q^{\delta_0 + \cdots + \delta_r} = q^{(\delta-1)(r-1)/2}.\end{align*} $$

Remembering that $\mathrm {dist}(x,F) = \Vert x-a \Vert = \Vert \pi _K^\delta u \Vert = q^{-\delta /r}$ , we finally find the formula given in the proposition.

Proof. Set $m = n'-r$ . Let $k_F$ (respectively, $k_K$ ) denote the residue field of F (respectively, K). Let $\xi \in k_K$ be the image of x. Our assumption implies that $k_F[\xi ] = k_K$ . Therefore, if an expression of the form $u_0 + u_1 x + \cdots + u_m x^m$ with $u_i \in {\mathcal O}_F$ vanishes in $k_K$ , then all $u_i$ have to vanish in $k_F$ . We deduce from this property that

$$ \begin{align*}\Vert u_0 + u_1 x + \cdots + u_m x^m \Vert = \max\big(\Vert u_0 \Vert, \, \Vert u_1 \Vert,\, \ldots, \, \Vert u_m \Vert \big)\\[-15pt]\end{align*} $$

for all $u_0, \ldots , u_m \in {\mathcal O}_F$ . It then follows from Definition 2.3 and our assumptions that

$$ \begin{align*}\rho_{n,K}(x) = \int_{{\mathcal O}_F^{m+1}} \max\big( \Vert u_0 \Vert^r, \, \Vert u_1 \Vert^r,\, \ldots, \, \Vert u_m \Vert^r \big) \: du_0 \cdots du_m.\\[-15pt]\end{align*} $$

Contrary to what we said at the end of Section 3.2, the latter integral can be easily computed. Let us fix a uniformiser $\pi $ of F and, for each nonnegative integer s, set $U_s = (\pi ^s {\mathcal O}_F)^{m+1}$ . The $U_s$ s then form a decreasing sequence of open subsets of ${\mathcal O}_F^{m+1}$ . Moreover, it is easy to check that $U_s$ is exactly the domain over which the integrand $\max \big (\Vert u_0 \Vert ^r, \, \ldots , \, \Vert u_m \Vert ^r \big )$ is less than or equal to $q^{-sr}$ . We deduce from this that

$$ \begin{align*}\rho_{n,K}(x) = \sum_{s=0}^\infty \big(\lambda_F^{\otimes{m+1}}(U_s) - \lambda_F^{\otimes{m+1}}(U_{s+1})\big) \cdot q^{-sr}.\\[-15pt]\end{align*} $$

Observing that $\lambda _F^{\otimes {m+1}}(U_s) = \lambda _F(\pi ^s {\mathcal O}_F)^{m+1} = q^{-s(m+1)}$ , we end up with

$$ \begin{align*}\rho_{n,K}(x) = \big( 1 - q^{-m-1}\big) \cdot \sum_{s=0}^\infty q^{-s(m+1+r)} = \frac{q^{m+1+r} - q^r}{q^{m+1+r} - 1},\end{align*} $$

which is what we wanted to prove.

Proof. A strict divisor of f cannot exceed $f/2$ . On the other hand, it is obvious from the definition that $G_m \leq q^m$ for all m. Hence, the sum of the lemma is bounded from above by

$$ \begin{align*}\sum_{m < f/2} q^{2m} \leq \frac{q^{f+2} - 1}{q^2 - 1} \leq \frac{q^f}{1 - q^{-2}} \leq 2 q^f\end{align*} $$

given that $q \geq 2$ .

Proof. Write $n = p_1^{\alpha _1} \cdots p_s^{\alpha _s}$ , where the $p_i$ s are pairwise distinct prime numbers. The divisors m of n are exactly the integers of the form $m = p_1^{\beta _1} \cdots p_s^{\beta _s}$ with $\beta _i \leq \alpha _i$ for all i. We then deduce from the definition of the Moebius function that

$$ \begin{align*} \sum_{m|n} \frac{\mu(m)} m = \sum_{\beta_1 = 0}^1 \cdots \sum_{\beta_s = 0}^1 \frac{(-1)^{\beta_1 + \cdots + \beta_s}}{p_1^{\beta_1}\ldots p_s^{\beta_s}} = \prod_{i=1}^s \left( 1 - \frac 1 {p_i}\right) = \frac{\varphi(n)}n, \end{align*} $$

which proves the lemma.

Proof. The lemma follows from the following sequence of equalities:

$$ \begin{align*} \begin{array}{@{}r@{\hspace{0.5ex}}l@{\qquad}l@{}} \displaystyle r \cdot \sum_{f|r} \frac{G_f}f & = \displaystyle r \cdot \sum_{m|f|r} \mu\left(\frac f m\right)\frac{q^m}f. & \text{by equation 4.2} \\ & = \displaystyle r \cdot \sum_{m|r} \, \sum_{n| \frac r m} \frac{\mu(n)} n \, \cdot \frac{q^m} m. & (\text{change of variables } n = \frac f m) \\ & = \displaystyle \sum_{m|r} \, \varphi\left(\frac r m\right) q^m & \text{by Lemma 4.2}. \end{array}\\[-45pt] \end{align*} $$

Proof. If $x \in {\mathcal O}_K$ generates ${\mathcal O}_K$ over ${\mathcal O}_F$ , its image in the residue field must generate $\mathbb F_{q^f}$ over $\mathbb F_q$ . This proves the lemma when $\alpha \not \in \mathcal G_f$ .

We now assume that $\alpha \in \mathcal G_f$ . First, it is clear that $\lambda _K(\mathcal G_K \cap U_\alpha ) \leq \lambda _K(U_\alpha ) = q^{-f}$ . Let $\bar Z$ be the minimal polynomial of $\alpha $ over $\mathbb F_q$ . Also let $a \in {\mathcal O}_K$ be a lifting of $\alpha $ and $Z \in \Omega _f$ be a polynomial lifting $\bar Z$ . We fix in addition a uniformiser $\pi $ of K. Then $Z(a)$ is a multiple of $\pi $ , and we write $Z(a) = \pi b$ . In addition, $\bar Z'(a)$ does not vanish given that $\mathbb F_{q^f}/ \mathbb F_q$ is a separable extension. For $x \in {\mathcal O}_K$ , the congruence

$$ \begin{align*}Z(a + \pi x) \,\equiv\, Z(a) + Z'(a) \pi x \,\equiv\, \pi \cdot \big(b + Z'(a)\:x\big)\quad \pmod{\pi^2} \end{align*} $$

shows that $Z(a + \pi x)$ is a uniformiser of K as soon as the image of x in the residue field is different from $\frac {-b}{Z'(a)}$ . When this occurs, ${\mathcal O}_F[x]$ then contains a uniformiser of K together with a generator of the residue field, implying that ${\mathcal O}_F[x] = {\mathcal O}_K$ . The lemma follows.

Proof. Set $V = K \backslash \mathcal G_K$ and $V_\alpha = V \cap U_\alpha $ for $\alpha \in \mathbb F_{q^f}$ . Also set $V_\infty = K\backslash {\mathcal O}_K$ . The set V then appears as the disjoint union of the $V_\alpha $ s for $\alpha $ varying in $\mathbb P^1(\mathbb F_{q^f})$ . We will bound the integral of $\rho _{K,n}$ on each $V_\alpha $ separately. When $\alpha \in \mathcal G_f$ , it follows from Lemma 4.4 that $\lambda _K(V_\alpha ) \leq q^{-2f}$ . In addition, coming back to Definition 2.3, we remark that $\rho _{K,n}$ is upper bounded by $\Vert D_K \Vert $ on ${\mathcal O}_K$ and so on $V_\alpha $ . We then deduce, in this case, that

$$ \begin{align*} \frac 1 {\Vert D_K \Vert} \cdot \int_{V_\alpha} \rho_{K,n}(x) \: dx \, \leq \, q^{-2f}. \end{align*} $$

We now consider the case where $\alpha \in \mathbb F_{q^f}$ , $\alpha \not \in \mathcal G_f$ . Write $\ell = {\mathbb F}_q[\alpha ]$ , and let $m(\alpha )$ be the degree of the extension $\ell /{\mathbb F}_q$ . By assumption, $m(\alpha ) < f$ . On the other hand, the reduction morphism ${\mathcal O}_K \to \mathbb F_{q^f}$ takes ${\mathcal O}_F[x]$ to $\ell $ . We deduce that the cardinality of the quotient ${\mathcal O}_K/{\mathcal O}_F[x]$ is bounded from below by the cardinality of $\mathbb F_{q^f}/\ell $ , which is $q^{f-m(\alpha )}$ . Injecting this bound into the definition of $\rho _{n,K}$ , we find

$$ \begin{align*} \frac 1 {\Vert D_K \Vert} \cdot \int_{V_\alpha} \rho_{K,n}(x) \: dx \, \leq \, \lambda_K(V_\alpha) \cdot q^{m(\alpha) - f} \, = \, q^{m(\alpha) - 2f}. \end{align*} $$

Finally, when $\alpha = \infty $ , we perform the change of variables $x \mapsto x^{-1}$ , which leaves us with

$$ \begin{align*}\int_{V_\infty} \rho_{K,n}(x) \: dx \,=\, \int_{V_0} \rho_{K,n}(x) \: dx \,\leq\, \Vert D_K \Vert \cdot q^{1 - 2f}.\end{align*} $$

Summing all the upper bounds, we find

$$ \begin{align*} \frac 1 {\Vert D_K \Vert} \cdot \int_{K\backslash\mathcal G_K} \rho_{K,n}(x) \: dx \,\leq\, q^{1-2f} \,+\, G_f q^{-2f} + \sum_{\substack{m|f\\m<f}} G_m q^{m - 2f} \end{align*} $$

given that the number of $\alpha \in \mathbb F_{q^f} \backslash \mathcal G_K$ for which $m(\alpha ) = m$ is equal to $G_m$ by definition. Remembering that $G_f \leq q^f$ and using Lemma 4.1, we end up with

$$ \begin{align*} \frac 1 {\Vert D_K \Vert} \cdot \int_{K\backslash\mathcal G_K} \rho_{K,n}(x) \: dx \,\leq\, q^{1-2f} + q^{-f} + 2 q^{-f} \,\leq\, 4 q^{-f}.\end{align*} $$

The proposition is proved.

Proof. Let $n = r$ or $n \geq 2r - 1$ . Recall that by definition

$$ \begin{align*}\rho_n(K) = \int_K \rho_{K,n}(x) \: dx \,=\, \int_{\mathcal G_K} \rho_{K,n}(x) \: dx \,+\, \int_{K\backslash\mathcal G_K} \rho_{K,n}(x) \:dx.\end{align*} $$

By Proposition 4.5, we know that the integral over $K \backslash \mathcal G_K$ is upper bounded by $4 q^{-f}$ . It is also obviously nonnegative since the integrand is nonnegative. Therefore, it is enough to prove that

$$ \begin{align*}0 \, \leq \, c_n \: \frac{G_f}{q^f} \,-\, \int_{\mathcal G_K} \frac{\rho_{K,n}(x)}{\Vert D_K \Vert} \: dx \, \leq\, q^{-f}\end{align*} $$

with $c_r = \frac {q^{r+1} - q^r}{q^{r+1} - 1}$ and $c_n = \frac {q^r}{q^r + 1}$ for $n \geq 2r -1$ . On the other hand, a computation similar to the one we carried out in Section 2.3.3 gives

$$ \begin{align*}\int_{{\mathcal O}_K} \Vert t \Vert^r \: dt = \frac{q^f}{q^f + 1}.\end{align*} $$

Hence, it follows from the explicit formulas of Theorem B that $\rho _{K,n}$ is constant equal to $c_n {\cdot } \Vert D_K \Vert $ on $\mathcal G_K$ . We are then reduced to checking that $0 \leq q^{-f} G_f - \lambda _K(\mathcal G_K) \leq q^{-f} c_n^{-1}$ . This follows from Lemma 4.4 after remarking that $0 \leq c_n \leq 1$ .

Proof. It is exactly the same as the proof of Corollary 4.6, except that the value of $\rho _{K,n}$ on $\mathcal G_K$ is now given by Proposition 3.6. (Also note that $f = r$ in the unramified case.)

Proof. See [Reference Serre18].

Proof. Let $F_f$ be the unique unramified extension of F of degree f in $\bar F$ . Its residue field is $\mathbb F_{q^f}$ and the normalised norm on $F_f$ is the fth power of $\Vert \cdot \Vert $ . Moreover, any extension K in $\mathbf {Ex}_{r,f}$ canonically contains $F_f$ and then appears uniquely as a totally ramified extension of $F_f$ . In addition, the discriminant of $K/F$ , still denoted by $D_K$ , is the fth power of the discriminant $D_{K/F_f}$ of $K/F_f$ . Hence $\Vert D_{K/F_f} \Vert ^f = \Vert D_K \Vert $ . The formula (4.4) applied with the base field $F_f$ then gives

$$ \begin{align*}\sum_{K \in \mathbf{Ex}_{r,f}} \Vert D_K \Vert = \sum_{K \in \mathbf{Ex}_{r,f}} \Vert D_{K/F_f} \Vert^f = \frac r f \cdot \frac 1 { (q^f)^{r\!/\!f\, -\,1}} = \frac r {q^r} \cdot \frac {q^f} f,\\[-16pt]\end{align*} $$

which establishes the proposition.

Proof. Let $e_i = (0, \ldots , 0, 1, 0, \ldots , 0) \in E$ with the $1$ in ith position. If x lies in the kernel of f, so does $e_i x$ for all i. This proves that $\ker f$ decomposed as $I_1 \times \cdots \times I_m$ , where $I_i$ is some ideal of $K_i$ . Since $K_i$ is a field, one must have $I_i = 0$ or $I_i = K_i$ . Moreover, since f is surjective, $\ker f$ is a maximal ideal. This shows that there exists a special index i such that $I_i = 0$ and $I_j = K_j$ for all $j \neq i$ . Hence f factors through $\text {pr}_i$ and the lemma follows.

Proof. Write $x = (x_{i,j})_{1 \leq i \leq m, 1 \leq j \leq a_i}$ . Given $i \in \{1, \ldots , m\}$ and $j \in \{1, \ldots , b_i\}$ , Lemma 5.1 tells us that the $(i,j)$ coordinate of $f(x)$ must be of the form $\varphi _{i,j}(x_{i,\sigma _i(j)})$ for some automorphism $\varphi _{i,j}$ of $K_i$ and some $\sigma _i(j) \in \{1, \ldots , b_i\}$ . This establishes the shape we have given for f. Finally, the fact that the $\sigma _i$ s are injective follows from the surjectivity of f.

Proof. It is obvious from the definition.

Proof. It follows from the fact that the image of $\varphi $ is the F-algebra generated by x.

Proof. Write $r = [K:F]$ , and pick a polynomial $P \in \Omega _n$ . Let $x = (x_1, \ldots , x_m) \in U^m$ . We claim that x is a new root of P in E if and only if $x_i$ is a new root of P in K for all i and the $x_i$ s are pairwise nonconjugate. To prove the claim, we let Z be the minimal polynomial of x over F; similarly, for all $i \in \{1, \ldots , m\}$ , we denote by $Z_i$ the minimal polynomial of $x_i$ over F. We then have $Z = \text {lcm} (Z_1,\ldots , Z_m)$ . Notice that all the $Z_i$ s are irreducible since we assume that K is a field. On the other hand, the fact that x is new in E (respectively, $x_i$ is new in K) is equivalent to the equality $\deg Z = rm$ (respectively, $\deg Z_i = r$ ). We deduce from this that x is new in E if and only if $x_i$ is new in K for all i and the $Z_i$ s are pairwise coprime. By irreducibility, the coprimality condition is equivalent to the fact that the $Z_i$ s are pairwise distinct, which is further equivalent to the fact that the $x_i$ s are pairwise nonconjugate. This establishes our claim.

We are now ready to count the number of new roots of P in $U^m$ . Based on the preceding, it is equivalent to count the number of tuples $(x_1, \ldots , x_m) \in U^m$ of new roots in K that are pairwise nonconjugate. By definition of $Z^{\mathrm {new}}_{U,n}$ , we have $Z^{\mathrm {new}}_{U,n}(P)$ possibilities for $x_1$ . The fact that $x_2$ cannot be conjugate to $x_1$ eliminates exactly a possibilities because we have assumed that U is stable under the action of $\mathrm {Aut}\:_{\!F\mathrm {-alg}}(K)$ . There then remain $Z^{\mathrm {new}}_{U,n}(P) - a$ possibilities for $x_2$ . Similarly, we have $Z^{\mathrm {new}}_{U,n}(P) - 2a$ possibilities for $x_3$ because it has to be nonconjugate to both $x_1$ and $x_2$ . Repeating this argument m times, we end up with the formula of the proposition.

Proof. It is entirely similar to that of Theorem 2.1.

Proof. The proof follows the same pattern than in the case of extensions. The first step is to extend Proposition 2.2 and show that

(5.3)

when x is new in E. As in the proof of Proposition 2.2, we write $I_s$ for the above integral. Let Z be the minimal monic polynomial of x; it does not need to be irreducible, but it has degree r since x is assumed to be new in E. Let $c \in F$ be the content of Z that is, by definition, the gcd of the coefficients of Z. Using that the content of a product is the product of the contents, we find that a product $QZ \in {\mathcal O}_F[x]$ if and only if $Q \in c^{-1} {\mathcal O}_F[X]$ . Moreover, we saw in the proof of Proposition 2.2 that there exists a positive constant $\gamma $ such that $\Vert R(x) \Vert \geq \gamma {\cdot } \Vert R \Vert $ for all polynomials R of degree at most $r{-}1$ . We deduce from these facts that for any fixed R with $\Vert R \Vert \leq \gamma $ , the property $QZ + R \in \Omega _n$ is equivalent to $Q \in c^{-1} \Omega _{n-r}$ . Remarking in addition that the mapping $(Q,R) \mapsto QZ + R$ preserves the measure (it is linear and has determinant one in the canonical bases), we obtain the equality

which holds true provided that s is sufficiently large. Repeating the argument presented in the proof of Proposition 2.2, we end up with

$$ \begin{align*} I_s & = \Vert D_E \Vert \cdot \lambda_E\big({\mathcal O}_F[x]\big) \cdot \int_{c^{-1} \Omega_{n-r}} \hspace{-1em} \Vert N_{E/F} (Q(x)) \Vert \cdot dQ \\ & = \Vert D_E \Vert \cdot \lambda_E\big({\mathcal O}_F[x]\big) \cdot \Vert c \Vert^{-n-1} \int_{\Omega_{n-r}} \hspace{-1em} \Vert N_{E/F} (Q(x)) \Vert \cdot dQ \end{align*} $$

for s large enough. Consider a decomposition $E = K_1 \times \cdots \times K_m$ , and write $x = (x_1, \ldots , x_m)$ accordingly. For each index i, set $r_i = [K_i:F]$ , and let $Z_i$ be the minimal monic polynomial of $x_i$ . Then Z is the lcm of the $Z_i$ s and comparing degrees, we get $Z = Z_1 \cdots Z_m$ . On the other hand, using relations between coefficients and roots, we find that the coefficient on $Z_i$ in $X^j$ has norm at most $\Vert x_i \Vert ^{r_i-j}$ , and equality is reached for $j \in \{0, r_i\}$ . Therefore, the content $c_i$ of $Z_i$ is $1$ when $x_i \in \mathcal O_{K_i}$ and is equal to $N_{K_i/F}(x_i)$ otherwise. Hence $\Vert c_i \Vert = \max \!\big (1,\, \Vert x_i \Vert ^{r_i}\big )$ in all cases. By the multiplicativity property of the contents, we conclude that $\Vert c \Vert = H(x)$ , which finally establishes equation (5.3). After this result, the proof of the first part of the theorem is totally similar to that of Theorem 2.6.