1 \(p\)-adic Fields
1.1 \(p\)-adic Equations
Let \(R\) be a ring. An element \((x_1, \cdots , x_n) \in R^n\) is primitive if at least one of the \(x_i\) is invertible.
Let \(f^{(i)} \in \mathbb {Z}_p[X_1, \cdots , X_m]\) be homogeneous polynomials with \(p\)-adic integer coefficients. For each \(n \ge 1\), denote by \(f_n^{(i)}\) the reduction of \(f^{(i)}\) modulo \(p^n\). The following are equivalent:
The \(f^{(i)}\) have a nontrivial common zero in \(\mathbb {Q}_p^m\).
The \(f^{(i)}\) have a common primitive zero in \(\mathbb {Z}_p^m\).
For all \(n \ge 1\), the \(f_n^{(i)}\) have a common primitive zero in \((\mathbb {Z}/ p^n \mathbb {Z}_p)^m\).
The implication (b) \( \implies \) (a) is trivial. Conversely, if \(x = (x_1, \cdots , x_m) \in \mathbb {Q}_p^m\) is a nontrivial common zero of the \(f^{(i)}\), put
Then \(y \in \mathbb {Z}_p^m\) is a common primitive zero of the \(f^{(i)}\). For the equivalence (b) \(\iff \) (c), Serre applies the first Lemma from [ . However, using Mathlib’s DirectLimit definition here might be tricky. It could be easier to just prove this specific result; note that the file Mathlib.NumberTheory.Padics.RingHoms contains useful results relating \(\mathbb {Z}_p\) and \(\mathbb {Z}/p^n\mathbb {Z}\) (without relying on DirectLimit).
1.2 Squares in \(\mathbb {Q}_p^{\times }\)
\({\mathbb {Q}_p^\times }^2\) is an open subgroup of \(\mathbb {Q}_p^\times \).
The only nontrivial part is to show that the squares form an open set. Since \(\mathbb {Q}_p\) is a field, the unit topology agrees with the subspace topology, so it suffices to check that the set of nonzero squares in \(\mathbb {Q}_p\) is open. Define \(r\) as \(1\) if \(p\) is odd, or as \(2^{-2}\) if \(p = 2\), and denote by \(B_{1, r}\) the open ball with center \(1\) and radius \(r\). For each \(c \in B_{1, r}\), by applying Hensel’s lemma with the polynomial \(X^2 - c\), we obtain that \(c\) is a square. Then, for each nonzero square \(s \in \mathbb {Q}_p\), \(s \cdot B_{1, r}\) is an open neighbourhood of \(s\) contained in the set of nonzero squares, which shows that the latter is open. See [ for more details.
Let \(p\) be a prime, let \(f \in \mathbb {Z}_p[X_1, \cdots , X_m]\) be a polynomial with \(p\)-adic integer coefficients, and let \(x \in (\mathbb {Z}/p\mathbb {Z})^m\) be a zero of \(f\) modulo \(p^n\), for some positive integer \(n\). Suppose that, for some \(j \in \{ 1, \cdots , m\} \), the partial derivative \(\frac{\partial f}{\partial X_j}(x)\) is nonzero modulo \(p^k\), where \(0 {\lt} 2k {\lt} n\). Then there exists a zero \(z\) of \(f\) in \(\mathbb {Z}_p^m\) whose reduction modulo \(p^n\) is \(x\), and such that \(z-x\) is divisible by \(p^{n-k}\).
This is in Serre’s [ .
1.3 Units and Legendre symbols
Let \(p\) be a prime, let \(x \in \mathbb {Q}_p^\times \), and let \(v_p\) denote the \(p\)-adic valuation. Then \(\| x \cdot p^{-v_p(x)}\| = 1\).
Straightforward.
Let \(p\) be a prime and \(x \in \mathbb {Q}_p^\times \). The unit part of \(x\) is defined as the element of \(\mathbb {Z}_p^\times \) corresponding to \(x \cdot p^{-v_p(x)}\).
Let \(p\) be a prime and \(u \in \mathbb {Z}_p^\times \). The Legendre symbol \(\left(\dfrac {u}{p}\right)\) is defined as
where \(\overline{u}\) is the image of \(u\) under the natural map \(\mathbb {Z}_p^\times \to (\mathbb {Z}/p\mathbb {Z})^\times \).