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Ultrametric Fields

Ultrametric Norms and Valuations

Remark 2.2
1. If $K$ is a field equipped with a ultrametric norm $\anorm{~}$ and if $\lambda<0$, then $v:K\ra\RR_+\cup\{+\infty\}$ defined by $v(x)=\lambda\log\anorm{x}$ is a valuation on $K$.
2. Reciprocally, if $v$ is a valuation on $K$ and $0<a<1,$ then $\anorm{x}=a^{v(x)}$ is a ultrametric norm on $K$.
3. The result of lemma 1.4 implies that we have $v(x+y)=\inf(v(x),v(y))$, if $v(x)\neq v(y)$

 We say that the valuation is discrete if $v(\tio{K})$ is a discrete subgroup of $\RR$ (and is thus of the form $a\Z$) and we say that it is normalized if $v(\tio{K})=\Z$.

 After the preceding remark, it is equivalent to reason in terms of ultrametric norms or in terms of valuations, and we define in an evident manner the completion $\hat{K}$ of the field $K$ equipped with valuation $v$. It is necessary however to pay attention to the fact that the inequalities get reversed. The formulas are generally much more agreeable in terms of valuations than norms.
Remark 2.3
1. If $K$ is equipped with a valuation $v$, from the result of lemma 1.1 (ii), $v(\widehat{\tio{K}})=v(\tio{K})$, and thus that $v(\widehat{\tio{K}})$ is an additive subgroup of $R$ that has no reason to be closed.
2. A sequence is Cauchy if and only if $v(u_{n+1}-u_n)$ tends to $+\infty$. Thus, if $K$ is complete, a sequence is convergent if and only if $v(u_{n+1}-u_n)$ tends to $+\infty$. Similarly, a series is convergent if and only if the valuations of its general term tends to $+\infty$.
Proposition 2.4 If $K$ is a field equipped with a valuation $v$, then $\Oo_K=\{x\in K\text{ such that }v(x)\geq 0\}$ is a local ring with maximal ideal $\id{m}_K=\{x\in K\text{ such that }v(x)>0\}$.
Proof The fact that $\Oo_K$ is a ring and $\id{m}_K$ is a maximal ideal is immediate consequence of the properties of valuation. On the other hand, if $x$ is an element of $\Oo_K\bs\id{m}_K$, then $v(x)=0$ and hence $x^{-1}$ is an element of $K$ with valuation zero and thus contained in $\Oo_K$, which proves that $\Oo_K\bs\id{m}_k$ is none other that the group of units $\tio{\Oo_K}$ of $\Oo_K$, and permits the conclusion.
Example 2.6 The ring of integers of $K((X))$ is $K[[X]]$ and its residue field is $K$.
Remark Let $K$ be a field complete with respect to valuation $v$. Let $K\{\{X\}\}$ be the set of Laurent series $f=\sum_{n\in\Z}a_nX^n$ such that the sequence ${(a_n)}_{n\in \Z}$ is bounded in $K$ and tends to zero when $n\ra-\infty$ (in other words, there exists $C\in\RR$ such that $v(a_n)\geq C$ for any $n\in\Z$ and $\lim_{n\ra-\infty}v(a_n)=+\infty$). If $f=\sum_{n\in\Z}a_nX^n\in K\{\{X\}\}$, we denote $v_G(f)=\inf_{n\in\Z}v(a_n).$
1. If $f,g\in K\{\{X\}\}$, then $v_G(fg)=v_G(f)+v_G(g)$ and $v_G$ is a valuation on $K\{\{X\}\}$ and $K\{\{X\}\}$ is complete for the valuation $v_G$.
2. If $v$ is a discrete valuation, then $K\{\{X\}\}$ is a field with discrete valuation and the residue field of $K\{\{X\}\}$ is $k_K((X))$.

The field $\Q_p$ and integers $\Z_p$

We denote $\Z_p$ as the ring of integers of $\Q_p$ (seen as the completion of $\Q$ for the valuation $v_p$). Its maximal ideal is $p\Z_p$ because $v_p(\tio{\Q_p})=v_p(\tio{\Q})=\Z$, and thus if $v_p(x)>0$, then $v_p(x)\geq 1$. It comes out from the general discussion that $\Q_p$ is complete ultrametric field and the series $\sum_{i\in I}a_i$ converge in $\Q_p$ if and only if $v_p(a_i)$ tends to $+\infty$.
Lemma 2.7 The natural map of $\Z/p^n\Z$ into $\Z_p/p^n\Z_p$ is an isomorphism.
Proof If $x\in\Z\cap p^n\Z_p$ we have $v_p(x)\geq n$, this signifies that $x$ is divisible by $p^n$ in $\Z$. We deduce injectivity. Proof of surjectivity: Let $\bar{x}\in\Z_p/p^n\Z_p$ and $x\in\Z_p$ giving image $\wbr{x}$ modulo $p^n$. Since $\Q$ is dense in $\Q_p$, there exists $r\in\Q$ verifying $v_p(x-r)\geq n;$ in particular $v_p(r)\geq 0$. Write $r=a/b$ with $a,b\in\Z$. Since $v_p(r)\geq 0$, we have $v_p(b)\leq v_p(a)$ and upto division by $p^{v_p(b)}$, we can suppose that $\gcd(b,p)=1$. Let $\bar{c}$ be the inverse of $b$ in $\Z/p^n\Z$ and $c\in\Z$ whose reduction modulo $p^n$ is $\bar{c}$. We have then $v_p(r-ac)=v_p(a)+v_p(1-bc)\geq n$ and thus $v_p(x-ac)\geq n$, which proves that $ac\in\Z$ has the image $\bar{x}\in\Z_p/p^n\Z_p$, permitting the conclusion.
Corollary 2.8 The residue field of $\Q_p$ is $\F_p=\Z/p\Z$.
We deduce from this lemma another, more algebraic description of the ring $Z_p$. We could have however, from this construction, obtained $\Q_p$ by rendering $p$ invertible. The two approaches have their advantages.
Proposition 2.9 The map that associates $x\in\Z_p$ a sequence ${(x_n)}_{n\in\N}$ the images modulo $p^n,n\in\N$ induces an isomorphism of ring topology of $\Z_p$ on the projective limit of $\Z/p^n\Z$.
The projective limit of $\Z/p\Z$ is the set of sequence ${(x_n)}_{n\in \N}$ where $x_n\in\Z/p^n\Z$ and $x_{n+1}$ has for its image $x_n$ by the natural map (reduction modulo $p^n$) of $\Z/p^{n+1}\Z$ in $\Z/p^n\Z$.
Proof It suffices to verify the bijectiveness of the map $\iota:{\Z}_p\ra\varprojlim\Z_p/p^n\Z_p$, which has $x$ associated to the sequence of the reductions modulo $p^n,n\in\N$. If $\iota(x)=0$, it is that $v_p(x)\geq n$ for all $n\in\N$, and thus $v_p(x)=+\infty$ and $x=0$, this gives injectivity. Now if ${(\bar{x}_n)}_{n\in\N}\in\varprojlim \Z_p/p^n\Z_p$, and if $x_n\in\Z_p$ is recovery of $\bar{x}_n$, then $v_p(x_{n+k}-x_n)\geq n$ for some $n,k\in\N$. The sequence ${(x_n)}_n\in\Z_p$ is cauchy in $\Q_p$ with limit $x$, which can be verified by passing to the limit the inequality $v_p(x-x_n)$ for any $n\in\N$. We have thus $\iota(x)={(\bar{x}_n)}_{n\in\N}$ which proves the surjectivity of $\iota$ and permits the conclusion.

Theorem 2.10
1. $\Z_p$ is compact and $\Q_p$ is locally compact.
2. All elements of $\Z_p$ can be written in a unique manner under the form $\sum_{n=0}^\infty p^n a_n$ where $a_n\in\F_p$.
3. $\N$ is dense in $\Z_p$, more generally, if $b\in\Z$ is prime with $p$ and $a\in\Z, a+b\N$ is dense in $\Z_p$.
Proof The first follows from the preceding proposition: $\Z_p$ is closed product of compacts (and even finite sets); it is thus compact and the balls of $\Q_p$ are isomorphic to $\Z_p$ and thus compact.
  It is not very difficult to that, if $x\in\Z_p$ and if $x_n$ is unique element of $\{0,1,\ldots, p^n-1\}$ giving the same image as that of $x$ in $\Z/p^n\Z$, then $\sum_{i=0}^na_ip^i$ is the expansion of $x_n$ in base $p$ (written in the opposite direction to the one we are used to ...); we thus deduce (ii).
  The (iii) is a consequence of (ii) ($x$ is the limit of the sequence of general term $\sum_{i=0}^np^ia_i$ all of its terms are in $\N$) and the fact that $x\ra bx+a$ is an isometry of $\Z_p$ if $(b,p)=1$.
Remark
1. If $x\in 1+p\Z_p$ and $n\in \N$ is prime to $p$, then the equation $y^n=x$ has a solution in $\Q_p$.
2. Inversely, if $x\in\Q_p\bs\{0\}$ is such that the equation $y^n=x$ has a solution in $\Q_p$ for all $n\in \N$ prime to $p$, then $x\in 1+p\Z_p$.
3. The group $\Aut(\Q_p)=\Ida$.

Hensel's Lemma

In this section we suppose that $K$ is complete for the valuation $v$ and $\Oo_K$ is the ring of integers.
If $s$ is an integer, we denote $P_s(K)$, the set of polynomials of degree $\leq s-1$ with coefficients in $K$ equipped with a canonical base $e_s=(1,X,\ldots, X^{s-1})$. If $g\in K[X]$ is of degree $\leq n$ and $h\in K[X]$ is of degree $\leq m$, we denote $\theta_{g,h}$ as the map of $P_m\oplus P_n$ in $P_{m+n}$ which takes $(u,v)$ to $ug+vh$ and denote by $R_{mn}(g,h)$ as the determinant of the matrix of $\theta_{g,h}$ expressed in the basis $(e_m,e_n)$ and $e_{m+n}$. (Resultant to polynomials of degree $m$ and $n$)
Lemma 2.11 $R_{m,n}(g,h)=0$ if and only if we are in one of two following cases (not exclusive)
1. $\deg g\leq n-1$ and $\deg h\leq m-1$
2. $g$ and $h$ are not relatively prime.
Proof If $\deg g\leq n$ and $\deg h\leq m-1$, then $\theta_{gh}(P_m\oplus P_n)\subset P_{m+n-1}$ and $\theta_{g,h}$ is not bijective, which implies that $R_{m,n}(g,h)=0$. If $g$ and $h$ are divisible by $w$ with $\deg w\geq 1$, we have $\theta_{g,h}\left(\dfrac{h}{w},-\dfrac{g}{w}\right)=0$ and $\theta_{g,h}$ is not injective, which implies that $R_{m,n}(g,h)=0$.
 Inversely if $g$ and $h$ are relatively prime, a solution to the equation $gu+hv=0$ must give $g|v$ and $h|u$, if $(u,v)\neq 0$, implying $\deg u\geq \deg h$ and $\deg v\geq \deg g$ and is impossible if $\deg g=n$ or $\deg h=m$ and $(u,v)\in P_m\oplus P_n$. This implies that $\theta_{g,h}$ is injective and thus bijective and hence $R_{m,n}(g,h\neq 0$.$\qquad\blacksquare$

  If $f=\sum_{i=0}^na_iX^i\in K[X]$, we define $v_G(f)$ by the formula $v_G(f)=\inf_{0\leq i\leq n}v(a_i)$.
Theorem 2.12 Let $C>0$ and let $f,g,h\in\Oo_K[X]$ verify
1. $\deg g\leq n, \deg h\leq m$ and $\deg (f-gh)\leq n+m-1$
2. $v_G(f-gh)\geq C+2v(R_{m,n}(g,h))$,
then there exist unique polynomials $\tilde{g},\tilde{h}\in\Oo_K[X]$ such that we have
1. $\deg(\tilde{g}-g)\leq n-1$ and $\deg(\tilde{h}-h)\leq m-1$,
2. $v_G(\tilde{g}-g)\geq C+v(R_{m,n}(g,h))$ and $v_G(\tilde{h}-h)\geq C+v(R_{m,n}(g,h))$
3. $f=\tilde{g}\tilde{h}$
Proof We find $u\in P_n$ and $v\in P_m$ such that we have $$ \begin{align*} f=(g+v)(h+u) &\quad\text{if and only if}\\ f-gh-uv=gu+hv&\quad\text{if and only if}\\ (u,v)=\theta_{g,h}^{-1}(f-gh-uv)& \end{align*} $$ Let $\varphi(u,v)=\theta^{-1}_{g,h}(g-gh-uv);$ we find a fixed point of $\varphi$. Let $$B=\{(u,v)\in P_n\oplus P_m,\inf(v_G(u),v_G(v))\geq C+v(R_{m,n}(g,h))\} $$ The theorem is an immediate consequence of the following lemma and the completeness of $K$.
Lemma 2.13 The map $\varphi$, from $B$ to itself, is a contraction (more precisely a $K$ Lipschitz map).
Proof If $(u,v)\in B,$ then
$$ \begin{align*} v_G (f - gh -uv) &\geq \inf(v_G (f -gh), v_G (uv))\\ &\geq \inf (C + 2v(R_{m,n} (g, h)), 2C + 2v(R_{m,n} (g, h)))\\ &=C+2v(R_{m,n}(g,h)). \end{align*} $$ On the other hand, $g$ and $h$ have coefficients in $\Oo_K$ and thus $\theta_{g,h}$ has all its coefficients in $\Oo_K$ and the matrix $\theta_{g,h}^{-1}$ has thus coefficients in $\dfrac{1}{R_{m,n}(g,h)}\Oo_K$ and thus $v_G(\theta_{g,h}^{-1})\geq v_G(x)-v(R_{m,n}(g,h)).$ We can deduce the fact that $\varphi$ sends $B$ in $B$. Finally, if $(u,v)$ and $(u',v')$ are two elements of $B$, $$ \begin{align*} v_G(\varphi(u,v)-\varphi(u',v'))&=v_G(\theta_{g,h}^{-1}(uv-u'v'))\\ &\geq v_G(u(v-v')+v'(u-u'))-v(R_{m,n}(g,h))\\ &\geq \inf(v_G(u)+v_G(v-v'),v_G(v')+v_G(u-u')-v(R_{m,n}))\\ &\geq C+\inf(v_G(u-u'),v_G(v-v')) \end{align*} $$ which proves that $\varphi$ is contraction on $B$.
Corollary 2.14 Let $f,g,h\in\Oo_K[X]$ verify the following conditions: $g$ is monic of degee $n$, $\deg h\leq m,\deg(f-gh)\leq n+m-1$, the reductions $\bar{g}=g$ modulo $\id{m}_K$ and $\bar{h}=h$ modulo $\id{m}_K$ are relatively prime, and $v_G(f-gh)>0$, then there exists $\tilde{g},\tilde{h}\in\Oo_K[X]$ uniquely verifying $\deg(\tilde{g}-g)\leq n-1,\deg(\tilde{h}-h)\leq m-1,v_G(\tilde{g}-g)>0,v_G(\tilde{h}-h)>0$ and $\tilde{g}\tilde{h}=f$.
Proof As $\bar{g}$ is of degree $n$ and $\bar{g}$ and $\bar{h}$ are relatively prime $R_{m,n}(\bar{g},\bar{h})\neq 0$, which implies that $v(R_{m,n}(g,h))=0$. We can apply the preceding theorem for $C\in\R_{>0}$ verfying $v_g(f-gh)\geq C>0$, and it suffices to take $C\ra 0$ to deduce the result.$\qquad\blacksquare$
Ultrametric analogue of Newton's algorithm for finding roots.
Theorem 2.15 If $f\in\Oo_K$ and $\alpha\in\Oo_K$ verifying $v(f(\alpha))>2v(f'(\alpha))$, then there exists $\tilde{\alpha}\in\Oo_K$ uniquely verifying the conditions $v(\tilde{\alpha}-\alpha)>v(f'(\alpha))$ and $f(\tilde{\alpha})=0$.
Proof Let $d$ be the degree of $f$ and $C\in(0,v(f(\alpha))-2v(f'(\alpha))])$; and the interval is non empty by assumption. Let $g$ and $h$ be the elements given by the formulae $$ \begin{align*} g(X)&=X-\alpha\\ h(X)&=\dfrac{f(X)-F(\alpha)}{X-\alpha}\\ &=f^{[1]}(\alpha)+f^{[2]}(\alpha)(X-\alpha)+\ldots+f^{[d]}(\alpha)(X-\alpha)^{d-1} \end{align*} $$ where $f^{[i]}$ designates the divided $i$ the derivative of $f$ (in $\Char 0$, we have $f^{[i]}=f^{(i)}/i!$).
 The matrix $\theta_{g,h}:{K[X]}_{d-2}\oplus {K[X]}_0\ra K{[X]}_{d-1}$ has as coefficients (in the basis constituting the powers of $X-\alpha$ instead of the powers of $X$) $1$ below the diagonal, $(f^{[1]}(\alpha),\ldots,f^{[d]}(\alpha))$ on the last column and $0$ everywhere else. Its determinant is thus $\pm f^{[1]}(\alpha)=f'(\alpha)$. The resultant $R_{d-1,1}(g,h)$ is thus the valuation $v(f'(\alpha))$. On the other hand we have $v_G(f-gh)=v(f(\alpha))\geq C+2v(R_{d-1,1}(g,h))$ by definition of $C$, and $\deg f-gh\leq d-1$. We have the conditions to apply theorem 2.12, and there exists $\tilde{g}$ uniquely dividing $f$ suhc that $\deg(\tilde{g}-g)=0$ and $v_G(\tilde{g}-g)\geq C+v(f'(\alpha))$. Thus, there exists unique $\tilde{\alpha}\in\Oo_K$ such that we have $\tilde{g}(X)=X-\tilde{\alpha}$ [which implies $f(\tilde{\alpha})=0$], and $v(\tilde{\alpha}-\alpha)\geq C+v(f'(\alpha))$. It suffices then to take $C\ra 0$ for deducing the result.
Corollary 2.16 Let $f\in\Oo_K[X]$ a monic polynomial and $\bar{f}$ the reduction of $f$ modulo $\id{m}_K$. If $\bar{\alpha}$ is a simple root of $\bar{f}$ in $k$, then there exists unique $\tilde{\alpha}\in\Oo_K$, whose reduction modulo $\id{m}_K$ is $\bar{\alpha}$ and it verifies $f(\tilde{\alpha})=0$.
Proof Let $\alpha\in\Oo_K$ whose reduction modulo $\id{m}_K$ is $\bar{\alpha}$. The hypothesis according to which $\bar{\alpha}$ is a simple root of $\bar{f}$ translated by the fact that $v(f(\alpha))>0$ and $v(f'(\alpha))=0$, which permits the utilization of the proposition 2.15 for the conclusion.
Proposition 2.17 Let $f\in K[X]$ be a monic polynomial verifying $f(0)\in\Oo_K$. Then all the coefficients of $f$ belong to $\Oo_K$.
Proof Suppose to the contrary and let $i$ be the biggest integer such that we have $v(a_i)=v_G(f)$. The polynomial $a_i^{-1}f$ is then of the form $$ b_nX^n+\ldots+b_{i+1}X^{i+1}+X^i+b_{i-1}X^{i-1}+\ldots+b_0 $$ where $b_k$ are the elements of $\Oo_K$ verifying $v(b_k)>0$ such that $k\in\{i+1,\ldots, n\}$. We can thus apply the preceding corollary to $g(X=X^i+\ldots+b_0)$ and $h=1+b_nX^n$ for showing that there exists polynomials $\tilde{g}$ and $\tilde{h}$ of degrees respectively $i$ and $J$ such that we have $f=a_i\tilde{g}\tilde{h}$, which is a contradiction with the hypothesis that $f$ is irreducible, and permits the conclusion.



A substantial part is a translation of LES NOMBRES p-ADIQUES, NOTES DU COURS DE M2 by Pierre Colmez.