Algebraic Closure of complete valued fields

Extension of Valuations

Theorem 3.1 Let $K$ be a complete field for a valuation $v$, and let $L$ a finite extension of $K$. Then there exists a unique extension of valuation $v$ to $L$. Moreover, if $x\in L$, then $$ v(x)=\dfrac{1}{[L:K]}v(N_{L/K}(x))$$ Proof We can see $L$ as a vector space of finite dimension $[L:K]$. If $v_1,v_2$ are two valuations of $L$ extending $v$, then $v_1$ and $v_2$ define the same topology on $L$, (use proposition 1.17). The proposition 1.7 then shows that there exists $s\in \tio{R}_+$ such that we have $v_2(x)=s\cdot v_1(x)$ for any $x\in L$, and since $v_1(x)=v_2(x),$ if $x\in K$, we deduce the uniqueness of extension of $v$ to $L$.
  For conclusion, it remains to show that the above formula is well defined for a valuation on $L$, in other words, if $x\in L$ verfies $v(N_{L/K}(x))\geq 0$, then $v(N_{L/K}(1+x))\geq 0$.
  Let the minimal polynomial of $x$ on $K$ be given as $f(X)=X^d+\ldots+a_1X+a_0$. This implies that $d$ divides $[L:K]$ and $N_{L/K}(x)=((-1)^da_0)^{[L:K]/d}$ and thus $v(N_{L/K}(x))\geq 1$ implies $a_0\in\Oo_K$ and irreducibility of $f$ implies that $f$ has coefficients in $\Oo_K$ (use proposition 2.17). On the other hand, the minimal polynomial of $1+x$ is $f(X-1)$ and thus $N_{L/k}(1+x)=((-1)^df(-1))^{[L:K]/d}\in\Oo_K$, which permits the conclusion.

Corollary 3.2 If $\bar{K}$ is an algebraic closure of $K$, there exists a unique extension of $v$ to $\bar{K}$, moreover, $\Aut(\bar{K}/K)$ acts on $K$ by isometries.
Proof The uniqueness of extension is a consequence of the preceding theorem. The rest of the statement follows from the fact that, if $x\in\bar{K}$, and if $\sigma\in\Aut(\bar{K}/K),$ then $N_{K(x)/K}(x)=N_{K(\sigma(x))/K}(\sigma(x))$.

Corollary 3.3 If $P\in K[X]$ is irreducible, then all the roots in $\bar{K}$ have the same valuation.
Proof The roots of the irreducible polynomial are permuted transitively by $\Aut(\bar{K}/K)$, and the preceding corollary permits the conclusion.

Newton Polygons

Lemma 3.4 Let $P(X)=a_nX^n+\ldots+a_0\in K[X]$, be a polynomial of degree $n$ and let $\alpha_1,\ldots,\alpha_n$, the roots (with their multiplicities) of $P$ in $\wbr{K}$ arranged by decreasing valuation. $$v(\alpha_1)\geq v(\alpha_2)\geq \ldots\geq v(\alpha_n) .$$ Then, if $i\in\{0,\ldots,n\}$, we have $$v(a_i)\geq v(a_n)+\sum_{k=0}^{n-i-1} v(\alpha_{n-k})$$ with equality if $v(\alpha_i)>v(\alpha_{i+1})$.
Proof The inequality is a consequence of the link between coefficients of $P$ and the symmetric functions of roots of $P$. The case of equality comes from $v(\alpha_i)>v(\alpha_{i+1}),$ then in the symmetric function of order $n-i$, all the other terms have a valuation strictly greater than that of $\prod_{k=0}^{n-i-1}\alpha_{n-k}$.

Let $u:[0,n)\ra\RR\cup\{+\infty\}$ be a function defined by $u(x)=-v(\alpha_i)$ if $x\in[i-1,i)$ and $\Newt_P(x)=v(a_n)+\int_n^xu(t)dt$. The function $u(x)$ is an increasing step function, and the function $\Newt_P(x)$ is thus convex and affine by pieces on [0,n]. The preceding lemma can also be translated as $v(a_i)\geq \Newt_P(i)$ with equality if $v(\alpha_i)>v(\alpha_{i+1})$, that is to say if the derivatives to the right and left of the $\Newt_P$ function at point $i$ are different (in other words, if $(i,\Newt_P(i))$ is a vertex (of graph) of $\Newt_P$). The set $\{(x,y)\text{ such that }x\in [0,1], y\geq \Newt_P(x)\}$ is called the Newton Polygon of $P$, it is also the convex envelope of the set $(i,v(a_i))$ and $[0,n]\times \{+\infty\}$. We call slope of $\Newt_P$ or slope of Newton Polygon of $P$ an element of ${\Newt'}_P([0,n])$ and if $\lambda$ is a slope of $\Newt_P$, we call segment of slope $\lambda$ of $\Newt_P$ as the set $\{(x,\Newt_P(x))\text{ such that }{\Newt'}_P(x)=\lambda\}$. The length of the segment is by definition the length of the projection on the $x$ axis.

  The following theorem is translation of the lemma 3.4, utilizing the language of Newton Polygons.

Theorem 3.5 There exists a root of $P$ of valuation $\lambda$ if and only if $-\lambda$ is a slope of $\Newt_P$. Moreover, the number of roots of $P$ (computed with multiplicity) of valuation $\lambda$ is the length of the segment of slope $-\lambda$ of $\Newt_P$.

Proposition 3.7 Suppose that $v$ is discrete and normalized. Let $P\in K[X]$
(i). The slope of Newton Polygon of $P$ is a rational number. Moreover, if $a/b$ with $(a,b)=1$, is slope of $\Newt_P$, then the length of the segment of slope $\lambda$ is a multiple of $b$.
(ii). If $\Newt_P$ only has slope $a/b$ with $(a,b)=1$, and if $\deg P=b$, then $P$ is irreducible.
Proof If the valuation of $K$ is discrete and normalized, the vertices of the Newton polygon of $P$ have integer coordinates. The result can be deduced from the Newton polygon of $P/Q$, where $Q\in K[X]$ divides $P$.

Corollary 3.8 (Criteria of Eisentein) If $P(X)=X^n+a_{n-1}X^{n-1}+\ldots+a_0$ verifies $v(a_i)\geq 1$ for any $0\leq i\leq n-1$ and $v(a_0)=1$, then $P$ is irreducible.
Proof Its Newton polygon is a segment of length $n$ and slope $-1/n$.

Condensed Galois Theory

Let $K$ be a field and $\bar{K}$ its algebraic closure. If $L$ is a subfield of $\bar{K}$ containing $K$, we denote $\sepa{L}$, the separable closure of $L$ in $\bar{K}$, that is to say the set of elements of $\bar{K}$ whose minimal polynomial is separable (i.e only have simple roots). If $K$ has $\Char 0$ or, more generally, if $K$ is perfect (If $K$ is $\Char p$, this signifies that $x\mapsto x^p$ is a surjection of $K$ in $K$), then $\sepa{L}=\bar{K};$ in the general case $\sepa{L}$ is a subfield of $\bar{K}$.
  If $L$ is a subfield of $\bar{K}$ containing $K$, we denote $\rad{L}$ as the radical closure of $L$ in $\bar{K}$, that is to say the set of elements $x$ of $\bar{K}$ such that there exists $n\in\N$ (depending on $x$) such that $x^{p^n}\in L$. It is a perfect subfield of $\bar{K}$, and we have $\sepa{L}\cap\rad{L}=L$ and $\sepa{L}\cdot\rad{L}=\bar{K}$.
  Let $G_K=\Aut(\bar{K}/K)$. If $L$ is an algebraic extension of $K$ contained in $\bar{K}$, we denote by $G_L$ the subgroup of elements of $G_K$ leaving $L$ fixed. We also have $G_L=\gal(\sepa{L}/L)$. If $L$ is a finite extension of $K$, then $G_L$ is of finite index in $G_K$ and $\anorm{G_K/G_L}\leq [L:K]$, with equality if and only if the extension $L/K$ is separable.
The group $G_K$ is equipped with profinite topology, a base of neutral elements is constituted by $G_L$, where $L$ traverses the set of finite extensions of $K$ in $\bar{K}$. In particular, the group $G_K$ is compact. If $H$ is a closed subgroup of $G_K$, we denote $\bar{K}^H$ as the subgroup of $\bar{K}$ fixed by $H$.
  The Galois Theory can then be condensed to the following:
Theorem 3.9
(i) If $H$ is a closed subgroup of $G_K$, then $(\sepa{K})^H$ (resp. $\bar{K}^H$) is a separable extension of $K$ (resp of $\rad{K}$), galois if $H$ is a normal subgroup in $G_K$, and we have $$G_{(\sepa{K})^H} =G_{\bar{K}^H}=H.$$
(ii) If $L$ is an algebraic extension of $K$, then $G_L$ is a closed subgroup of $G_K$, and we have $$(L\cdot\sepa{K})^{G_L}=L\text{ and }\bar{K}^{G_L}=\rad{L}.$$

Completion of algebraically closed field

After corollary 3.2, there exists a unique way of extending the valuation to the algebraic closure of complete valued field. This algebraic closure has no reason to be complete(and in general it is not), thus we can complete it, take algebraic closure, complete again... The following theorem shows that in fact the process converges very quickly.

Theorem 3.10 If $K$ is an algebraically closed field equipped with a valuation, its completion $\hat{K}$ is algebraically closed.
Proof Let $P(X)=X^n+a_{n-1}X^{n-1}+\ldots+a_0$ be an irreducible monic polynomial of $\hat{K}[X]$. Our goal is to prove that $P$ has a root in $\hat{K}$. Leaving $P$ and $\alpha^nP(X/\alpha)$, which multiplies $a_i$ with $\alpha^{n-i}$, we can suppose that $P$ has integer coefficients. Begin by supposing $P$ is separable, that is to say that $P$ and $P'$ are relatively prime. Thus, there exists polynomials $U$ and $V$ such that we have $UP+VP'=1$.
  As usual, $v_G$ is the Gauss valuation on $\hat{K}[X]$. Let $C>\sup(0,-v_G(U), -2v_G(V)),$ and if $0\leq i\leq n-1,$ let $b_i\in K$ such that we have $v(b_i-a_i)\geq C$. Let $x_0\in K$ a root of polynomial $Q(X)=X^n+b_{n-1}X^{n-1}+\cdots+b_0$. We have $v(x_0)\geq 0$ because $Q$ has integer coefficients. This implies that $$v(U(x_0)P(x_0))\geq C+v_G(U)>0\Rightarrow v(P'(x_0)V(x_0))=0,$$ from where we get $$v(P'(x_0))\leq v_G(V)\text{ and }v(P(x_0))\geq C>2v(P'(x_0)).$$ The Hensel's Lemma permits the conclusion of the fact that the equation $P(x)=0$ is a solution in $\hat{K}$.
 If $P$ is irreducible but not separable, we are in $\Char p\neq 0$, and there exists $Q$ irreducible and separable and $m\in \N$ such that we have $P(X)=Q(X^{p^m})$. If $x$ is a root of $Q$ and $x_n$ a sequence of elements of $K$ tending to $x$ in $\hat{K}$, then $x_n^{1/p^m}$ is a sequence of elements of $K$ tending to a root of $P$. It is Cauchy because $v(x^{1/p^m}-y^{1/p^m})=(1/p^m)v(x-y)$.

Residue Field of Algebraically closed Fields

Lemma 3.11 Let $K$ be a complete ultrametric field and $L$ a finite extension of $K$, then $k_L$ is an algebraic extension of $k_K$ of degree $\leq [L:K]$.
Proof. We have $\Oo_K\cap\id{m}_L=\id{m}_K$ and thus $k_K$ injects into $k_L$. Let $\bar{\alpha}_1,\ldots,\bar{\alpha}_d$ be elements of $k_L$ forming a free family on $k_K$. Choose for each $i\in\{1,\ldots,d\}$ an element $\alpha_i$ of $\oo_L$ whose image in $k_L$ is $\bar{\alpha}_i$. Suppose that the $\alpha_i$ form a related family on $K$ and let $(\lambda_1,\ldots,\lambda_d)$ be a family of non zero elements of $K$ such that we have $\lambda_1\alpha_1+\ldots+\lambda_d\alpha_d=0$. Divide all the all the $\lambda_i$ by the one with the highest norm, we can suppose they are all elements of $\oo_K$ and that one of them equals $1$, which leads to a contradiction when we reduce modulo $p$.

Lemma 3.12 If $K$ is a algebraically closed ultrametric field, then $k_K$ is algebraically closed.
Proof Let $\bar{P}(X)\in k_K[X]$ a monic polynomial of degree $n\geq 1$ and let $P(X)\in\oo_K[X]$ be the monic polynomial of degree $n$ , a lift of $\bar{P}$. Let $\alpha\in K$ be the root of $P$. We have $\alpha\in\oo_K$ (the slope of the Newton polygon of $P$ are negative, thus the roots have valuations $\geq 0$), and the image of $\alpha$ in $k_K$ is a root of $\bar{P}$, this permits the conclusion.

Lemma 3.13 If $K$ is a ultrametric field and $\hat{K}$ denotes its completion, then $k_K=k_\hat{K}$.
Proof The natural map of $k_K$ in $k_\hat{K}$ is injective because $$\oo_K\cap\id{m}_\hat{K}=\{x\in\oo_K\text{ such that }v(x)>0\}=\id{m}_K.$$ On the other hand, as $\oo_K$ is dense in $\oo_\hat{K}$, this map is surjective, which permits the conclusion.

Corollary 3.14 If $K$ a complete valued field, then the residue field of $\hat{\bar{K}}$ is algebraic closure of $k_K$.

The theorem of Ax-Sen-Tate

Let $K$ be a complete field for a valuation. As we know from corollary 3.2 the valuation on $K$ extends in a unique manner to a valuation in $\bar{K}$ and $\sigma\in G_K=\Aut(\bar{K}/K)$ acts by an isometry; we can thus extend the action of $G_K$ by continuity to an action on $\hat{\bar{K}}$ which is an algebraically closed field after theorem 3.10.

Theorem 3.15 Let $H$ be a closed subgroup of $G_K$; then $(\hat{\bar{K}})^H$ is completion of $(\bar{K})^H$. In other words, $(\bar{K})^H$ is dense in $(\bar{K})^H$.
Let $L=(\bar{K})^H$; it is a perfect subfield of $\bar{K}$. If $\alpha\in\bar{K}$, we define diameter $\Delta_L(\alpha)$ of $\alpha$ with respect to $L$ by $\Delta_L(\alpha)=\inf_{\sigma\in H}v(\sigma(\alpha)-\alpha)$. Note that $\alpha\in L$ if and only if $\Delta_L(\alpha)=+\infty$.
The proof of theorem 3.15 rests on the following proposition.
Proposition 3.16(Ax) There exists a constant $C$ such that if $\alpha\in\bar{K}$, then there exists $a\in L$ verifying $v(\alpha-a)\geq \Delta_L(\alpha)-C$.
Proof If $\alpha$ is almost fixed by $H$, then $\alpha\in\bar{K}$ is close to an element of $L$. Let $x\in\hat{\bar{K}}$ be fixed by $H$. If $\alpha_n$ is a sequence of elements of $\bar{K}$ tending to $x$, then $\Delta_L(\alpha_n)$ tends to $+\infty$ because $$v(\sigma(\alpha_n)-\alpha_n)=v(\sigma(\alpha_n)-\sigma(x)+x-\alpha_n)\geq \inf(v(\sigma(x-\alpha_n)),v(x-\alpha_n))=v(x-\alpha_n), $$ and thus if $a_n$ is an element of $L$ verifying $v(a_n-\alpha_n)\geq \Delta_L(\alpha_n)-C,$ the sequence $a_n$ is a sequence of element of $L$ tending to $x$ and $x\in\hat{L}, $ which was needed for proof to deduce theorem 3.15 from proposition 3.16.
  Proceeding to the proof of propositoin 3.16. The proof is a bit different following if it is equal or inequal Characteristic.
Case of equal Characteristic: Denote $M$ by the field $L(\alpha)$; it is a finite extension of $L$ which is separable because $L$ is perfect.
  The case where $K$ is of $\Char 0$ and its residual field is obvious: we can take $C=0$ and denote $a=\tr_{M/L}(y\alpha),$ with $y=1/[M:L]$; we have then $v(y)=0,$ and $$ a-\alpha=\sum_{\sigma\in\Hom_L(M,\bar{K})} (\sigma(y\alpha)-y\alpha)=y\cdot\sum_{\sigma\in\Hom_L(M,\bar{K})} (\sigma(\alpha)-\alpha) $$ a valuation $\geq \inf_{\sigma\in\Hom_L(M,\bar{K})}v(\sigma(\alpha)-\alpha)=\Delta_L(\alpha)$.
If $\Char K=p$, the trace map $\tr_{M/L}$ is surjective (because $M/L$ is separable), and there exists $x\in M$ such that we have $\tr_{M/L}(x)=1$. But, then $\tr_{M/L}(x^{1/p^n})=1$ for any $n\in \N$; we cam decide the fact tjat fpr any $\delta>0$, there exists $y\in M$ verifying $\tr_{M/L}(y)=1$ and $v(y)>-\delta$. Now, we have $a=\tr_{M/L}(y\alpha)\in L$ and $$ a-\alpha=\sum_{\sigma\in\Hom_L(M,\bar{K})} (\sigma(y\alpha)-\sigma(y)\alpha)= \sum_{\sigma\in\Hom_L(M,\bar{K})}\sigma(y)(\sigma(\alpha)-\alpha) $$ a valuation $\geq \inf_{\sigma\in\Hom_L(M,\bar{K})}v(\sigma(y)) +v(\sigma(\alpha)-\alpha)=-\delta+\Delta_L(\alpha)$. We can thus take for $C$ any strictly positive number.
Case of inequal Characteristic: Suppose now that $\Char K=0$ and its residue field is of $\Char p$. Upto renormalizing $v$, we can suppose that $v=v_p$ (that is $v(p)=1$). We will need the following Lemma
Lemma 3.17 Let $P\in\bar{K}[X]$ monic of degree $n$ all of its roots satisfy $v_p(\alpha)\geq u$.
(i) If $n=p^kd$ with $(d,k)=1$ and $d>0$ and if $q=p^k$, the the polynomial $P^{(q)}$, the $q$th derivative of $P$, has atleast one root $\beta$ satisfying $v_p(\beta)\geq u$.
(ii) If $n=p^{k+1}$ and $q=p^k$, then $P^{(q)}$ has atleast one root $\beta$ satisfying $v_p(\beta)\geq u-\dfrac{1}{p^{k+1}-p^k}$.
Proof We have $P(X)=X^n+a_{n-1}X^{n-1}+\ldots+a_0$ with $v_p(a_{n-i})\geq iu$ after the theory of newton polygons (which amounts to the same, by a brutal computation ). We have $\dfrac{1}{q!}P^{(q)}(X)=\sum_{i=0}^{n-q}\binom{n-i}{q}a_{n-i}X^{n-i-q}$, and the product of roots of $P^{(q)}$, is upto a sign $\dfrac{a_q}{\binom{n}{q}}$. We have thus $$\sum_\beta v_p(\beta)=v_p(a_q)-v_p\left(\binom{n}{q}\right)\geq (n-q)u-v_p\left(\binom{n}{q}\right),$$ and there exists $\beta$ verifying $v_p(\beta)\geq u-\dfrac{1}{n-q}v_p\left(\binom{n}{q}\right)$.
  On the other hand, we have $\binom{n}{q}=\dfrac{n}{q}\prod_{i=1}^{q-1}\dfrac{n-i}{i}$ and as $q=p^k$ and $v_p(n)\geq k$, we have $v_p\left(\dfrac{n-i}{i}\right)=0$ and $v_p\left(\binom{n}{q}\right)=v_p(n)-v_p(q)$, and we deduce the result.
The following proposition shows that we can take $C=p/(p-1)^2$.
Proposition 3.18 If $[L(\alpha):L]=n$ and $\ell(n)$ is the biggest integer $\ell$ such that $p^\ell\leq n$, there exists $a\in L$ verifying $v_p(a-\alpha)\geq \Delta_L(\alpha)-\sum_{i=1}^{\ell(n)}\dfrac{1}{p^i-p^{i-1}}$.
Proof By induction on $n$, the case $n=1$ being evident. We can apply the preceding lemma with $P=Q(X+\alpha)$, where $Q$ is the minimal polynomial of $\alpha$ on $L$. Notice that roots of $P$ are the $\sigma(\alpha)-\alpha,$ for $\sigma\in H$, and thus the $u$ of preceding lemma can exist and taken equal to $\Delta_L(\alpha)$. There are two cases

1. If $n$ is not a power of $p$, there exists $q\in \N$ such that the polynomial $P^{(q)}$ has a roots $\beta$ verifying $v_p(\beta-\alpha)\geq\Delta_L(\alpha)$. On the other hand, if $\sigma\in H$, then $$ \begin{align*} v_p(\sigma(\beta)-\beta)&=v_p(\sigma(\beta)-\sigma(\alpha)+\sigma(\alpha)-\alpha+\alpha-\beta) \\ &\geq \min(v_p(\sigma(\beta-\alpha)),v_p(\sigma(\alpha)-\alpha), v_p(\beta-\alpha)) \end{align*} $$ and since, $$ \begin{align*} v_p(\sigma(\beta-\alpha))&=v_p(\beta-\alpha)\geq\Delta_L\text{ and}\\ v_p(\sigma(\alpha)-\alpha)&\geq\Delta_L\text{ by definition} \end{align*} $$ we get the inequality $\Delta_L(\beta)\geq\Delta_L(\alpha)$, and since $[L(\beta):L]<n$, this permits to conclude by induction hypothesis.
2. If $n=p^{k+1}$, we can find a root $\beta$ of $P^{(p^k)}$ verifying $v_p(\beta-\alpha)\geq\Delta_L(\alpha)-\dfrac{v_p(p)}{p^{k+1}-p^k}$ and we obtain by the same reasoning the existence of $\beta\in\bar{K}$ verifying $\Delta_L(\beta)\geq \Delta_L(\alpha)-\dfrac{1}{p^{k+1}-p^k}$ and $[L(\beta):L]<n=p^{k+1}$. We draw from the induction hypothesis the existence of $a\in L$ verifying $$ v_p(\beta-a)\geq\Delta_L(\beta)-\sum_{i=1}^k\dfrac{1}{p^i-p^{i-1}} \geq\Delta_L(\alpha)-\sum_{i=1}^{k+1}\dfrac{1}{p^i-p^{i-1}}$$ and since $v_p(\alpha-\beta)\geq \Delta_L(\alpha)-\sum_{i=1}^{k+1}\dfrac{1}{p^i-p^{i-1}} $, this permits the conclusion.



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