# 数学代写|代数数论代写Algebraic number theory代考|MTH2106

## 数学代写|代数数论代写Algebraic number theory代考|Extensions of valuations

Let $(K,|\cdot|)$ be a complete non-archimedean valuation field, $\mathcal{O}_K, \mathfrak{m}_K$ and $k$ be as in the previous section.

Theorem 8.5.1. – Let $L / K$ be an algebraic extension of fields. Then there exists a unique extension of $|\cdot|$ to a non-archimedean valuation on $L$. If $L / K$ is finite of degree $n$, then this extension is given by
$$|x|L=\left|\mathrm{N}{L / K}(x)\right|^{1 / n}, \quad \text { for any } x \in L$$
and $L$ is complete with respect to this unique extension of $|\cdot|$.
Proof. – Since an algebraic extension is a union of finite extensions, we may assume that $n=[L: K]$ is finite. Let $|\cdot|L$ be as in the statement. It is clear that $|x|_L=|x|$ for $x \in K$. We show first that $|\cdot|_L$ is indeed a non-archimedean valuation on $L$. Let $\mathcal{O}_L$ denote the integral closure of $\mathcal{O}_K$ in $L$. We claim that $\mathcal{O}_L$ is exactly the subring of $\alpha \in L$ with $|\alpha|_L \leq 1$. Actually, if $x \in \mathcal{O}_L$, then $\mathrm{N}{L / K}(x) \in \mathcal{O}L$, hence $|x|_L \leq 1$. Conversely, if $|x|_L \leq 1$, then $\mathrm{N}{L / K}(x) \in \mathcal{O}K$. Let $f(x)$ be the minimal monic polynomial of $\alpha$ over $K$. Then by Corollary 8.5.6, the Gauss norm of $f$ is $$|f|=\max \left{1,\left|\mathrm{~N}{L / K}(\alpha)\right|\right}=1 .$$
Therefore, one has $f(x) \in \mathcal{O}_K[x]$ and $\alpha \in \mathcal{O}_L$.
It is clear that $|x y|_L=|x|_L|y|_L$ and $|x|_L=0$ if and only if $x=0$. It remains to show that $|x+y|_L \leq \max \left{|x|_L,|y|_L\right}$. We may assume that $|x|_L \leq|y|_L$. Up to dividing by $|y|_L$, it suffices to show that $|x+1|_L \leq 1$ for $|x|_L \leq 1$. By the discussion above, this is equivalent to saying that $x+1 \in \mathcal{O}_L$ for $x \in \mathcal{O}_L$, which is obvious since $\mathcal{O}_L$ is ring.

Now assume that $|\cdot|L^{\prime}$ is another non-archimedean norm on $L$ extending $|\cdot|$. We have to show that $|\alpha|_L=|\alpha|_L^{\prime}$ for all $\alpha \in L$. We first prove that $|\alpha|_L \leq 1$ if and only if $|\alpha|_L^{\prime} \leq 1$. Let $f(x)=x^d+a{d-1} x^{d-1}+\cdots+a_0$ denote the minimal polynomial of $\alpha$ over $K$ with some $d \mid n$. Assume $|\alpha| \leq 1$. Then one has $a_i \in \mathcal{O}_K$ for all $i$. If $|\alpha|_L^{\prime}>1$, then $\left|\alpha^d\right|_L^{\prime}>\left|a_i \alpha^i\right|_L^{\prime} \geq|\alpha|_L^{\prime}$ for all $0 \leq i \leq d-1$. By Lemma $8.2 .6$, one has $0=|f(\alpha)|=|\alpha|^d$, which is absurd. Hence, $|\alpha|_L \leq 1$ implies that $|\alpha|_L^{\prime} \leq 1$. Assuming moreover that $|\alpha|_L=\left|a_0^{n / d}\right|<1$, we prove that $|\alpha|_L^{\prime}<1$ as well. We claim that $a_i \in \mathfrak{m}_K$ for all $0 \leq i \leq n-1$.

## 数学代写|代数数论代写Algebraic number theory代考|Unramified extensions

If $\iota: L \hookrightarrow L^{\prime}$ is a $K$-embedding of two finite extensions of $K$, then $\iota$ sends $\mathcal{O}L$ into $\mathcal{O}{L^{\prime}}$ and $\mathfrak{m}L$ into $\mathfrak{m}{L^{\prime}}$ respectively, hence it induces an embedding of residues fields $k_L \hookrightarrow k_{L^{\prime}}$.
Theorem 9.2.1. – Assume $k^{\prime} / k$ is a finite separable extension. Then the following assertions hold:
(1) There exists an unramified extension $K^{\prime} / K$ with residue field $k^{\prime}$. Moreover, this extension is unique up to isomorphisms, and it is Galois if and only if $k^{\prime} / k$ is Galois.
(2) For any finite extension $L / K$ with residue field $k_L$, there exists a natural bijection (induced by reduction) between the set of $K$-embeddings of $K^{\prime}$ into $L$ and the set of $k$-embeddings of $k^{\prime}$ into $k_L$. In particular, if $k^{\prime} / k$ is Galois, we have $\operatorname{Gal}\left(K^{\prime} / K\right) \cong$ $\operatorname{Gal}\left(k^{\prime} / k\right)$.

Proof. – (1) We prove first the existence of $K^{\prime}$. We may assume that $k^{\prime} \cong k[x] /(\bar{f}(x))$ for some irreducible monic polynomial $\bar{f}(x) \in k[x]$ of degree $n$. Take a monic polynomial $f(x) \in \mathcal{O}K[x]$ of degree $n$ such that $f \bmod \mathfrak{m}_K=\bar{f}$. Then $f(x)$ is necessarily irreducible, and we claim that $K^{\prime}=K[x] /(f(x))$ satisfies the required property. Let $\alpha$ denote the image of $x$ in $K^{\prime}$, and $\bar{\alpha}$ be its reduction modulo $\mathfrak{m}_K$. First, we note that $v(\alpha)=0$, i.e. the image of $\alpha$ in the residue filed of $\mathcal{O}{K^{\prime}}$ is non-zero. Hence, the residue field of $K^{\prime}$ contains $k[\bar{\alpha}]=k^{\prime}$. Thus, we get $f\left(K^{\prime} \mid K\right) \geq n$. By Proposition 9.1.4, we see that $k_{K^{\prime}}=k^{\prime}$ and $e\left(K^{\prime} \mid K\right)=1$. Note that $1, \bar{\alpha}, \cdots, \bar{\alpha}^{n-1}$ are linearly independent over $k$. For rank reasons, we have
$$\mathcal{O}{K^{\prime}} /\left(\pi_K\right) \cong \sum{i=1}^n k \bar{\alpha}^{i-1}$$
By Remark 9.1.2, we have $\mathcal{O}_{K^{\prime}}=\mathcal{O}_K[\alpha]$.
(2) We now prove that $K^{\prime}$ satisfies the property in (2) for any finite extension $L / K$. The rest part of the Theorem will be an easy consequence of this property. Let $L / K$ be as in Statement (2). Let $S(L)$ denote the set of roots of $f(x)$ in $L$, and $S\left(k_L\right)$ the set of roots of $\bar{f}(x)$ in $k_L$. Then the set of $K$-embeddings of $K^{\prime}=K[\alpha]$ into $L$ is in natural bijection with $S$, while the set of $k$-embeddings of $k^{\prime}$ into $k_L$ is in bijection with $S\left(k_L\right)$. Since $\bar{f}(x)$ is separable, Hensel’s Lemma 8.4.1 implies that the reduction map from $S(L)$ to $S\left(k_L\right)$ is bijective. Now the assertion follows immediately.

# 代数数论代考

## 数学代写|代数数论代写Algebraic number theory代考|Extensions of valuations

$$|x| L=|\mathrm{N} L / K(x)|^{1 / n}, \quad \text { for any } x \in L$$

$$|\mathrm{ff}|=\backslash \max \backslash \mathrm{left}{1, \backslash \text { left } \mid \backslash \mathrm{mathrm}(\sim N}{L / K} \backslash \text { alpha) } \backslash \text { right } \mid \backslash \text { right }}=1 \text {. }$$

## 数学代写|代数数论代写Algebraic number theory代考|Unramified extensions

（1）存在一个末分支的扩展 $K^{\prime} / K$ 有残差场 $k^{\prime}$. 此外，这个扩展对于同构是唯一的，并且它是 Galois 当且 仅当 $k^{\prime} / k$ 是伽罗华。
(2) 对于任何有限扩展 $L / K$ 有残差场 $k_L$ ，在集合之间存在自然双射 (由归约引起) $K$-嵌入 $K^{\prime}$ 进入 $L$ 和 一组 $k$-嵌入 $k^{\prime}$ 进入 $k_L$. 特别是，如果 $k^{\prime} / k$ 是伽罗华，我们有 $\operatorname{Gal}\left(K^{\prime} / K\right) \cong \operatorname{Gal}\left(k^{\prime} / k\right)$.

$$\mathcal{O} K^{\prime} /\left(\pi_K\right) \cong \sum i=1^n k \bar{\alpha}^{i-1}$$

(2) 现在证明 $K^{\prime}$ 对于任何有限扩展满足 (2) 中的性质 $L / K$. 定理的其余部分将是该属性的简单结果。让 $L / K$ 如声明 (2) 中所述。让 $S(L)$ 表示根的集合 $f(x)$ 在 $L$ ，和 $S\left(k_L\right)$ 根的集合 $\bar{f}(x)$ 在 $k_L$. 然后是一组 $K$ 嵌入 $K^{\prime}=K[\alpha]$ 进入 $L$ 与 $S$ ，而集合 $k$-嵌入 $k^{\prime}$ 进入 $k_L$ 与 $S\left(k_L\right)$. 自从 $\bar{f}(x)$ 是可分离的，Hensel 的引理 8.4.1 意味着从 $S(L)$ 至 $S\left(k_L\right)$ 是双射的。现在断言紧随其后。

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