Fraction Field of Non-Archimedean Valuation Ring and Relate the Divisibility to the Comparison of Absolute Value for Elements in this Ring Let $K$ be a field with $|\ \ |$ a non-archimedean absolute value on it. Define the valuation ring of $|\ \ |$ to be $$R:=\{a\in K:|a|\leq 1\}.$$ I want to prove the following two claims:


*

*$K$ is the fraction field of $R$


*For all $a,b\in R$, $a|b$ if and only if $|b|\leq |a|$.

For the notation of field of fraction, we denote $\frac{s}{t}$ to be an equivalence class in $\text{Frac}(R)$ by the equivalence relation: for all $s,c\in R$ and $t,d\in R\setminus\{0\}$, $$(s,t)\sim (c,d)\iff sd=tc.$$

For 1, firstly, $R$ is an integral domain. Indeed, as $K$ is a field, by definition $1_{K}\neq 0_{K}$. If $0\neq x$ is a zero divisor in $R$, then there exists $y\neq 0$ in $R$ such that $xy=0$. But since $x,y\neq 0$, we have $|x|,|y|\neq 0$, so we have a contradiction $0\neq |x|\cdot|y|=|xy|=|0|=0.$
Now, I understand that the reason of why $K=\text{Frac}(R)$ is that for every $a\in K$, it is either $a\in R$ or $a^{-1}\in R$. Hence, I want to construct a field isomorphism as follows. Let $\phi:K\longrightarrow\text{Frac}(R)$ be a map defined by
$$\phi(x):=\left\{
   \begin{array}{ll}
   \frac{x}{1},\ \ \text{if}\ \ x\in R\\
   \frac{1}{x^{-1}},\ \ \text{if}\ \ x\notin R.
   \end{array}
\right.$$
This map is well-defined because if $x\notin R$, then $|x|>1$, and thus $x\neq 0$, so $x^{-1}$ exists and $x^{-1}\in R$. But then I have trouble showing that it is a homomorphism.
Let $x,y\in K$. If $x,y\in R$, then $x+y\in R$, and $$\phi(x+y):=\dfrac{x+y}{1}=\dfrac{x}{1}+\dfrac{y}{1}=:\phi(x)+\phi(y).$$ If $x\in R$ but $y\notin R$, then $|x|<|y|$ and thus $|x+y|=\max\{|x|,|y|\}=|y|>1$, which means that $x+y\notin R$, and thus $$\phi(x+y)=\dfrac{1}{(x+y)^{-1}}.$$ On the other hand, $$\phi(x)+\phi(y):=\dfrac{x}{1}+\dfrac{1}{y^{-1}}=\dfrac{xy^{-1}+1}{y^{-1}}.$$ In this case, how could I show that $\phi(x)+\phi(y)=\phi(x+y)$? Let alone we have a possibly more complicated case when $x,y\notin R$. I am starting to believe that the map $\phi$ is not a good one. But I don't know what alternative I can use.

For 2. I can prove that $a|b\implies |b|\leq |a|$. This follows from $|n|\leq 1$ for any $n\in\mathbb{N}$ (because $|\ \ |$ is non-archimedean) --- if $b=an$ then $|b|=|an|\leq |a|.$
However, I don't know how to prove the converse. I have several attempted to start the proof by using $|b|\leq |a|$, but it turned out hard to get $a|b$. So, I assumed that $a\nmid b$ and tried to prove that $|b|>|a|$. But a weird thing happened. Suppose that $a\nmid b$ and $a<b$, then let's say $b=na+m$ for some $n,m\in\mathbb{N}$ and $m<a$. Then, we have $$|b|=|na+m|\leq\max\{|na|,|m|\}\leq \max\{|a|,|m|\}= |a|,$$ so somehow I proved that $a\nmid b\implies |b|\leq |a|$.
What is wrong with my proof here? Thank you so much!
 A: As suggested by Chilote, I am going to answer my own question. I am reading the book "Algebra Volume II: Fields with Structure, Algebras and Advanced Topics" by Falko Lorenz. This is not a widely-used book to learn field theory, but I found it accidentally and it has everything I need --- especially a good illustration of the extension of absolute value to any number field and to any purely inseparable extension field, and a clear introduction of local field theory. The good part of this book is that it doesn't require too many prerequisites.
I never formally learnt the theory about absolute values on fields, so I decided to work through Chapter 23 before going to local field theory.
My question related to Page 45 of this book when he introduced the non-archimedean valuation.
Let $K$ be a field, and let $|\ \ |$ be a non-archimedean absolute value on it. Then, he gave the following proposition:


*

*$R:=\{a\in K:|a|\leq 1\}$ is a subring of $K$, called the valuation of ring of $K$


*$K$ is the field of fration of $R$


*$\mathfrak{p}:=\{a\in K:|a|<1\}$ is an ideal of $R$, called the valuation ideal of $K$.


*An element $a\in K$ is a unit of $R$ if and only if $|a|=1$. Thus, $R^{\times}=R\setminus\mathfrak{p}$. It then follows that $R$ is a local ring and $\mathfrak{p}$ is its maximal ideal.


*$R/\mathfrak{p}$ is a field, called the residue field of $K$ with respect to $|\ \ |$.


*For $a,b\in R$, $a|b$ if and only if $|b|\leq|a|$. (As suggested by Chilote, we say that $a|b$ if and only if there exists $k\in R$ such that $b=ak.$)

My questions is 2 and 6. As most of these results are immediate, I will give a full proof of this proposition for future reference if anyone is interested in this.


*

*$R$ clearly contains the multiplicative identity $1_{K}$ because $|1_{K}|=1$. Let $a,b\in R$, then $|a+b|\leq\max\{|a|,|b|\}\leq 1,$ and thus $a+b\in R$. Since $|-a|=|a|$, $-a\in R$ also holds. Hence, $R$ is an additive subgroup of $K$. Moreover, $|ab|=|a||b|\leq 1$ since both $|a|$ and $|b|$ are at most $1$. Hence, $ab\in R$, and thus $R$ is a subring.




*Firstly, $R$ is an integral domain. Indeed, as $K$ is a field, by definition $1_{K}\neq 0_{K}$. If $0\neq x$ is a zero divisor in $R$, then there exists $y\neq 0$ in $R$ such that $xy=0$. But since $x,y\neq 0$, we have $|x|,|y|\neq 0$, so we have a contradiction $0\neq |x|\cdot|y|=|xy|=|0|=0.$ Hence, it makes sense to consider the field of fraction $\text{Frac}(R)$ of $R$. Let $\phi:K\longrightarrow\text{Frac}(R)$ be a map defined by
$$\phi(x):=\left\{
   \begin{array}{ll}
   \frac{x}{1},\ \ \text{if}\ \ x\in R\\
   \frac{1}{x^{-1}},\ \ \text{if}\ \ x\notin R.
   \end{array}
\right.$$
This map is well-defined because if $x\notin R$, then $|x|>1$, and thus $x\neq 0$, so $x^{-1}$ exists and $x^{-1}\in R$.
Let $x,y\in K$. If $x,y\in R$, then the strong triangle inequality implies that $x+y\in R$, and $$\phi(x+y):=\dfrac{x+y}{1}=\dfrac{x}{1}+\dfrac{y}{1}=:\phi(x)+\phi(y).$$
If $x\in R$ but $y\notin R$, then $|x|<|y|$ and thus $|x+y|=\max\{|x|,|y|\}=|y|>1$, which means that $x+y\notin R$, and thus $$\phi(x+y)=\dfrac{1}{(x+y)^{-1}}.$$ On the other hand, $$\phi(x)+\phi(y):=\dfrac{x}{1}+\dfrac{1}{y^{-1}}=\dfrac{xy^{-1}+1}{y^{-1}}.$$ But $$y^{-1}(x+y)=y^{-1}x+1\implies y^{-1}=(x+y)^{-1}(y^{-1}x+1)\implies (1,(x+y)^{-1})=(xy^{-1}+1,y^{-1}),$$ which means that $\phi(x+y)=\phi(x)+\phi(y).$ A similar proof can be applied to the case of $x\notin R$ and $y\in R$.
If $x\notin R$ and $y\notin R$, then $$\phi(x)+\phi(y)=\dfrac{1}{x^{-1}}+\dfrac{1}{y^{-1}}=\dfrac{y^{-1}+x^{-1}}{x^{-1}y^{-1}}.$$ If $x+y\notin R$, then $$\phi(x+y)=\dfrac{1}{(x+y)^{-1}}.$$ But
\begin{align*}
x^{-1}y^{-1}(x+y)=y^{-1}+x^{-1}&\implies x^{-1}y^{-1}=(y^{-1}+x^{-1})(x+y)^{-1}\\
&\implies (1,(x+y)^{-1})=(y^{-1}+x^{-1},x^{-1}y^{-1})\\
&\implies \phi(x+y)=\phi(x)+\phi(y).
\end{align*}
If $x+y\in R$, then $$\phi(x+y)=\dfrac{x+y}{1},$$ but we have seen that $$x^{-1}y^{-1}(x+y)=y^{-1}+x^{-1}$$ and thus $\phi(x+y)=\phi(x)+\phi(y).$
Hence, $\phi$ preserves addition. To prove the perseverance of multiplication, let $x,y\in K$, and if $x,y\in R$, then $xy\in R$, and it is clear that $\phi(xy)=\phi(x)\phi(y)$. If $x,y\notin R$, then $xy\notin R$, and thus once again it is clear that $\phi(xy)=\phi(x)\phi(y)$.  If $x\in R$ but $y\notin R$, then $$\phi(x)\phi(y)=\dfrac{x}{y^{-1}},$$ but $x=(xy)y^{-1}$ ensures that $\phi(x)\phi(y)=\phi(xy)$ when $xy\in R$, and $(xy)^{-1}x=y^{-1}x^{-1}x=y^{-1}$ ensures that $\phi(x)\phi(y)=\phi(xy)$ when $xy\notin R$.
Finally, $\phi(1):=\dfrac{1}{1}$ which is the identity of $\text{Frac}(R)$, and thus $\phi$ is a field homomorphism.
It is surjective, because for any $\frac{s}{t}\in\text{Frac}(R)$, let $(a,b)$ be a representative of $\frac{s}{t}$. Note that $b\neq 0$ and $b^{-1}\notin R$, so we have $$\phi(ab^{-1})=\phi(a)\phi(b^{-1})=\dfrac{a}{(b^{-1})^{-1}}=\dfrac{a}{b}=\dfrac{s}{t}.$$
It is injective, because for any $x\in K$ such that $\phi(x)=\frac{0}{s}$ for any $s\in R\setminus\{0\}$, then $x\in R$ must hold otherwise $\phi(x)=\frac{1}{x^{-1}}=\frac{0}{s}$ implies that $s=0$, contradiction. So $x\in R$, which means that $\phi(x)=\frac{x}{1}=\frac{0}{s}$ implying that $xs=0$, since $s\neq 0$ and $R$ is an integral domain, it implies that $x=0$.
Hence, $\phi$ is a field isomorphism and $K\cong \text{Frac}(R)$.



*Immediate. Closure under addition follows from the strong triangle inequality.




*Let $a\in R$ be a unit in $R$. Then, there exists $b\in R$ such that $ab=1$, and thus $|a|\cdot|b|=1$. If $|a|<1$, then $|b|>1$ must hold, but this is a contradiction because $b\in R$. Hence, $|a|=1$ must hold. Conversely, let $a\in R$ be such that $|a|=1$. Since $a\in K$ and $K$ is a field, it must have a multiplicative inverse $b\in K$ such that $ab=1$ and thus $|a|\cdot|b|=1$. But $|a|=1$, which means that $|b|=1$, and thus $b\in R$, and thus $R$ contains the multiplicative inverse of $a$, and thus $a$ is a unit in $R$.

Hence, $R^{\times}=R\setminus\mathfrak{p}$. It is clear that $\mathfrak{p}\neq (1)$, because $1\notin\mathfrak{p}$. We just prove that every element $x\in R\setminus\mathfrak{p}$ is a unit in $R$, which means that $R$ is a local ring and $\mathfrak{p}$ is its (only) maximal ideal. (Atiyah-MacDonald Proposition 1.6.)



*Immediate because $\mathfrak{p}$ is the maximal ideal of $R$.




*Immediate. If $b=ak$, then since $k\in R$, we have $|b|=|a|\cdot|k|\leq |a|.$ Conversely, if $|b|\leq |a|$, then $|b a^{-1}|\leq |a|\cdot|a^{-1}|=1,$ and thus $ba^{-1}\in R$, and thus there exists some $k\in R$ such that $ba^{-1}=k$, which means that $b=ka$.


Please let me know if you find any mistakes. I am really grateful for the help from Chilote!

As suggested by reuns, I believe that the following field isomorphism is much easier to deal with: $$\phi_{2}:\text{Frac}(R)\longrightarrow K\ \ \text{defined by}\ \ \dfrac{s}{t}\mapsto st^{-1}.$$
For this map $\phi_{2}$, it is much easier to prove that it preserves the addition, multiplication and  multiplicative identity. It is also easy to show that it is injective. But one needs to verify that $\phi_{2}$ is independent of the choice of the representative of $\frac{s}{t}$.
Indeed, let $(a,b)\in\frac{s}{t}$, then we know that $(a,b)=(s,t)$ which means that $bs=at$ and thus $ab^{-1}=st^{-1}$.
I believe that this works in any case as long as $R$ is a subring of the field $K$. (And I believe that $R$ needs to be an integral domain to make $\phi_{2}$ injective.)
In our special case, this map is surjective. Because for any $a\in K$, if $a\in R$, then $\phi_{2}(a/1)=a\in K$. If $a\notin R$, then $a^{-1}\in R$, and thus $\phi_{2}(1/a^{-1})=a\in K$.
Hence, $K\cong\text{Frac}(R).$
