About the definition of local fields I read in a book that the definition of local fields is: A field $K$ for which with respect to a discrete valuation $v$, the residue field is finite and $K$ is complete with respect to $v$.
However, I have a few questions:


*

*Is it the field $K$ which we call a local field? Or the combination of $K$ and $v$, such as $(K,v)$?

*Or is it true that if with respect to one discrete valuation $v$ the residue field is finite and $K$ is complete, then with respect to any other discrete valuation, we also have that the residue field is finite and $K$ is complete?
Please help me with these questions.
Thank you very much!
 A: It is irrelevant whether you call $K$ the local field or $(K,v)$ the local field since there are not multiple options for $v$: from the field $K$ we will  reconstruct $\mathcal O_v$ and its maximal ideal $\mathfrak m_v$ below by purely algebraic properties (although the justification for those properties does use limits).  And once we know what $\mathcal O_v$ and $\mathfrak m_v$ are as subsets of $K$, the valuation $v$ on $K$ can be reconstructed: if $x \in \mathcal O_v$ with $x \not= 0$ then there is a unique $n \geq 0$ such that $x \in \mathfrak m_v^n$ and $x \not\in \mathfrak m_v^{n+1}$.  That $n$ is $v(x)$. If $x \in K - \mathcal O_v$ then $1/x \in \mathfrak m_v$ and $v(x) = -v(1/x)$.
In order to reconstruct $\mathcal O_v$ and $\mathfrak m_v$, first I'll reconstruct $\mathcal O_v^\times$. By Hensel's lemma (analysis!), each element of $1 + \mathfrak m_v$ is an $n$th power in $K$ for infinitely many positive integers $n$ (purely algebraic property!). Each element of $K^\times$ that is an $n$th power in $K$ for infinitely many $n$ must be in $\mathcal O_v^\times$, so if we set
$$
S = \{x \in K^\times : x \text{ is an } n\text{th power in } K \text{ for infinitely many } n\}
$$
then $1 + \mathfrak m_v \subset S \subset \mathcal O_v^\times$.  The gap between $1 + \mathfrak m_v$ and $\mathcal O_v^\times$ is filled by roots of unity: writing $\mu(K)$ for the set of all roots of unity in $K$, we have $\mathcal O_v^\times = \mu(K)(1+\mathfrak m_v)$, so $$\mathcal O_v^\times = \mu(K)S.$$  Both $\mu(K)$ and $S$ are defined by purely algebraic properties, so the above formula shows that we can reconstruct $\mathcal O_v^\times$ from $K$ purely algebraically.
For $x \in K$, check that $x \in \mathfrak m_v$ if and only if $\zeta + x \in \mathcal O_v^\times$ for all $\zeta \in \mu(K)$ (For $(\Leftarrow)$: from $1 + x \in \mathcal O_v^\times$ we get $x \in \mathcal O_v$, and if $x \not\in \mathfrak m_v$ then $x \equiv \omega \bmod \mathfrak m_v$ for some root of unity $\omega$, so $x + (-\omega) \in \mathfrak m_v$.)  Since $\mathcal O_v^\times$ and $\mu(K)$ are determined from $K$ purely algebraically, so is $\mathfrak m_v$, and thus so is $\mathcal O_v^\times \cup \mathfrak m_v = \mathcal O_v$. This completes the proof that $\mathcal O_v$ and $\mathfrak m_v$ are determined by $K$ purely algebraically, so $v$ is as well by the reasoning given in the first paragraph.
This proof did not use anywhere that $K$ has a finite residue field, but only that the residue field is an algebraic extension of a finite field (so that the nonzero elements of $\mathcal O_v/\mathfrak m_v$ are represented by roots of unity). I don't think this is particularly important, but I wanted to point it out anyway.
