Uniqueness of residue field of a complete field 
Suppose that we have a field $F$ which is complete under an absolute value $|\cdot|_1$, and its residue field is $k$. My question is, can we come up with some other non-trivial absolute value $|\cdot|_2$ on $F$, such that $F$ remains complete under it, but instead produces a residue field $k_2$ that is not isomorphic to $k$.
Now, clearly this is not possible if we drop the assumption of $F$ being complete, because on $\mathbb Q$ we have the $p-$adic and the $q-$adic absolute values which produce different residue fields.

The reason I am asking such a question, is because I notice a tendency in literature to refer to complete discretely valued fields without referring to the absolute values, which are also part of the datum. I was wondering if there was a reason for doing that, besides, say that the choice of absolute value would be not be relevant for the statements they are trying to prove.
 A: The completion $\mathbb{C}_p$  of $\overline{\mathbb{Q}_p}$ with respect to its usual absolute value has the same cardinality as $\mathbb{C}$.  Moreover, these fields are both algebraically closed and of characteristic zero.  These three pieces of information tell us that they are isomorphic as fields.  This means that all the complete fields $\mathbb{C}_p : p$ prime are isomorphic, but they have nonisomorphic residue fields, since the residue field of $\mathbb{C}_p$ is $\overline{\mathbb F_p}$.
Edit: Here is a sketch of the main results of transcendence degrees.  Let $k \subseteq L$ be a field.  A subset $S \subseteq L$ is algebraically independent over $k$ if for any finite subset $\{s_1, ... , s_n\} \subseteq S$, the condition $f \in k[X_1, ... , X_n]$, $f(s_1, ... , s_n) = 0$ implies $f = 0$.
A transcendence basis of $L/k$ is a maximal algebraically independent set.  Transcendence bases exist, and any two transcendence bases have the same cardinality.  If $S$ is a transcendence basis, then up to isomorphism the field $k(S)$ depends only on the cardinality of $S$, because $k(S)$ is isomorphic to quotient field of the polynomial ring over $k$ in $|S|$ indeterminates.  The extension $L/k(S)$ is algebraic.  
Let $L, K$ be two algebraically closed fields of characteristic zero.  They each contain a copy of $\mathbb{Q}$.  Suppose $S, T$ are transcendence bases of $L/\mathbb{Q}$ and $K/\mathbb{Q}$ with the same cardinality.  Then $L$ and $K$ are isomorphic: there is an isomorphism $\mathbb{Q}(S) \rightarrow \mathbb{Q}(T)$.  Since $L$ is algebraically closed, and algebraic over $\mathbb{Q}(S)$, it is an algebraic closure of $\mathbb{Q}(S)$.  The same for $K$ over $\mathbb{Q}(T)$.  Any two algebraic closures of a given field are noncanonically isomorphic, so the isomorphism $\mathbb{Q}(S) \rightarrow \mathbb{Q}(T)$ extends to an isomorphism $L \rightarrow K$.
Suppose moreover that $L$ is an uncountable algebraically closed field.  In that case, $L$ is classified up to isomorphism by its cardinality and its characteristic.  This is because if $S$ is a transcendence basis of $L$ over $\mathbb{Q}$, then the cardinality of $L$ is equal to the cardinality of $\mathbb{Q}(S)$, which is equal to the cardinality of $S$.
Thus any algebraically closed field of characteristic zero and cardinality that of $\mathbb{R}$ must be isomorphic to $\mathbb{C}$.
A: Information that may be helpful:
In such fields $F$, describing $|\cdot|$ is the same as describing the maximal ideal $\mathfrak m$ of the ring of $|\cdot|$-local integers. I usually think of this ideal as being the things whose powers converge to $0$, but if there’s an equivalent definition of this ideal that depends only an algebraic properties of elements, the topology will be unique.
For disconnected local fields, i.e. for $\Bbb Q_p$ and its finite extensions and for the fields $\kappa((t))$ for which $\kappa$ is finite, there is a purely algebraic definition of the maximal ideal, so there is no other non-trivial absolute value.
I suppose an equivalent characterization of fields with only the one absolute value would be that every field automorphism is necessarily continuous.
