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.


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 $\mathbb{C}$.

  • $\begingroup$ It's a general result from field theory about transcendence degrees. $\endgroup$ – D_S May 20 '17 at 3:27
  • $\begingroup$ $\overline{\mathbb{Q}_p}$ contains some non-algebraic elements (over $\mathbb{Q}$) so it can't be isomorphic as a field to $\overline{\mathbb{Q}}$. Did you mean complete algebraically closed characteristic zero fields ? $\endgroup$ – reuns May 20 '17 at 4:19
  • $\begingroup$ What does $\overline{\mathbb{Q}}$ have to do with anything? $\endgroup$ – D_S May 20 '17 at 14:08
  • $\begingroup$ See my edit. Sorry I realize my answer was misleading, in addition to algebraically closed and characteristic zero, you also need them to have the same the same transcendence degree over $\mathbb{Q}$. $\endgroup$ – D_S May 20 '17 at 14:25
  • $\begingroup$ Yes it is clear now, except what would be $S$ exactly (axiom of choice ?..) in the case of $\mathbb{C}$ or $\mathbb{C}_p$ $\endgroup$ – reuns May 20 '17 at 17:13

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.

  • $\begingroup$ What would be the algebraic definition of $\mathbb{Z}_p$ (why is there only one possible "ring of integers" in $\mathbb{Q}_p$) ? For a number field $K$ it is obvious because there is only one embedding on $\mathbb{Z}$ into $K$, and $\mathcal{O}_K$ are the integral (over $\mathbb{Z}$) elements of $K$. But it doesn't work for $\mathbb{Q}_p$ (which is not contained in $\overline{\mathbb{Q}}$) $\endgroup$ – reuns May 20 '17 at 20:31
  • $\begingroup$ I was hoping nobody would ask, @user1952009. There are other characterizations of the maximal ideal of $\Bbb Z_p$, but the one I use is something like this: $\alpha\in\Bbb Q_p$ is in $p\Bbb Z_p$ if and only if for every natural number $m$ prime to $p$, $1-\alpha$ has an $m$-th root in the field. (In this formulation, you need to know $p$ beforehand.) $\endgroup$ – Lubin May 20 '17 at 20:56
  • $\begingroup$ It seems this might help for showing what you said, using the $p$-adic binomial series of $(1+p a)^{1/m}, a \in \mathbb{Z}_p$. $\endgroup$ – reuns May 20 '17 at 21:32
  • $\begingroup$ Right you are, @user1952009. Or, the fact that $(1+t)^{1/m}$ has no $p$’s in the denominators. Or, an application of Hensel’s Lemma. There’s also a way of applying the series for $\log(1+t)$. $\endgroup$ – Lubin May 21 '17 at 12:56

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