3
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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 $\mathbb{C}$.

$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.