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The AKE priciple states that two finitely ramified Henselian field (this means that the field is either of residual caracteristic 0 or is of characteristic $p$ and there is only a finite number of elements x of the value group such that $0< x < v(p)$) are elementary equivalent if and only if their value groups are elementary equivalent (in the language of group) and their residue fields elementary equivalent (in the language of ring).

I'm looking for application of this principle to prove non trivial results.

I also have a side question which I fear is a dumb one. Being algebraically closed is a first order property. So it should be equivalent to properties on the residues field and on the value group. What could be these properties ?

Thanks to any one who might answer !

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    $\begingroup$ I don't think a finitely ramified Henselian field can ever be algebraically closed, as it won't contain the $n$-th root of a prime element for $n$ large. $\endgroup$
    – Felipe Voloch
    Commented Oct 6, 2012 at 0:25
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    $\begingroup$ Thank you very much ! But for unramified field of characteristic 0 ? What prevent them from being algebraically closed ? $\endgroup$
    – elvis
    Commented Oct 6, 2012 at 13:50
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    $\begingroup$ I think the version of the transfer principle stated above is not entirely correct: one has to require that the valued fields themselves have characteristic 0; otherwise $\mathbb{Q}_p$ vs. $\mathbb{F}_p((t))$ is a counter-example. $\endgroup$
    – Immi Halupczok
    Commented Mar 15, 2013 at 15:17
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    $\begingroup$ I’m afraid the AKE principle does not hold the way it is stated here. For example, $\mathbb Q_p$ and $\mathbb Q_p(\sqrt p)$ are finitely ramified, henselian, and have the same value groups and residue fields, but are not elementarily equivalent. See encyclopediaofmath.org/index.php/Model_theory_of_valued_fields for some variants that do hold. $\endgroup$
    – Emil Jeřábek
    Commented Aug 14, 2013 at 13:56
  • $\begingroup$ @Immi Halupczok: Finitely ramified fields always have characteristic $0$. In your example, the value group of $\mathbb F_p((t))$ has infinitely many elements between $0$ and $v(p)=\infty$. $\endgroup$
    – Emil Jeřábek
    Commented Aug 14, 2013 at 14:01

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The above AKE principle implies the transfer principle between $\mathbb{Q}_p$ and $\mathbb{F}_p((t))$, and that one has some "real" applications.

The transfer principle states that for every first order statement $\phi$ in the language of valued fields, there exists a bound $N$ such that for every $p > N$ we have $\mathbb{Q}_p \models \phi \iff \mathbb{F}_p((t)) \models \phi$. To prove it, suppose the equivalence is violated for an inifinite set $P$ of primes. Let $K$ be a (non-principal) ultraproduct of the $\mathbb{Q}_p$ with $p \in P$ and let $L$ be the corresponding ultraproduct of the $\mathbb{F}_p((t))$. Then $K$ and $L$ have characteristic $0$ and they have the same value group and residue field, so AKE implies $K \models \phi \iff L \models \phi$, which yields a contradiction to the choice of $P$.

Here is one important application of the transfer principle:

The "Fundamental Lemma of the Langlands program" (which, actually, was a conjecture for a long time) is a statement about linear algebraic groups over non-archimedean local fields $K$. Ngo [arXiv:0801.0446] proved this "Lemma" in the case $K = \mathbb{F}_p((t))$ (and obtained the Fields Medal for that proof). Cluckers-Hales-Loeser [arXiv:0712.0708] showed that the transfer principle can be applied to that result, yielding the Fundamental Lemma for $K = \mathbb{Q}_p$ when $p$ is big. (Moreover, Hales had already shown before that it is sufficient to know the Fundamental Lemma for big $p$.)

Note that the Fundamental Lemma is an equality of integrals of functions from definable sets in $K$ to $\mathbb{C}$, so a priori, it doesn't seem to be a first order statement where the transfer principle could be applied. To apply it nevertheless, one has to encode the functions appearing in the Fundamental Lemma using model theoretic objects living purely in $K$ and then check that integration can be carried out purely on these encodings. In this way, the equality of integrals becomes a first order statement. (This can be seen as a very short explanation of what motivic integration is about.)

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AKE prinicpal alows us to deduce conclusions about the theory of the valuaed field from the theories of the residue field and value group. Via this you prove, for example, that such and such theories are model complete, which supplies you with the tool of transfer argument (as in the model theoretic proof for hilbert's Nullstellensatz).

As for your side question, if I have understood correctly, the answer is that the residue fiels is algebraically closed and the value group is divisible.

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  • $\begingroup$ Thank you very much. I'm sorry if my second question was unclear. $\endgroup$
    – elvis
    Commented Oct 6, 2012 at 17:25

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