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Fix a prime number $p$. Suppose that I have a valuation $v_p: \mathbb{Q} \to \mathbb{Q}$ on the rationals $\mathbb{Q}$. That is, $v_p( p^n(\frac{a}{b})) = p^{-n}$ where each of $a,b$ is not divisible by $p$.

How can I extend $v_p$ to $v$ on the reals $\mathbb{R}$ such that $v|_\mathbb{Q} = v_p$? I am looking for an explicit description of $v$, if that is possible. I know for a fact that one can extend valuation on any field extension.

Thank you,

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You mean, $v|_\mathbb Q = v_p$ ? –  Abhishek Parab Jul 5 '10 at 5:37
correct... sorry about misspelt. –  Tharatorn Supasiti Jul 5 '10 at 6:09
have corrected the typo –  Yemon Choi Jul 5 '10 at 6:22
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4 Answers

up vote 11 down vote accepted

As you point out, it follows on general grounds that there is an extension of $v_p$ to a valuation on ${\mathbb R}$ (in fact, there are uncountably many such extensions), but it is impossible to give an "explicit" description. Indeed, not only will any such extension by discontinuous with respect to the usual Euclidean topology on the reals, but it will not be a measurable function.

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Indeed, $t\mapsto v(\exp t)$ would be a surjective group morphism from $\mathbb R$ to $\mathbb Q$, which is non-measurable both in Lebesgue and in Baire sense. By results of Robert Solovay and Saharon Shelah, such a thing cannot be constructed (if ZFC is consistent, of course) using only countably many choices, more precisely the axiom of dependent choice, which suffices for most classical analysis. See wikipedia "Solovay model". –  BS. Jul 5 '10 at 15:07
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You can extend $v_p$ to $\overline{\mathbb{Q}}_p$ in a unique way, and then the general theory of fields tells you that $\overline{\mathbb{Q}}_p$ is isomorphic to $\mathbb{C}$ which gives you the extension you want. The fields $\overline{\mathbb{Q}}_p$ and $\mathbb{C}$ are isomorphic because they are both algebraically closed and have the same transcendence degree over $\mathbb{Q}$. However as Thomas said, this construction is not explicit, indeed it uses the axiom of choice.

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Just in case you don't know this, the extension of $v$ to $\overline{\mathbb{Q}_p}$ is done as follows: If $x$ is algebraic over $\mathbb{Q}_p$, let $x^n + a_{n-1} x^{n-1} + \cdots a_0=0$ be the minimal polynomial of $x$. Then $v(x) = (1/n) v(a_0)$. Turning this into a valuation on $\mathbb{C}$ requires noncanonical choices of two kinds.

For the first kind of choice, consider let $p=7$. The equation $x^2-6x+7=0$ has two roots in $\mathbb{R}$: one is roughly $4.414\ldots $ and the other is roughly $1.585\ldots $. It also has two roots in $\mathbb{Q}_7$: one is roughly $(\ldots 60)_7$ and the other is roughly $(\ldots16)_7$. You have to decide which one will be identified with which. For the second kind, there are transcendental elements of $\mathbb{C}$, such as $\pi$. You can choose their valuation freely. (More precisely, you can choose the valuations of a transcendence base freely.)

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I guess everybody knows this, I am just mentioning it: in Atiyah-MacDonald's Commutative Algebra, the proof for the extension of valuation is done by using maps $f$ from the original ring A (in our case $\mathbb{Z}_p$, corrected to(p)) to an algebraic closure (say $\mathbb{C}$ or as Laurent Berger mentioned $\overline{\mathbb{Q}_p}$, or $\overline{\mathbb{F}_p}$ in our case), and then for any $a\not\in A$, there is a way to decide whether we can extend the valution to $A[a]$ or $A[a^{-1}]$, using the maps previously defined. To me, this method seems a little bit more explicit, which is actually not.

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What Atiyah and MacDonald are talking about is extending homomorphisms from a subring of a field $K$ into an algebraically closed field to a valuation ring of $K$. This is not the same as the extension of valuations being discussed here. –  Keenan Kidwell Jul 5 '10 at 15:33
Well, Keenan, in my opinion, this is how one shows the existence of a extension of a valuation to the bigger field; you can always treat the smaller valuation ring as a subring of the bigger field. There is also a disguised version of it called Chevalley's Extension Theorem, in Engler & Prestel's <i>Valued Fields</i>, where they prove that the extension of the valution on the bigger field always exists. I would still say that they are the same thing, although I don't know if situation discussed here is usually done this way. –  Jizhan Hong Jul 5 '10 at 15:58
I mean, any finite extension of a complete valued field admits a unique absolute value extending the one on the smaller field. These are compatible by uniqueness, so we get a unique extension to any algebraic extension, i.e., to an algebraic closure of $\mathbb{Q}_p$. This is all constructive. At this point Zorn's lemma is needed (any two algebraically closed fields of characteristic zero with the same tr.deg over $\mathbb{Q}$ are isomorphic as fields). It isn't obvious to me how one would use the results in A and M to get this result. Would you mind explaining it? –  Keenan Kidwell Jul 5 '10 at 16:11
Are you asking about the uniqueness or the existence of the extension? It is also not obvious to me that the uniqueness could be easily obtained by A&M's argument. If anybody knows, please go ahead. About the extension, A&M says that starting from a subring A of a field K(or any field extension L), with a homomorphism f from A to the alg.closure of K, one would obtain a chain of "compatible" pairs (A, f) with a maximal element, say (B, h); then B is a valuation ring of K (or L) extending A. But I guess that's not what you are asking? –  Jizhan Hong Jul 5 '10 at 18:26
The original question is: how does one extend the $p$-adic valuation on $\mathbb{Q}$ to $\mathbb{R}$. Several answers were given, all citing the fact that it's essentially a consequence of abstract nonsense with transcendence degrees. You mentioned A&M's results on valuation rings and said that you found this method for "extending valuations" more explicit. My first comment was to the effect that I don't see what their results on valuation rings have to do with extending valuations, and I was wondering if you could explain this to me. –  Keenan Kidwell Jul 5 '10 at 19:05
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