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Let $A$ be a Dedekind domain, and let $K$ be its field of fractions. In Serre's Local Fields, the following Lemma is stated.

Approximation Lemma Let $k$ be a positive integer. For every $i$, $1\leq i \leq k$, let $\mathfrak p_i$ be prime ideals of $A$, $x_i$ elements of $K$, and $n_i$ integers. Then there exists an $x\in L$ such that $v_{\mathfrak p_i}(x - x_i)\geq n_i$ for all $i$, and $v_{\mathfrak q} \geq 0$ for $\mathfrak q \neq \mathfrak p_1,\ldots, \mathfrak p_k$.

To get a better feel for this I would like to be able to actually find the $x$ stated in Lemma. I have tried to follow the proof to do this, and there is one crucial step that I don't understand at all.

At the start, after assuming that the $x_i$ are in $A$, it is stated that "by linearity, one may assume that $x_2 = \ldots = x_k = 0$". I don't see why we can assume this, and I don't even know what is meant by linearity here.

Any suggestion as to what it means in the proof, or alternative (preferably constructive) proofs would be much appreciated.

As a further, related question, this lemma is often stated as showing that we can find an element that is in one collection of ideals, and not in some other ideal (for example, the proof of proposition 19). I don't see how the Lemma controls an element NOT being in an ideal - it has no control over how high the valuation at an ideal is. Any hints about this would be appreciated too.

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I also thought this was really confusing the first time I read it. – Dylan Moreland May 10 '12 at 21:16
I am glad to know I am not the only one :) – Joe Tait May 11 '12 at 9:26
up vote 4 down vote accepted

I agree with you that this is a (very rare!) slightly obscure moment in Serre's exposition.

This result however is found in absolutely every treatment of valuation theory / local fieds. It is a very famous and useful result, usually called either Weak Approximation or Artin-Whaples Approximation.

For instance, I give (what I want to be) a quite careful, detailed proof of this result in $\S 1.4$ of these notes. I took the proof from a classic text of Artin, so it's a good bet that it is the original Artin-Whaples proof. Also, this proof is "constructive".

Note that what I (and most others) call Artin-Whaples approximation is slightly more general than the result you quoted above in that it concerns norms rather than (discrete) valuations: in particular, Archimedean norms are allowed. This extra generality doesn't make the proof any harder but really does come up in algebraic number theory: e.g. I have often needed an element $x$ of a number field which jumps through various $v_{\mathfrak{p}}$-adic hoops (for a finite set of prime ideals of the ring of integers) and is positive with respect to every real Archimedean place.

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Thank you, that is very useful. I had looked for approximation lemmas, but there seem to be a number of things that go by the same name. I think I didn't spot the names you mentioned as being the same due to the different notation. I will look through the link, but skimming it looks like that will make a lot of sense. – Joe Tait May 11 '12 at 9:26

As for "by linearity": if you want to find $x$ that is $p$-adically close to $x_1$ and $q$-adically close to $x_2$, then you can do this as follows: suppose you can find a $y$ that is $p$-adically close to $x_1$ and $q$-adically close to $0$, and a $z$ which is $p$-adically close to 0 and $q$-adically close to $x_2$; then set $x = y + z$. That (or the $n$-variable version) is all that's meant.

For your other question, you can force an element not to be in an ideal by making it congruent to 1 modulo that ideal, for instance.

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Just to check, when you say "$p$-adically close to 0", in this context do you mean it has zero valuation at $p$? And similarly, by "y that is $p$-adically close to $x_1$" do you mean has the same valuation as $x_1$ at $p$? – Joe Tait May 11 '12 at 8:58
By "x is p-adically close to y" I mean "x - y has large p-adic valuation". – David Loeffler May 11 '12 at 12:40
@JoeTait Keep in mind that later on we'll define a norm on $K$ by $|x| = c^{v(x)}$ where $0<c<1$, so "small" will mean "large valuation". – Dylan Moreland May 11 '12 at 19:25
Thanks - after reading Pete Clarks comment it all makes a lot more sense - I think the order partly threw me off, so I mentally disassociated the norm and the valuations. Thank you for clarifying :) – Joe Tait May 11 '12 at 22:10

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