5
$\begingroup$

I understand the mechanics of the proof of Ostrowski's Theorem, but I'm a little unclear on why one should expect valuations to be related to primes. Is this a special property of number fields and function fields, or do primes of K[x,y] correspond to valuations on K(x,y) in the same way?

I'm hoping for an answer that can explain what exactly are the algebraic analogs of archimedian valuations, and how to use them - for example, I've heard that the infinite place on K(x) corresponds to the "prime (1/x)" - how does one take a polynomial in K[x] "mod (1/x)" rigorously?

Thanks in advance.

  • 2
    Given a prime p, the p-adic valuation is, well, a valuation. The intuition you should have is that the p-adic valuation tells you the "order of the pole" (or zero) at p, which is exactly what it does when the prime is a prime of a finite integral extension of C[x].2010-07-29
  • 1
    As for the prime 1/x, the problem here is that unlike a number field, a function field doesn't have a distinguished choice of "ring of integer." Making such a choice corresponds to focusing attention on an affine open, and when you choose K[x] you focus attention away from 1/x. So you should choose K[1/x] instead.2010-07-29
  • 0
    I'm a little confused - the order of the pole/zero of what function? The first part of your second comment, that's an interesting way of thinking about it - thanks! But in extensions of K(x), there might be more than one "infinite valuation", right? How do we find the analog of (1/x) for them?2010-07-29
  • 1
    The element you're taking a valuation of. In the function field case, this is a meromorphic function on a Riemann surface, and this is a good intuition about number fields as well. As for the second question, write the extension as K(x, t)/f(x, t) for some f and take the homogenization of f; the points at infinity are the infinite valuations.2010-07-29
  • 0
    You might find it helpful to see how valuations are employed in Hardy-Rosenlicht fields to measure "orders of infinity" in algebraic approaches to asymptotics ("transseries"). For example, see Shackell, Rosenlicht Fields, TAMS, 355, 2, 1993, 829-836. http://www.jstor.org/stable/19991102010-07-29

2 Answers 2

3

I couldn't divine much information on your background (e.g. undergraduate, master's level, PhD student...) from the question, but I recently taught an intermediate level graduate course which had a unit on valuation theory. Sections 1.6 through 1.8 of

http://math.uga.edu/~pete/8410Chapter1.pdf

address your questions. In particular, if your field $K$ is the fraction field of a Dedekind domain $R$, then you can always use each prime ideal $\mathfrak{p}$ of $R$ to define a valuation on $K$, essentially the "order at $\mathfrak{p}$". There is also a converse result, Theorem 13: if you have a valuation on $K$ which has the additional property that it is non-negative at every element of the Dedekind domain $R$, then it has to be (up to equivalence) the $\mathfrak{p}$-adic valuation for some $\mathfrak{p}$. I felt the need to give this additional condition a name, so I called such a valuation R-regular.

The point is that (as Qiaochu says in his comments), in case $K$ is a number field and $R$ is its ring of integers, every valuation on $K$ is $R$-regular. However, in the function field setting this is not true and this leads to a discussion of "infinite places". Note that I do describe the analogues of Ostrowski's Theorem for finite extensions both of $\mathbb{Q}$ and of $F(t)$ for any field $F$ (in the latter case, one restricts to valuations which are trivial on $F$; when $F$ is finite, this condition is automatic).

I would be interested to know whether you find the notes helpful. If not, I or someone else can probably recommend an alternative reference.

3

Discrete valuations <-> points on a curve

For a nonsingular projective curve over an algebraically closed field, there is a one-one correspondence between the points on it, and the discrete valuations of the function field (i.e. all the meromorphic functions of the curve). The correspondence is point P -> the valuation that sends a function f, to the order of zero/pole of f at P.

Maximal ideals <-> points on a curve

At least for varieties (common zeros of several polynomials) over an algebraically closed field, there is a one-one correspondence between points on it, and the maximal ideal in $k[x_1,\cdots,x_n]$. The correspondence is point $P = (a_1,\cdots,a_n)$ -> the polynomials vanishing at P, which turns out to be $(x_1-a_1,\cdots,x_n-a_n)$. This is something true not only for curves, but for varieties. (Hilbert's Nullstellensatz)

So putting these together, for nonsingular projective curves over an algebraically closed field, you know that there is a one-one correspondence between the maximal ideals (think them as points) and the discrete valuations of the function field. Now the situation here is analogous. You consider a "curve", whose coordinate ring is $\mathbb{Z}$, with function field $\mathbb{Q}$. The nonarchimedean valuations correspond to discrete valuations in this case. So they should capture order of zeros/poles at some "points". What are the points? They should correspond to the maximal ideals of $\mathbb{Z}$, which are exactly the primes here.

As for $K(x)$, look at it as the function field of $K\mathbb{P}^1$. Just like the usual real/complex projective spaces, you should have two pieces here. Let's say $K[x]$ corresponds to the piece where the second coordinate is nonzero. So the corresponding homogeneous coordinates here is like $[x,1]$. We know there is one point missing, which is $[1,0]$. For this, we change our coordinates $[x,1] \to [1,1/x]$, so the piece where the first coordinate is nonzero should be $K[1/x]$. The missing point corresponds to the ideal $(1/x - 0) = (1/x)$, so this is why the infinite place corresponds to (1/x). Of course, a more straight forward interpretation is that for a rational function, you divide both numerator and denominator sufficiently high power of $x$ so that they both become polynomials in 1/x, have nonzero constant term, with an extra term (x to the some power). The infinite place measures this power.