## Valuations

One way to use algebraic methods in number theory is via ideal theory. That’s how the discipline began, and that’s how I’ve learned it. But for more advanced things, mathematicians need other approaches – most elementarily, valuation theory (which many textbooks make the annoying mistake of beginning with, without talking about ideal theory first).

A valuation on a field K is just a function from K to the real numbers that satisfies the usual conditions of absolute values:

1. For all a and b in K, |ab| = |a||b|

2. |a| >= 0, with equality iff a = 0

3. |a + b| <= |a| + |b|

The most obvious example of a valuation is the normal absolute value on Q, which motivates the abstract definition. There are others, though – for example, instead of taking |a|, we can take |a|^k, where k is any constant between 0 and 1. It’s not hard to see that all three conditions are satisfied. But this also holds for every valuation, so it’s in a sense trivial.

For a less trivial example, we can take |.|(p), the p-adic valuation. That valuation is defined by factoring a/b uniquely as (p^n)c/d, where n is any integer and c and d are not divisible by p; then |a/b|(p) is defined to be p^(-n). Conditions 1 and 2 are obviously satisfied. Instead of condition 3, look at condition 3′, which is stronger:

3′. |a + b| <= max{|a|, |b|}

By multiplying by a suitable integer n, we can make a and b integers, so it’s enough to prove the condition for all a and b in Z. If a = cp^n and b = dp^m where c and d are not divisible by p and n >= m, then a + b = (d + cp^(nm))p^m, so it’s enough to prove that |d + cp^(nm)| <= 1. But d + cp^(nm) is an integer, and every integer can be written as cp^n‘ with n‘ >= 0, and |cp^n‘| = p^(-n‘) <= 1.

In fact, if K is a number field, then every valuation satisfying condition 3’ arises from some prime ideal P – that is, if a is in O(K) then |a|(P) is some positive constant c < 1, to the power of the largest power of P that a is contained in; the valuation can then be extended to K via |a/b| = |a|/|b|.

To see why, first note that |a| <= 1 for all a in O(K); if a is a positive integer, then |a| <= max{|a-1|, 1} <= max{|a-2|, 1} <= … <= 1, while if a is a negative integer, then |a| = |-a| since |-1| is the unique positive square root of |1| = 1. Now, if a is merely an algebraic integer, then for some n we have a^n + c(n-1)a^(n-1) + … + c0 = 0, so |a^n| = |c(n-1)a^(n-1) + … + c0| <= max{|c(n-1)a^(n-1)|, …, |c0|} <= max{|a^(n-1)|, …, |a|, 1} which is impossible if |a| > 1.

Now, look at the set of all elements such that |a| < 1. By condition 3′, |a1 + a2| < 1 if |a1| < 1 and |a2| < 1; by condition 1, if |a| < 1 and b is in O(K), then |ab| < 1; by condition 1 again, if |b1| = |b2| = 1, then |b1b2| = 1. Therefore the set is a prime ideal of O(K), say P. Even if P is not principal, for every two elements a1 and a2 in P that are not in P^2 there exist some r1 and r2 not in P with a1r1 = a2r2 (for example, if K = Q(SQRT(-5)), P = (2, 1 + SQRT(-5)), a1 = 1 + SQRT(-5), and a2 = 1 – SQRT(-5), then let r1 = 3, r2 = -2 + SQRT(-5)). So |a| is a constant c for all a in P but not in P^2; similarly, |a| = c^2 for a in P^2 but not in P^3, etc.

Finally, note that P is allowed to be the zero ideal, i.e. |a| = 1 for all a != 0. In that case the valuation is called trivial, since every field has it (in particular, if a field is finite, then it only has the trivial valuation).

Valuations that don’t satisfy 3′ are a lot harder to compute from first principles, but using a theorem of Ostrowski that has an annoyingly tedious proof, it’s possible to show that every valuation on Q that doesn’t satisfy 3′ is some power of the real absolute value, and that every valuation on a number field K that doesn’t satisfy 3′ is some power of the real or complex absolute value arising from one of the ring homomorphisms from K to C.

Therefore, if we treat valuations that are powers of each other as identical, there is one valuation for each prime of O(K), and one for each homomorphism into C up to complex conjugation. There is more: considering complex-conjugate valuations to be different, we have |w1(a)|…|w(n)(a)| = |w1(a)…w(n)(a)| = |N(a)| and, letting c in |.|(P) be 1/N(P), we have |a|(P1)*|a|(P2)*… = 1/|N(a)| (the product is finite since |a|(P(i)) is 1 for all prime ideals except those containing a, which are finite in number). So in some sense the product of all valuations on K is always 1.

The same holds for function fields. Condition 3′ is always satisfied for function fields, since if |a + b| > |a| >= |b|, then dividing by a, we get |1 + b‘| > 1 >= |b‘|; we can take it to the pth power, which will yield (1 + b‘)^p = 1 + b‘^p because all other terms are divisible by p so they get killed off. Then |1 + b‘^p| = |1 + b‘|^p > 1; after sufficiently many iterations of taking pth powers, we get |1 + b‘^(p^m)| > 2 = 1 + 1 >= 1 + |b‘^(p^m)|, a contradiction.

The problematic condition is the one that says that on O(K) the valuation is at most 1. If it’s true for a given valuation of O(K), then the same proof applies and it’s generated by some prime ideal of K. But it’s true iff |x| <= 1, where O(K) = F(p)[x]: the only if direction is clear; for the if direction, every polynomial in x satisfies |a(n)x^n + … + a0| <= max{|a(i)x^i|} = max{|x^i|} <= 1 since a(i) is in F(p), on which every valuation is trivial. If O(K) is the integral closure of F(p)[x] in some finite extension of F(p)(x), then the same argument as for number fields applies.

If |x| > 1, then |a(n)x^n + … + a0| <= max{|a(i)x^i|} = max{|x^i|} = |x|^n, and equality holds throughout or else |x|^n = |x^n| <= max{|a(n-1)x^(n-1) + … + a0|, |a(n)x^n + … + a0|} <= max{|x|^(n-1), |a(n)x^n + … + a0|} < |x|^n. At least in F(p)[x], then, a similar formula to the one in number fields holds, since if P = (f(x)) and we take c to be p^(-deg(f)) then the product of |g|(P) over all primes P is p^(-deg(g)), while we can take |x| = p in the degree valuation and make the product of all valuations 1, again.