The field R has exactly one proper algebraic extension, C. Further, C has exactly one valuation extending the valuation on R, and in fact |z| = |Re(z) + iIm(z)| = SQRT(Re(z)^2 + Im(z)^2) = SQRT(N(z)). Although elements of C that are transcendental over Q have infinitely many conjugates, here we take the norm to refer to the norm with respect to R, i.e. the product of all conjugates fixing R. In that case, an element of C has at most 2 = [C:R] conjugates, and everything works normally.
This process is generalizable to local fields: if K is a local field, and L is any finite extension of K, then there exists a unique extension of the valuation on K to L, defined by |a| = |N(a)|^1/[L:K]. This function is clearly multiplicative, and satisfies |a| >= 0, equality iff a = 0.
In fact, it’s also independent of the choice of L, in the sense that a has the same valuation in every extension of K(a). In K(a), N(a) is always (-1)^[K(a):K] times the constant term of the minimal polynomial of a over K. In an arbitrary finite extension L, the distinct K-conjugates of a are still the roots of its minimal polynomial over K, and the ring homomorphisms from L to a suitable bigger field – C in the number field case – are such that every root occurs exactly [L:K(a)] times. So in fact, |a| = |N(L:K)(a)|^1/[L:K] = |N(K(a):K)(a)|^[L:K(a)]/[L:K] = |N(K(a):K)(a)|^1/[K(a):K] which is independent of L.
Now, to show that this defines a valuation, it remains to be shown that the triangle inequality holds. In fact since local fields always satisfy the ultrametric inequality, |a + b| <= max{|a|, |b|}, I’ll show the stronger statement that this extension satisfies the ultrametric inequality.
By division, it’s enough to show that |1 + a| <= max{1, |a|}. Clearly, K(a) = K(1 + a), and the minimal polynomial of 1 + a over K is just m(a)(x – 1), where m(a)(x) is the minimal polynomial of a. If you think about this right, the constant term in m(a)(x – 1) is (-1)^n + (-1)^(n – 1)*a(n – 1) + … – a1 + a0 where n = deg(m(a)) and a(i) is the x^i coefficient of m(a)(x), with a0 = N(a). But then |1 + a| = |(-1)^n + (-1)^(n – 1)*a(n – 1) + … – a1 + a0|^1/n <= max{1, |a(n – 1)|, …, |a0|}^1/n = max{1, a0}^1/n since m(a) is irreducible over K. But max{1, a0}^1/n = max{1, |a|}, proving that the extension satisfies the ultrametric inequality.
For uniqueness, I’m going to assume a certain result from analysis, namely that all norms on K^n are equivalent. A norm on a vector space is similar to a valuation, except that |a||b| = |ab| is replaced with a similar condition for scalar multiplication, i.e. |kv| = |k||v| where |.| on K is a fixed valuation.
There are various conditions for norms to be equivalent, which for finite-dimensional vector spaces over complete fields turn out to be the same. The most general one is that the open sets are the same in both valuations, i.e. every D1(x, e) = {y in K^n: |x – y|1 < e} contains some D2(x, d) = {y in K^n: |x – y|2 < d}. It’s pretty easy to show that it’s always equivalent to the condition that a sequence converges to x with respect to |.|1 iff it converges to x with respect to |.|2, so let’s use that.
Given two valuations on L, we have equivalence, i.e. sequences converge to the same limits under both valuations. That means that |a|1 < 1 iff (a, a^2, a^3, …) converges to 0 in |.|1, iff it converges to 0 in |.|2, iff |a|2 < 1. By taking multiplicative inverses, |a|1 > 1 iff |a|2 > 1.
Now, suppose that |a|1, |b|1 > 1, and let |b|1 = (|a|1)^r, |b|2 = (|a|2)^s. Then r = s. Otherwise, suppose r < s, and note that we can find integers p and q with r < p/q < s, so that qr – p < 0 < qs – p; then |b^q/a^p|1 = (|a|1)^(qr – p) < 1 and |b^q/a^p|2 = (|a|2)^(qs – p) > 1, a contradiction. If |a|2 = (|a|1)^c, then |b|2 = (|a|1)^rc = (|b|1)^c, i.e. |.|2 is a fixed positive real power of |.|1.
Going back to L, if |.|1 and |.|2 are two valuations on L, then they’re equivalent, which means |.|2 = (|.|1)^c for some c > 0. But |.|1 and |.|2 agree on K, so c = 1 and |.| is unique. Also, the unique class of equivalent norms on the vector space K^n is complete, which makes the extended valuation constructed in this post complete.
Most of the above works even if L is merely algebraic rather than finite. Uniqueness relies on the fact that although L may not be a finite extension, K(a) is a finite extension for every a in L, and L is the union of all K(a) with a in L. However, this piecemeal approach does not show completeness.
Indeed, if on K, the valuation is of the form c^n, where c is a fixed real number other than 1 and n is an integer, then every element algebraic over K will have valuation c^d where d is rational. By taking roots of an element of valuation c, every such c^d arises as a valuation of some element in the algebraic closure. But then the image of the function |.| from L to R is not complete, so L isn’t complete.
(Note: I’ll prove either of my analysis handwaves – the one about equivalent valuations on K^n, and the one about complete valued fields mapping to complete spaces under continuous functions – if anyone asks me to).
Abstract Nonsense 