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*) + *i*Im(*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) + … – *a*1 + *a*0 where *n* = deg(*m*(*a*)) and *a*(*i*) is the *x*^*i* coefficient of *m*(*a*)(*x*), with *a*0 = N(*a*). But then |1 + *a*| = |(-1)^*n* + (-1)^(*n* – 1)**a*(*n* – 1) + … – *a*1 + *a*0|^1/*n* <= max{1, |*a*(*n* – 1)|, …, |*a*0|}^1/*n* = max{1, *a*0}^1/*n* since *m*(*a*) is irreducible over K. But max{1, *a*0}^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).