## Valuation Rings, Induced Metric

### Induced Metric

In an earlier section we placed a topology on the valuation group G. In this section we will place a topology on the field F. In fact F becomes a metric space. All we need do is define a valid metric.

This process assumes the valuation group G can be embedded in the reals. This is usually the case, since G is linearly ordered. It certainly holds when G = Z.

Let c be any real number between 0 and 1, and establish the following metric.

|x,y| = cr(v(x-y))

In other words, subtract x and y, find the valuation of the difference, map that to a real number, and raise c to that power. If the difference is 0, let the metric equal 0. Otherwise the metric will be positive.

As you can see, |x,y| = 0 iff x = y. And since the valuation does not depend on the sign, |x,y| = |y,x|. We only need prove the triangular inequality.

Let x y and z be elements of the field F. We want to show |x,z| ≤ |x,y| + |y,z|. Verify by hand that this is true when any two of the three variables are equal.

Note that z-x = z-y + y-x. Use the property of sums to show that v(z-x) is at least as large as the lesser of v(z-y) and v(y-x). Since c is less than 1, larger valuations lead to smaller metrics. Thus the metric on the left is bounded by one of the metrics on the right. It is certainly bounded by the sum of the metrics on the right, and that proves the triangular inequality. We have a valid metric space.

In this space, every triangle is isosceles. If z-y and y-x have different valuations, then their sum, z-x, has the lesser of the two valuations. Two of the three lengths are always the same.

As usual, a circle is the locus of points a fixed distance from a given center. The unit circle is the elements of F with metric 1, having valuation 0. These are the units of R. The unit disk is all of R.

Now consider any circle with center c and radius t. Let p be a point inside the circle and let q be any point on the circle. Draw the triangle cpq. We know that the distance from c to p is less than the distance from c to q. That's what it means to be "inside" the circle. So cq has a smaller valuation. The valuation of the sum, from p to q, has to equal this lesser valuation. Thus the distance pq is the same as the distance cq. The conclusion: every point inside a circle is at the center of the circle.

Return to our favorite example, the ring of fractions without p in the denominator. As you recall, the valuation is the exponent on p, which embeds nicely into the reals. Hence the metric is well defined. Set c = ½ for illustration. The points that are ½ distance from the origin have a factor of p in the numerator. Points that are ¼ distance from the origin have a factor of p2 in the numerator, and so on. Points like p100, which use to be far away under the traditional distance metric, are just a hair's breath from the origin. Conversely, the point p-100 has been hurled light-years away.

The rationals have definitely been rearranged, but the result is still a metric space. This is called the p-adic topology on the rationals.

### Continuous Operators

Addition is a continuous map from F cross F into F. This makes sense in the p-adic topology. If you change x and y by adding high powers of p, their sum will differ from the original by a high power of p. Numbers close to x and y add up to something close to x+y. Let's prove this in general.

Let v be any valuation that is larger than the valuation of x or y. Add s to x and t to y, where s and t have valuation at least v. This gives x+y+(s+t). The valuation of s+t is at least v, so (x+s)+(y+t) is within ε of x+y, as long as s and t are less than ε.

Multiplication is also continuous. Consider the valuation of (x+s)×(y+t)-xy. This is at least the valuation of xt or the valuation of ys or the valuation of st. Since s is under our control, make sure its valuation is at least v - the valuation of y. Thus the valuation of ys is at least v. Do the same for t, and the valuation of xt is at least v. Finally, make sure s has a valuation at least v, and t has a valuation at least 0. Now st has a valuation at least v, and the same is true of the sum. When the factors differ by s and t, where s and t are less than ε, the product is within ε of xy.

Next look at the inverse map 1/x. If x is changed by s, look at the difference between 1/x and 1/(x+s). This is s over x*(x+s). Select s so that its valuation is higher than x. The denominator has the same valuation as x2, which is twice the valuation of x. Add v to this, and make sure s has an even higher valuation. Now the valuation of s/x2 is at least v, and we are within ε of 1/x. Obviously this fails when x = 0.

Put this together and division is a continuous operator from F cross F into F, provided the divisor is not 0.