## Valuation Rings, x+y

### x+y

Let x and y be nonzero elements of the field F. How does the valuation of x+y compare with the valuation of x or y? If x ≤ x+y, then 1+y/x belongs to R. Similarly, if y ≤ x+y than 1+x/y belongs to R. We know that R contains either x/y or y/x, and it certainly contains 1, so either x or y is less than x+y. The valuation of the sum is at least as large as the lesser valuation.

Assume the valuation of x is strictly greater than the valuation of y. This means 1+x/y is in R, but not 1+y/x. Hence y ≤ x+y, but not so for x.

Since R does not contain 1+y/x it contains the inverse, x/(x+y). It also contains 1, hence R contains y/(x+y). The sum is both ≤ and ≥ y, hence it equals y. If x and y have different valuations their sum has the lesser valuation.

Recall the example of the last section, fractions without p in the denomiinator. The valuation is the exponent of p. Sure enough, p2 + p4 has exactly two powers of p in its factorization. The valuation of the sum is 2. When the operands have the same valuation the sum could have the same valuation or larger. Add p+p to get the same valuation, or p + (p-1)p to get a larger valuation.