## 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.