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Let $O$ be a discrete valuation ring with valuation $v$. We normalize by $v(\pi) =1$, with $\pi$ a prime element in $O$.

By definition, for all $x,y$ in $O$, we have $v(x+y) \geq \min (v(x),v(y))$. For some reason, the texts (Neukirch and Serre) suggest that one has equality when $v(x) \neq v(y)$. Where does this come from?

Does one also have an upper bound for $v(x+y)$?

I actually have elements $x_1,\ldots,x_n$ and I want to show that $v(x_1+\ldots+x_n) = \min(v(x_i))$. Does it suffice to show that $v(x_i) \neq v(x_j)$ for all $i\neq j$?

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Suppose $v(x)\ne v(y)$, and without loss of generality assume $v(x) > v(y)$, so that $v(x+y) \ge v(y)$ and you're trying to prove $v(x+y) = v(y)$. Apply the ultrametric inequality to $y = (x+y)-x$ to obtain $v(y) \ge \min\{v(x+y),v(x)\}$, which allows you to conclude that $v(y) \ge v(x+y)$.

(The intuitive explanation is: if $y$ is a multiple of 4 (but not 8) and $x$ is a multiple of 32, then $x+y$ is a multiple of 4 but definitely not a multiple of 8. If it were a multiple of 8, then $y=(x+y)-x$ would also be a multiple of 8.)

There's no upper bound for $v(x+y)$ as the extreme example $x=-y$ shows. Alternatively, take $x=\pi^k-y$, which for $k>v(y)$ has $v(x)=v(y)$ by your first question, yet $v(x+y)=k$ can be arbitrarily large.

To conclude that $v(x_1+\cdots+x_n) = \min\{v(x_1),\dots,v(x_n)\}$, your condition suffices by your first question and induction; in fact, even a weaker condition suffices, namely that there is exactly one $i$ such that $v(x_i) = \min\{v(x_1),\dots,v(x_n)\}$.

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    Ahh I forgot that $v(-x) = v(x)$. I was stuck on that (silly) point and didn't succeed at deducing the lower bound for $v(y)$. Thnx for clarifying things for me and your last remark on the weaker condition! That's going to be helpful probably.2012-01-24