From Dummit and Foote problem about the Discrete Valuation Ring $R$ with field $K$
Prove that for each nonzero element $x \in K$ that either $x$ or $x^{-1}$ is in $R$
My attempt:
$v(1) = v(x\cdot x^{-1}) = v(x) + v(x^{-1}).$ I show $v(1) = 0$ via $v(1) = v(1\cdot 1) = v(1) + v(1) \implies v(1) = 0$. So we have $v(x) = -v(x^{-1})$ Then we can conclude if either $v(x) > 0$ or $v(x^{-1}) > 0$ then the other can't be in $R$.
But my problem comes if $v(x) = 0$, then both $x, x^{-1} \in R$. What's wrong with this?
English is an awkward language for expressing logical connectives. Despite the use of "either ... or ...", the problem did not intend for $x \in R$ and $x^{-1} \in R$ to be exclusive; it merely wants you to show at least one of the two statements is true.