My question appeared while I was solving a problem in Abstract Algebra of Dummit and Foote. That is problem 26.
My solution is almost the same as the proof in this link.
However, I'm still confused with the solution because it does not use the surjectivity hypothesis of discrete valuation. Maybe that hypothesis is necessary for some further applications of discrete valuations?
Thanks for any help.
On
If $v$ is a non-surjective valuation, then so is any scalar multiple of $v$. Also $v$ is non-surjective iff $v^{-1}(1)$ doesn't exist, in which case we can take $k=\min\{n: n\in v(R)\}$, and it's easy to check that $v(K)\subseteq k\mathbb Z$.
This is because if $v(a)=k$ and we have $x\in R$ with $v(x)=n$, then use division algorithm to find $q,r$ with $n=kq+r$. Then $v(\frac x{a^q})=r<k$ implying $v(\frac x{a^q})=0$ ie $(x)=(a^q)$.
So the surjectivity condition just ensures a pre-image of $1$.
Surjectivity is basically to ensure that $v$ is not trivial, that is $v(a)\ne0$, for some $a\in K^\times$.
If all properties are satisfied, except (ii) replaced by the hypothesis above, the image of $v$ is a non trivial subgroup of $\mathbb{Z}$, so it is of the form $k\mathbb{Z}$, for a unique $k>0$. Then the map $$ v'\colon K^\times\to\mathbb{Z} $$ defined by $v'(a)=v(a)/k$ has the same properties and is surjective. The valuation ring is the same.
Note that a valuation ring in $K$ (that is, a subring $R$ such that for $a\in K^\times$ either $a\in R$ or $a^{-1}\in R$) determines a valuation on the group $K^\times/D^\times$ that can be totally ordered. The valuation is discrete if and only if $K^\times/D^\times$ is isomorphic to the integers. So assuming $v$ surjective is not really a restriction.