Definition of valuation ring:
Let $R$ be an integral domain with $frac(R)=K$. Then $R$ is said to be a valuation ring if

(1) $R \neq K$

(2) $\forall x \in K, x \in R$ or $x^{-1} \in R$.

Now my problem is:

Let $A, B$ be two valuation rings with $K=frac(A)=frac(B)$. Now suppose $A⊂B$, can we conclude that $A=B$?

I can prove that this is not true if $A$ is a discrete valuation ring:

For suppose the maximal ideal in $A$ is generated by $t$, then the only non units in $A$ are of the form $ut^{n}$, with $u$ a unit in $A$, $n$ an integer $\geq 1$.

If $A \subsetneq B$, then $B$ must contains the inverse of some element of the above form, i. e. $u^{-1}t^{-n}=(ut^n)^{-1} \in B$, but then $ \frac 1 t = (u t^{n-1})(u^{-1}t^{-n}) \in B$. Hence $B\supset A[\frac 1 t ]= K$.

However, I can't think of anything useful in the general case that $A, B$ may not be discrete.

Any help will be appreciated.

  • $\begingroup$ For DVR's see here. You can also find there a proof for the the claim in the answer below. $\endgroup$ – user26857 Mar 21 '16 at 11:57

If $A$ is a valuation ring with quotient field $K$, then any ring $B$ between $A$ and $K$ is a valuation ring.

  • $\begingroup$ Oh, so now the question becomes: If A is a valuation ring, can we find another ring C such that $A\subsetneq C \subsetneq K$, I believe it is not true, but I cannot think of an example $\endgroup$ – chan kifung Mar 21 '16 at 3:24
  • $\begingroup$ Do you know of any non-discrete valuation rings you could play around with? $\endgroup$ – D_S Mar 21 '16 at 3:45
  • $\begingroup$ Hint: there exist valuation rings with finite Krull dimension greater than one. Let $A$ be such a ring. Then there exists a prime ideal $\mathfrak p$ of $A$ which is not zero and which is not the maximal ideal of $A$. What can you do with $\mathfrak p$? $\endgroup$ – D_S Mar 21 '16 at 3:50
  • $\begingroup$ So we would localize $A$ at $\mathfrak p$, I have found on the Internet that the ring $k[x,x/y,x/y^2,\cdots ]_{(y)}$ is an example of valuation ring with krull dimension $>1$, thank you so much $\endgroup$ – chan kifung Mar 21 '16 at 4:23
  • $\begingroup$ You're welcome. Actually your question got me thinking about very similar question which I just asked (math.stackexchange.com/questions/1706740/…). I'm too tired to do it now, but maybe you can solve it before I can :) $\endgroup$ – D_S Mar 21 '16 at 4:28

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