valuation ring is a field?

Question

suppose $a$ and $a'$ are units of $B$ ,$b$ and $b'$ are the elements of any ideal of $B$. $x$ is a element of $K$.

$K$ consist of $a/a,a/b,b/a,b/b$

$\color{green} x=a/a' \Rightarrow x\in B~and~x^{-1}\in B$

$\color{green} x=a/b \Rightarrow x^{-1}\in B $

$\color{green} x=b/a \Rightarrow x\in B $

$\color{red} x=b/b' \Rightarrow x\notin B~and~x^{-1}\notin B $

So there no ideal in $B$,if $B$ is a valuation ring

Something seems wrong.

Can someone fix it and tell me what is valuation rings?

Answer 1

Another good way to understand a valuation ring is "an integral domain whose ideals are linearly ordered."

As such, it has a unique maximal ideal. A trivial example would be a field F, which clearly has that property in its own field of fractions (which is itself.)

If it's not a field, then its unique maximal ideal is nonzero. One example of this is the ring of formal power series $F[[x]]$ inside its field of fractions $F((x))$, the Laurent series.

Given an arbitrary $f\in F((x))$, either $f$ or its inverse has to have lower degree greater than or equal to $0$, and no nonzero coefficients on negative powers of $x$.

Answer 2

I don't follow your reasoning. I think you may be arguing that $K$ has no (nonzero) ideals.

Here is a simple nondegenerate example: let $B$ the the ring of all rational numbers with odd denominator. It is a valuation ring of $\mathbf{Q}$.

$B$ is also a local ring, whose maximal ideal is the one generated by $2$; it is the set of all rational numbers with even numerator and odd denominator.