valuation ring is a field? 
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?
 A: 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.
A: 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 $x$.
