The exercise is the following:



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> Let $A$ be a valuation ring, $K$ its field of fractions. Show that every subring of $K$ which contains $A$ is a local ring of $A$.
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Does anyone know what is meant by "to be a local ring of a valuation ring"?



Atiyah-MacDonald: exercise 5.29 - "local ring of a valuation ring"
Let B be the subring of K containing A. I am going to prove that B is the localization of A at a prime ideal $P\subset A$, which seems a reasonable interpretation of the statement that B is a local ring of A.


First B is a local ring [by Atiyah-MacDonald Prop 5.18 i)] ; let $M\_B$ be its maximal ideal.
Similarly let $M\_A$ be the maximal ideal of A. 
Define $P=A\cap M\_B$. Then of course $P\subset M\_A$ is prime and if we localize A at P, I claim that we get B:


a) If $\pi \in A-P$ then $\pi \notin M\_B$ and so $\pi$ is invertible in B. Hence for all $ a \in A$ we have $a/ \pi \in B$.
This shows $A\_P \subset B$. 


b) Now we see that B dominates $A\_P$: this means we have an inclusion of local rings $A\_P \subset B$ and of their maximal ideals: $PA\_P \subset M\_B$. But $A\_P$ is a valuation ring of K because A is (This is the preceding exercise 28 : these guys know where they are heading!)
and valuations rings are maximal for the relation of domination (Exercise 27: ditto !). 
Hence we have the claimed equality $B=A\_P$.


