The exercise is the following:
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 $A$.
Does anyone know what is meant by "to be a local ring of a valuation ring"?
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Sign up to join this communityThe exercise is the following:
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 $A$.
Does anyone know what is meant by "to be a 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 $B=A_P$.
I agree with TG's answer. I think more specifically it means it's a localization at a prime ideal. If you think of $A$ as an affine scheme, then the phrase "local ring of $A$" sounds more familiar.