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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asked Nov 16, 2009 at 11:51
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3 Answers
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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 $B=A_P$.
answered Nov 16, 2009 at 14:41
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4$\begingroup$ A minor comment: as someone who is teaching from Atiyah-Macdonald at the moment (and setting exercises from it for homework), I think there might be something to be said for not working out exercises in full "in public". $\endgroup$– TSGCommented Nov 16, 2009 at 21:38
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1$\begingroup$ Dear TG: you are absolutely right, and in principle I wouldn't solve an exercise from a textbook in public. I made an exception here because non-discrete valuation rings are often seen as esoteric creatures and I have never heard of a course in commutative algebra in which they are taught: I would be very happy if you decided to be an exception! Incidentally this is rather strange and might be due to Grothendieck's hatred of valuation rings, which is alleged to be one of the causes of his leaving Bourbaki. Se non è vero... $\endgroup$ Commented Nov 16, 2009 at 22:45
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$\begingroup$ This is well-explained in Matsumura's Commutative Ring Theory (not to be confused with Commutative Algebra), section 10, until theorem 10.1 (ii). $\endgroup$ Commented Sep 20, 2023 at 12:10
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I believe it just means "a localisation of A".
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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.
