MA3G6 Commutative Algebra 2012 Q4 part e

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1. A guest user

   #1 15:11, Fri 16 May 2014

   Hi could you possibly explain the solution to question 4e of the 2012 paper, proving R/P is a valuation ring? I understand the correspondence that I/P is contained in J/P if and only if I is contained in J. So if we show that for all I and J containing P that I contains J or vice versa then we must have by (d) that R/P is a valuation ring. But the solution then just seems to say that this must be true for all such I and J in a valuation ring R with prime ideal P and I don't understand why. Cheers
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2. John Cremona Staff Maths

   #2 15:40, Fri 16 May 2014

   Isn't that exactly what part (d) tells you? The original definition of valuation ring is that for any two elements $x,y$ either $x$ divides $y$ or vice versa, so that either $(x)\supseteq(y)$ or vice versa, or in other words that any for two nonzero/principal/ ideals one contains the other. Part (d) of the question was to prove that for/any/ two ideals $I$ amd $J$, one must contain the other, and also prove the converse, that this implies that $R$ is a valuation ring. (The converse was the easier part, since if something is true for any two nonzero ideals it is certainly true for two principal ideals). In words one could say that the result of part (d) is that an integral domain is a valuation ring if and only if the lattice of nonzero ideals is "linearly ordered" by inclusion. Passing from $R$ to $R/P$ or to $R_P$ we cuts down from the lattice of ideals of the new ring in a way which preserves this ordering by inclusion, hence the result. I hope I haven't made it less clear by saying too much!
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3. A guest user

   #3 15:51, Fri 16 May 2014

   I am still a bit confused, because in part (d) did we not have to use that R was an integral domain? Whereas in part (e) we don't necessarily have that R is an integral domain.
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4. John Cremona Staff Maths

   #4 15:55, Fri 16 May 2014

   Yes we did (use that $R$ is an integral domain in part (d)). It refers throughout to the field of fractions, which is only defined when $R$ is an integral domain. The question not how to characterise valuation rings amongst all rings, just how to characterise them amongst integral domains. In (e) you do have to say why $P/P$ and $R_P$ are both integral domains (as the solution does).
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5. A guest user

   #5 16:03, Fri 16 May 2014

   Oh! I had just completely neglected that for R to even be a valuation ring it must be an integral domain, hence the confusion when I got to part (e). Sorry that it wasn't obvious which bit I got wrong but thanks for pointing it out!
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6. John Cremona Staff Maths

   #6 17:39, Fri 16 May 2014

   I'm glad it is now sorted. Thanks for posting this on the forum, I'm sure there will be others who will find this useful.
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