Mathematics » MA3G6 Commutative Algebra » 2012 Q4 part e
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The latest posts to Mathematics » MA3G6 Commutative Algebra » 2012 Q4 part een-GB(C) 2023 University of WarwickFri, 16 May 2014 14:11:16 GMThttp://blogs.law.harvard.edu/tech/rssSiteBuilder2, University of Warwick, http://go.warwick.ac.uk/sitebuilder2012 Q4 part e
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<p>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.</p>
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<p> </p>Fri, 16 May 2014 16:39:51 GMTJohn Cremona8a1785d781f6ae40018216a9c73f14352012 Q4 part e
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<p>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!</p>Fri, 16 May 2014 15:03:22 GMTA guest user8a1785d781f6ae40018216a9c73f14372012 Q4 part e
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<p>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.</p>
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<p>In (e) you do have to say why $P/P$ and $R_P$ are both integral domains (as the solution does).</p>
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<p> </p>Fri, 16 May 2014 14:55:54 GMTJohn Cremona8a1785d781f6ae40018216a9c73f14392012 Q4 part e
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<p>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.</p>Fri, 16 May 2014 14:51:25 GMTA guest user8a1785d781f6ae40018216a9c73f14362012 Q4 part e
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<p>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 <em>principal</em> ideals one contains the other. Part (d) of the question was to prove that for <em>any</em> 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).</p>
<p>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.</p>
<p>I hope I haven't made it less clear by saying too much!</p>
<p> </p>Fri, 16 May 2014 14:40:14 GMTJohn Cremona8a1785d781f6ae40018216a9c73f14382012 Q4 part e
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<p>Hi could you possibly explain the solution to question 4e of the 2012 paper, proving R/P is a valuation ring?</p>
<p>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.</p>
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<p>Cheers</p>Fri, 16 May 2014 14:11:16 GMTA guest user8a1785d781f6ae40018216a9c73f143a