User lennart galinat - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T13:15:48Z http://mathoverflow.net/feeds/user/9710 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/40713/an-elementary-lemma-of-commutative-algebra/40746#40746 Answer by Lennart Galinat for An elementary lemma of commutative algebra Lennart Galinat 2010-10-01T13:45:12Z 2010-10-01T14:08:02Z <p>If you want to avoid the use of Ext-groups, you could prove it like this (which is basically the same proof):</p> <p>Let 0→A→B→C→0 be a short exact sequence of R-modules with C finitely presented and assume it splits after localisation at every maximal ideal.</p> <p>Use the natural isomorphism Hom$_{R_m}(C_m,A_m)$=R$_m$⊗Hom$_R$(C,A) (which uses the flatness of localisation and the finitely presentedness of C) and the assumption to see that the map Hom$_R$(C,B)→Hom$_R$(C,C) is surjective since all of its localisations are.</p> <p>Since the giving a splitting is equivalent to this map being surjective we are done.</p> <p>P.S.: Of course one could prove "if and only if" in the statement like this.</p> <p>Counterexample in the general case: Take R=$\prod_{\mathbb N} \mathbb F_2$, I=$\sum_{\mathbb N} \mathbb F_2$ and C the cokernel of the inclusion.</p> <p>Now I claim three things: 1.) R has dimension zero</p> <p>Proof: Every element of R is an idempotent and every prime ideal in R has to contain exactly one of e or 1-e for every idempotent e in R. If we had a chain of prime ideals, the larger one would nescessarily have to contain e and 1-e for one such idempotent and so couldn't exist.</p> <p>2.) The localisation of R at every maximal ideal is a field.</p> <p>Proof: It is a zero-dimensional local ring by 1.) and reduced since R doesn't contain nilpotent elements. Thus this localisation is a field.</p> <p>3.) I is not a direct summand of R.</p> <p>Proof: Direct Summands correspond to idempotents in R. Since every element in R is idempotent we need to analyze all principal ideals. If an element has only finitely many non-zero entries the ideal created by it has only finitely many elements, thus can't be I. If on the other hand it contains infinitely many non-zero entries, the ideal created by it has uncountably many elements. Since I contains countably many elements, it can't be a direct summand of R.</p> <p>This establishes the counterexample, since the short exact sequence 0→I→R→C→0 splits in every localisation at a maximal ideal.</p> http://mathoverflow.net/questions/109544/is-being-principal-a-local-property/109545#109545 Comment by Lennart Galinat Lennart Galinat 2012-10-13T18:19:42Z 2012-10-13T18:19:42Z To help you with your problem: There are two steps one needs to consider (whose proof I leave to you). 1.) In a UFD any prime ideal of height one is principal. 2.) An ideal maximal among non-principal ideals is automatically a prime ideal. http://mathoverflow.net/questions/78423/clean-proofs-of-properties-of-projective-space/78496#78496 Comment by Lennart Galinat Lennart Galinat 2011-10-19T04:58:30Z 2011-10-19T04:58:30Z @Anton: For a reduced algebraic variety over a field one can show by hand that its global sections are a finite dimensional vector space, Liu does it this way in his book. Then the fact that P^n is geometrically integral shows that its global sections are a one-dimensional. http://mathoverflow.net/questions/59981/global-dimension-and-localization Comment by Lennart Galinat Lennart Galinat 2011-03-29T16:41:20Z 2011-03-29T16:41:20Z @Martin: Global dimension can be measured by only considering cyclic R-modules in the first variable of Ext and simple R-modules in the second. But then we can swap localisations and Ext and the result is clear. http://mathoverflow.net/questions/59981/global-dimension-and-localization Comment by Lennart Galinat Lennart Galinat 2011-03-29T15:59:00Z 2011-03-29T15:59:00Z I don't know if this is what you're looking for, but being noetherian guarantees the result you ask for. http://mathoverflow.net/questions/48865/if-a-map-restricts-to-an-isomorphism-on-a-closed-subscheme-and-its-open-complemen/48885#48885 Comment by Lennart Galinat Lennart Galinat 2010-12-10T23:08:31Z 2010-12-10T23:08:31Z Thinking about the injectivity of the map, I realised that the nilradical is the problem. For example, if K is a field, R=K[X]/(X^3) and S=K[X]/(X^2) with the canonical projection as the map from R to S, then R/(X) is isomorphic to S/(X) via this map and all localisations at elements inside (X) are isomorphic, since they are zero because we are inverting nilpotent elements. But clearly R and S are not isomorphic.