Construction of a maximal ideal - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T20:39:19Z http://mathoverflow.net/feeds/question/50680 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/50680/construction-of-a-maximal-ideal Construction of a maximal ideal unknown (google) 2010-12-29T21:48:23Z 2013-05-06T21:33:21Z <p>Hello,</p> <p>Let R denote the ring of continuous functions defined on the real line, let I in R be the ideal consisting of functions with compact support. Obviously, I is not maximal, and by Zorn's Lemma there exists a maximal ideal M in R which contains I. Is there now an explicit construction or characterization for M, i.e. when given any function in R, can one decide whether it lies in M or not?</p> http://mathoverflow.net/questions/50680/construction-of-a-maximal-ideal/50967#50967 Answer by Łukasz Grabowski for Construction of a maximal ideal Łukasz Grabowski 2011-01-02T23:23:01Z 2011-01-02T23:23:01Z <p>This is an expansion of the comment of Qiaochu Yuan. </p> <p>As mentioned in the comments there can be no "constructive" description. However, maybe you'll find this (tautological) construction useful: take any sequence $m_n$, $n=1,2,\ldots$ of real points which diverges to infinity. Choose an ultrafilter $U$, and define $M$ to be the set of those functions $f$ for which there exists an element $K$ of $U$ (so in particular $K$ is a subset of natural numbers) such that $f(m_k)=0$ for $k\in K$.</p> <p>It's obvious that $M$ contains your set $I$ and that it is closed under multiplication by elements of $R$. That it is additively closed follows from the fact that $U$ is a filter, and that it is maximal follows from the fact that $U$ is an ultrafilter.</p> <p>So $M$ is "as explicit" as $U$ is. In particular all functions which vanish on all but finitely many points of the sequence $m_n$ are in $M$.</p> <p>Also, you can take your favourite explicit filter $U'$ and define the ideal $M'$ as above, and have a non-maximal but explicit ideal which contains $I$. </p> http://mathoverflow.net/questions/50680/construction-of-a-maximal-ideal/50989#50989 Answer by Ashutosh for Construction of a maximal ideal Ashutosh 2011-01-03T04:11:10Z 2013-05-06T21:33:21Z <p>This may be relevant: </p> <p>Call an ideal on your ring free if there is no point where all functions in this ideal vanish. Then the following holds:</p> <p>The existence of a definable (in the language of Zermelo Fraenkel set theory with choice) free maximal ideal on the ring of continuous functions from reals to reals is independent.</p> <p>The consistency of the existence of such an ideal follows from fact that the axiom of constructibility implies that the universe has a definable well ordering.</p> <p>The consistency of the non existence of a definable maximal free ideal follows from an old result of Solovay which says that there is a model of set theory in which every definable set of reals is Lebesgue measurable (one also needs to assume that inaccessibles are consistent). The reason why Solovay's model doesn't have definable maximal free ideals is that every such ideal can be used to get a non principal ultrafilter over natural numbers which corresponds to a non measurable subset of the unit interval. This passage from the ideal to the non measurable set is definable hence there cannot be a definable maximal free ideal in Solovay's model.</p> <p>Edit: I wrote a short note on this <a href="http://www.math.wisc.edu/~akumar/MAXIMAL_IDEALS.pdf" rel="nofollow">here</a>.</p> http://mathoverflow.net/questions/50680/construction-of-a-maximal-ideal/50999#50999 Answer by Greg Graviton for Construction of a maximal ideal Greg Graviton 2011-01-03T09:33:27Z 2011-01-03T09:33:27Z <p>As <a href="http://mathoverflow.net/questions/50680/construction-of-a-maximal-ideal/50989#50989" rel="nofollow">Ashutosh</a> points out, it's impossible to describe $M$ constructively.</p> <p>However, from a practical point view, the "untouchability" of ultrafilters and maximal ideals in $\mathcal C(\mathbb R)$is not necessarily a problem. Usually, you only care about a finite, countable, or otherwise bounded cardinality of "properties" of $M$, so you can choose any ideal $J \supseteq I$ that has everything you want and pick $M$ maximal containing $J$. In other words, $J$ approximates $M$, and that's often good enough.</p>