Maximal ideal and Zorn's lemma - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T17:34:28Z http://mathoverflow.net/feeds/question/53523 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/53523/maximal-ideal-and-zorns-lemma Maximal ideal and Zorn's lemma expmat 2011-01-27T19:10:04Z 2011-01-28T19:42:56Z <p>It is known that any ring A (say commutative with 1) has a maximal ideal. The proof uses Zorn's lemma.</p> <p>Now I heard some people saying that if we assume A to be noetherian, then we don't need to use Zorn's lemma. The argument would basically be as follows:</p> <p>"Suppose it doesn't have a maximal ideal. Then we can build an ascending chain of distinct ideals."</p> <p>But, as far as I know it, we have to use Zorn's lemma in order to construct such an ascending chain. Am I right?</p> <p>If I am right, is it still true (via some other argument) that we don't need to use Zorn's lemma to prove the result?</p> <p>(EDIT: My definition of noetherian ring is that any ascending chain of ideals stabilizes.)</p> http://mathoverflow.net/questions/53523/maximal-ideal-and-zorns-lemma/53529#53529 Answer by Joel David Hamkins for Maximal ideal and Zorn's lemma Joel David Hamkins 2011-01-27T19:38:43Z 2011-01-27T19:38:43Z <p>With your definition of Noetherian, then you don't need full AC to carry out the argument, but only the weaker principle known as Dependent Choices (DC), which asserts that one can make countably many choices in succession. In your argument, if there is no maximal ideal, then by DC you could successively pick larger and larger ones, violating the Noetherian property. </p> http://mathoverflow.net/questions/53523/maximal-ideal-and-zorns-lemma/53550#53550 Answer by Andrej Bauer for Maximal ideal and Zorn's lemma Andrej Bauer 2011-01-27T23:14:12Z 2011-01-27T23:14:12Z <p>This is an elaboration of Joel's answer. Suppose \$R\$ is a ring in which every ascending chain \$I_0 \subseteq I_1 \subseteq I_2 \subseteq \cdots \$ of ideals stabilizes.</p> <p>We show that every (non-trivial) ideal can be extended to a maximal ideal. Given an ideal \$J\$, define a sequence of ideals \$I_n\$ as follows:</p> <ol> <li>Let \$I_0 = J\$.</li> <li>For \$n \geq 0\$, if \$I_n\$ is a maximal ideal then let \$I_{n+1} = I_n\$, otherwise choose an ideal \$I_{n+1}\$ which is strictly larger than \$I_n\$.</li> </ol> <p>We used Dependent Choice to construct the sequence of ideals \$I_n\$. Because \$I_n\$ is an ascending chain of ideals it stabilizes. When it does, it reaches a maximal ideal that extends \$J\$.</p> <p>So we used Dependent Choice to construct the ascending chain of ideals and Excluded Middle to decide whether an ideal is maximal or not.</p> <p>In a particular case it may be possible to extend a non-maximal ideal to a larger one in a canonical way (for example, if we know how to well-order \$R\$ then we generate \$I_{n+1}\$ by adjoining to \$I_n\$ the first element which is not in \$I_n\$ yet). In this case we do not even need Dependent Choice.</p>