Nilradicals without Zorn's lemma - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T03:37:50Z http://mathoverflow.net/feeds/question/27163 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/27163/nilradicals-without-zorns-lemma Nilradicals without Zorn's lemma Daniele Turchetti 2010-06-05T15:50:04Z 2010-06-05T19:29:41Z <p>It's well known that the nilradical of a commutative ring with identity $A$ is the intersection of all the prime ideals of $A$.</p> <p>Every proof I found (e.g. in the classical "Commutative Algebra" by Atiyah and Macdonald) uses Zorn's lemma to prove that $x \notin Nil(A) \Rightarrow x \notin \cap_{\mathfrak{p}\in Spec(A)} \mathfrak{p}$ (the other way is immediate). Does anybody know a proof that doesn't involve it?</p> http://mathoverflow.net/questions/27163/nilradicals-without-zorns-lemma/27165#27165 Answer by Eric Rowell for Nilradicals without Zorn's lemma Eric Rowell 2010-06-05T16:06:52Z 2010-06-05T16:06:52Z <p>This seems difficult as Krull's theorem (existence of maximal ideals) implies the Axiom of Choice. This is due to <a href="http://jlms.oxfordjournals.org/cgi/pdf_extract/s2-19/2/285" rel="nofollow">W. Hodges</a> I think.</p> http://mathoverflow.net/questions/27163/nilradicals-without-zorns-lemma/27166#27166 Answer by Ryan Reich for Nilradicals without Zorn's lemma Ryan Reich 2010-06-05T16:11:28Z 2010-06-05T16:20:02Z <p>I assume the argument you have in mind is the folowing: suppose $f \in \cap \mathfrak{p}$; then to show that $f$ is nilpotent, it suffices to show that the localized ring $A_f$ is zero. And indeed, if $f$ is in every prime ideal of $A$, then $A_f$ has no prime ideals at all; since every nonzero ring has a maximal ideal by Zorn's Lemma, we must have $A_f = 0$.</p> <p>This line of reasoning easily adapts to show that in fact, the statement that $\operatorname{Nil}(A) = \cap \mathfrak{p}$ implies that every nonzero ring has a prime ideal. Indeed, suppose that $A$ were nonzero with no prime ideals; then $\cap \mathfrak{p} = A$, so every element of $A$ is nilpotent. In particular, $1 = 1^n = 0$, so $A = 0$.</p> <p>Following Eric Rowell's answer, this is very close to being equivalent to the axiom of choice (however, it does not obviously imply the existence of <em>maximal</em> ideals).</p> http://mathoverflow.net/questions/27163/nilradicals-without-zorns-lemma/27172#27172 Answer by Chris Phan for Nilradicals without Zorn's lemma Chris Phan 2010-06-05T17:23:06Z 2010-06-05T17:59:58Z <p>Y. Rav proved this using the Ultrafilter Principle ("Every filter on a set can be extended to an ultrafilter"), which is weaker than the Axiom of Choice. Theorem 4.1 of <a href="http://www.ams.org/mathscinet-getitem?mr=476530" rel="nofollow">Variants of Rado's selection lemma and their applications, Math. Nachr. 79 (1977), 145--165</a> states:<blockquote><b>Theorem 4.1.</b> Let <em>R</em> be a ring, $\mathfrak{a}$ a proper ideal in <em>R</em>, and suppose that <em>S</em> is multiplicative subsemigroup of <em>R</em> which does not meet $\mathfrak{a}$. Then it follows from the Ultrafilter Principle that their exists a prime ideal $\mathfrak{p}$ in <em>R</em> such $\mathfrak{a} \subseteq \mathfrak{p}$ and $\mathfrak{p} \cap S= \emptyset$. </blockquote></p> <p>Rav also showed:</p> <blockquote> <b>Corollary 4.4.</b> The following statements are mutually equivalent in ZF set theory:<br /> (a) Every filter on a set can be extended to an ultrafilter.<br /> (b) In every commutative associative ring with identity, every proper ideal is included in some prime ideal.<br /> (c) In every Boolean algebra, every proper ideal (resp. filter) is included in some prime ideal (resp. ultrafilter). </blockquote> http://mathoverflow.net/questions/27163/nilradicals-without-zorns-lemma/27187#27187 Answer by François G. Dorais for Nilradicals without Zorn's lemma François G. Dorais 2010-06-05T19:05:40Z 2010-06-05T19:29:41Z <p>Since you asked for a proof, let me complement Chris Phan's answer by outlining a proof that relies only on the <a href="http://en.wikipedia.org/wiki/Compactness_theorem" rel="nofollow">Compactness Theorem</a> for propositional logic, which is yet another equivalent to the <a href="http://en.wikipedia.org/wiki/Boolean_prime_ideal_theorem" rel="nofollow">Ultrafilter Theorem</a> over ZF. </p> <p>Let A be a commutative ring and let x &notin; Nil(A). To each element a &isin; A associate a propositional variable p<sub>a</sub> and let T be the theory whose axioms are</p> <ul> <li>p<sub>0</sub>, &not;p<sub>1</sub>, &not;p<sub>x</sub>, &not;p<sub>x<sup>2</sup></sub>, &not;p<sub>x<sup>3</sup></sub>,...</li> <li>p<sub>a</sub> &and; p<sub>b</sub> &rarr; p<sub>a+b</sub> for all a, b &isin; A.</li> <li>p<sub>a</sub> &rarr; p<sub>ab</sub> for all a, b &isin; A.</li> <li>p<sub>ab</sub> &rarr; p<sub>a</sub> &or; p<sub>b</sub> for all a, b &isin; A.</li> </ul> <p>Models of T correspond precisely to prime ideals that do not contain x. Indeed, if P is such an ideal, then setting p<sub>a</sub> to be true iff a &isin; P satisfies all of the above axioms, and conversely. So it suffices to show that T has a model.</p> <p>Since x<sup>n</sup> &ne; 0 for all n, one can verify using ideals over finitely generated subrings of A that the theory T is finitely consistent, i.e. any finite subset of T has a model. (What I just swept under the rug here is a constructive proof of the theorem for quotients of Z[v<sub>1</sub>,...,v<sub>n</sub>].) The Compactness Theorem for propositional logic then ensures that T has a model.</p>