What does the semiring of ideals of a ring R tell us about R? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T13:28:48Z http://mathoverflow.net/feeds/question/26607 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/26607/what-does-the-semiring-of-ideals-of-a-ring-r-tell-us-about-r What does the semiring of ideals of a ring R tell us about R? Timothy Chow 2010-05-31T17:36:37Z 2011-12-14T20:55:21Z <p>Here is something I've wondered about since I was an undergraduate. Let $R$ be a ring (commutative, let's say, although the generalization to noncommutative rings is obvious). Ideals of $R$ can be multiplied and can be added (the ideal $I+J$ is the ideal generated by $I$ and $J$), and multiplication distributes over addition. Therefore we can consider the semiring $S$ of ideals of $R$. The question is, does the structure of $S$ tell us anything interesting about the structure of $R$? Or vice versa?</p> <p>I asked this question on sci.math.research last year and got a few replies but nothing very substantive.</p> <p><a href="http://mathforum.org/kb/thread.jspa?messageID=6599151" rel="nofollow">http://mathforum.org/kb/thread.jspa?messageID=6599151</a></p> <p>For a more concrete question: Give an interesting sufficient condition for $S$ to be finitely generated. Conversely, if $S$ is finitely generated, does that imply anything interesting about $R$?</p> http://mathoverflow.net/questions/26607/what-does-the-semiring-of-ideals-of-a-ring-r-tell-us-about-r/26642#26642 Answer by Manny Reyes for What does the semiring of ideals of a ring R tell us about R? Manny Reyes 2010-06-01T00:37:40Z 2010-06-01T00:37:40Z <p>Here are a few observations. None of them require our ring to be commutative.</p> <p>First, notice that one can recover the natural partial ordering of the ideals via addition, because for any two ideals $I$ and $J$ of $R$, $I\subseteq J$ if and only if $I+J=J$. (More generally, $I+J$ is the smallest ideal containing both $I$ and $J$.)</p> <p>Second, this allows us to recover the prime ideals of $R$. This is because an ideal $P$ of $R$ is prime if and only if, for any ideals $I$ and $J$ of $R$, $IJ\subseteq P$ implies $I\subseteq P$ or $J\subseteq P$. (The same can be said for the semiprime ideals of $R$, which are the radical ideals of $R$ in case $R$ is commutative.)</p> <p>Third, we can recover the Zariski topology on the prime spectrum of $R$ because it is defined using the natural partial ordering on the ideals of $R$.</p> http://mathoverflow.net/questions/26607/what-does-the-semiring-of-ideals-of-a-ring-r-tell-us-about-r/26668#26668 Answer by Pietro KC for What does the semiring of ideals of a ring R tell us about R? Pietro KC 2010-06-01T05:11:16Z 2010-06-01T20:32:28Z <p>This is pretty simple too, but here it goes. I take R to be commutative and have 1.</p> <p>If S is generated by just one ideal P, then all the ideals of R are of the form P^k. Thus R is local. If P^2 is not all of P then any p in P\P^2 generates P, and each P^k is generated by p^k. Hence the only prime ideal is P, and it is exactly the ideal of nilpotents (since these are the intersection of all prime ideals). It follows that some P^k = 0.</p> <p>If P^2 is all of P then again there is just the one prime ideal P, but now P = P^k = 0, so R is a field.</p> <p>So either R is a field or there is a nilpotent p in R s.t. all x in R are of the form u*p^k for some unit u and non-negative integer k. (Just consider the biggest k for which x is a multiple of p^k.) Sometimes, but not always (see below), we can identify R with a quotient of the polynomial algebra (R/P)[X] (note that R/P is a field), namely (R/P)[X]/(X^k) where k is the smallest s.t. P^k = 0.</p> <p>Conversely, any quotient F[X]/(X^k), F a field, has its semiring of ideals generated by (X).</p> http://mathoverflow.net/questions/26607/what-does-the-semiring-of-ideals-of-a-ring-r-tell-us-about-r/26671#26671 Answer by Pete L. Clark for What does the semiring of ideals of a ring R tell us about R? Pete L. Clark 2010-06-01T05:35:14Z 2010-06-01T05:35:14Z <p>This is sort of a sideways answer, but: in many ways the monoid $\operatorname{Prin}(R)$ of principal ideals carries more information. If $R$ is a domain $\operatorname{Prin}(R)$ is a cancellative monoid so injects into its group completion, the <strong>group of divisibility</strong> $K^{\times}/R^{\times}$ of $R$. Many of the factorization properties of $R$ can be gracefully rephrased in terms of $\operatorname{Prin}(R)$ and/or $K^{\times}/R^{\times}$. </p> <p>See for instance Section 4.1 of</p> <p><a href="http://www.math.uga.edu/~pete/factorization2010.pdf" rel="nofollow">http://www.math.uga.edu/~pete/factorization2010.pdf</a></p> <p>for more on this point of view. </p> http://mathoverflow.net/questions/26607/what-does-the-semiring-of-ideals-of-a-ring-r-tell-us-about-r/83466#83466 Answer by rschwieb for What does the semiring of ideals of a ring R tell us about R? rschwieb 2011-12-14T20:20:34Z 2011-12-14T20:55:21Z <p>I've been interested in this lately. Hopefully you have seen this?</p> <p>Golan, Jonathan S.(IL-HAIF)</p> <p>Semirings for the ring theorist. </p> <p>Rev. Roumaine Math. Pures Appl. 35 (1990), no. 6, 531–540. </p> <p>Golan cautions that treating semirings as 'poor man's rings' is not always good. They can really be different animals altogether. I think someone noted above that the semiring of ideals is additively idempotent. In a sense, this is as far as you can get from an additive group.</p> <p>The compensation for the loss of the additive group is the complete lattice structure compatible with the multiplication. That is, if A\leq B, then AC\leq BC and CA\leq CB.</p> <p>Ring theorists have been saying things about rings via the lattice of one sided ideals for years! The lattices of onesided ideals are almost as nice, except you lose compatibility of multiplication with the order, and there is no longer a twosided identity for the semiring. These are called <em>quantales</em> in some places.</p>