# What does the semiring of ideals of a ring R tell us about R?

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?

I asked this question on sci.math.research last year and got a few replies but nothing very substantive.

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 $R$?

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Just one example which comes into my mind: If R is a direct product of n fields, then the semiring of ideals is the power set of $1,...,n$ with union as sum and intersection as product (zero element is the empty subset, unit element is the whole set). You cannot recover the fields. –  Martin Brandenburg Jun 1 '10 at 0:05
So basically you want to know which properties can be covered? –  Martin Brandenburg Jun 1 '10 at 0:08
Sorry you didn't like my answers... –  Pete L. Clark Jun 1 '10 at 2:31
@Pete: I appreciated your answers, but am still hoping for something more substantive. –  Timothy Chow Jun 1 '10 at 4:35
To multiply ideals shouldn't they be two-sided ideals? There are interesting (noncommutative) rings with no nontrivial two-sided ideals (for example, the first Weyl algebra, which is differential operators in 1 variable with polynomial coefficients), so in this generality I believe not very much can be said. –  Peter Samuelson Jul 29 '10 at 17:57

Here are a few observations. None of them require our ring to be commutative.

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 $J$.)

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.)

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 $R$.

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The Zariski topology doesn't require the ring to be commutative? –  Pete L. Clark Jun 1 '10 at 3:48
No, the Zariski topology makes sense for any ring! This is because, for ideals $I$ and $J$, $V(I)\cap V(J) = V(IJ)$ and $\bigcup_j V(I_j) = V(\sum I_j)$. What you lose, however, is the fact that ring homomorhpisms determine continuous functions between prime spectra. –  Manny Reyes Jun 1 '10 at 4:13
For a noncommutative ring and in the case of $\textit{primitive}$ ideals, this is known as "Jacobson topology". (Primitive ideals are proper analogues of maximal ideals in the noncommutative setting.) It's a terminological issue, I simply cannot remember whether the same term applies to the whole prime spectrum, but certainly, "Zariski topology" implies that the ring is commutative. –  Victor Protsak Jun 1 '10 at 5:15

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 group of divisibility $K^{\times}/R^{\times}$ of $R$. Many of the factorization properties of $R$ can be gracefully rephrased in terms of $\operatorname{Prin}(R)$ and/or $K^{\times}/R^{\times}$.

See for instance Section 4.1 of

http://www.math.uga.edu/~pete/factorization2010.pdf

for more on this point of view.

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I've been interested in this lately. Hopefully you have seen this?

Golan, Jonathan S.(IL-HAIF)

Semirings for the ring theorist.

Rev. Roumaine Math. Pures Appl. 35 (1990), no. 6, 531–540.

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.

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.

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 quantales in some places.

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This is pretty simple too, but here it goes. I take R to be commutative and have 1.

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.

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.

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.

Conversely, any quotient F[X]/(X^k), F a field, has its semiring of ideals generated by (X).

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The identification of $R$ with a quotient of $(R/P)[X]$ is a little shaky. How does $\mathbb{Z}/9\mathbb{Z}$ fit into your classification scheme? About the best you can say, I think, is that if $S$ is one-generated, then $R \cong V/t^k$ for some discrete valuation ring $(V,t)$. –  Graham Leuschke Jun 1 '10 at 19:00
You're right, of course. I've edited accordingly. I wasn't sure myself but I wanted it to be true so badly! –  Pietro KC Jun 1 '10 at 20:31
From an ideal-theoretic point of view, DVRs are just as good as F[X]/(X^k) -- in fact better, since you don't have to worry about silly arithmetic issues with the field. Might as well just say that $S$ is one-generated iff $R$ is a non-trivial image of a DVR. Cleaner that way. –  Graham Leuschke Jun 1 '10 at 22:44