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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    $\begingroup$ 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. $\endgroup$ Jun 1, 2010 at 0:05
  • $\begingroup$ So basically you want to know which properties can be covered? $\endgroup$ Jun 1, 2010 at 0:08
  • $\begingroup$ Sorry you didn't like my answers... $\endgroup$ Jun 1, 2010 at 2:31
  • $\begingroup$ @Pete: I appreciated your answers, but am still hoping for something more substantive. $\endgroup$ Jun 1, 2010 at 4:35
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    $\begingroup$ 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. $\endgroup$ Jul 29, 2010 at 17:57

6 Answers 6


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 $R$.

  • $\begingroup$ The Zariski topology doesn't require the ring to be commutative? $\endgroup$ Jun 1, 2010 at 3:48
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    $\begingroup$ 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. $\endgroup$ Jun 1, 2010 at 4:13
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    $\begingroup$ 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. $\endgroup$ Jun 1, 2010 at 5:15

Just thought that I'd add for anyone who comes across this thread: as mentioned in this thread, the set of ideals of a commutative ring $R$ forms a semiring (also known as a 'rig') with additive identity ${0}$ and mulitiplicative identity $R$. A similar situation is true for modules over a commutative ring; the set of submodules of a unital $R$-module $M$ forms a unital 'module' over the semiring of ideals of $R$. That is, the four unital module axioms are satisfied:

Distributivity of 'scalar' multiplication over 'vector' addition: $I(N+L) = IN + IL$ for any ideal $I$ of $R$ and any submodules $N, L$ of $M$,

Distributivity of 'scalar' multiplication over 'scalar' addition: $(I+J)N = IN + JN$ for any ideals $I, J$ of $R$ and any submodule $N$ of $M$,

Compatibility of 'scalar' multiplication with 'ring multiplication': $I(JN) = (IJ)N$ for any ideals $I, J$ of $R$ and any submodule $N$ of $M$,

Unitary law: $RN = N$ for any submodule $N$ of $M$.


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


for more on this point of view.


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.


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).

  • $\begingroup$ 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 $(V,t)$. $\endgroup$ Jun 1, 2010 at 19:00
  • $\begingroup$ You're right, of course. I've edited accordingly. I wasn't sure myself but I wanted it to be true so badly! $\endgroup$
    – Pietro
    Jun 1, 2010 at 20:31
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    $\begingroup$ 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. $\endgroup$ Jun 1, 2010 at 22:44

In my question, I linked to a sci.math.research archive that is now defunct, so I am re-posting (most of) that content here.

Phil wrote:

What's the Grothendieck completion of this semiring? That might be something interesting.

Olivier wrote:

$I+J$ is linked with gcd—these beasts do not have a good structure, but, but, some investigations is surely called for. One usually sees the ordered lattice with $I \subset J$ and $I+J$ and $I \cap J$. It seems it behaves nicely with respect to multiplication, so one should be able to prove something akin to "this lattice is a product lattice over all prime ideals". This result in itself would not be very sexy. However, when it fails would most probably be more attractive. The distinction between maximal and prime ideal is of major incidence in ring theory, and it is clear such a distinction will have impact on the core "result" I mentioned.

Pete Clark wrote:

By coincidence a student asked me the same question about a month ago. It seemed promising at first, until we noticed that for all ideals $I$, $I + I = I$. Therefore $I = 0$ in the ring completion—i.e., the completion is the zero ring.

I responded:

That's a good point. I'm not quite ready to give up yet, though. There does seem to be some interesting structure. For example, in the integers, multiplication is multiplication, and addition is gcd. I think that the "ideal semiring" of any Dedekind domain is isomorphic to that of the integers. In general, though, one gets something else—I'm not sure what.

Experience with tropical semirings suggests that simply trying to complete a semiring to a ring isn't always a good idea. I don't really know what tools are appropriate here though.

Pete Clark pointed out that my statement about Dedekind domains was wrong:

No, that's not right. For instance, if $R$ is the univariate polynomial ring over the complex numbers, then its ideal semiring is uncountable.

Perhaps what you meant to say is the following? The ideal semiring of any Dedekind domain $R$ is isomorphic to the direct sum of copies of the semiring $(\mathbb{N} \cup \{\infty\},\min,+)$, where $\mathbb{N}$ is the natural numbers (positive integers plus 0), the addition law is given by the minimum, and the multiplication law is given by addition. The direct sum extends over all nonzero prime ideals of $R$. (And yes, this is strongly reminsicent of tropical geometry….)

Agreed that for a non-Dedekind domain the structure will be much different. As someone else already noted, it has a lattice-theoretic feel to it, but I don't know what work has been done in this direction.

I wrote:

I asked a friend of mine this question and one of his first instincts was to ask, when is this semiring finitely generated?

The answer might be "not very often" since even the integers don't give an example, but if there are some nontrivial examples, these might be the most tractable to analyze.

Dave Cullen wrote (I hope I'm interpreting his $*$ notation correctly; originally he wrote J* and A*B and A*X):

Well, one thing I notice from this conversation is that the set of ideals of a ring with ideal sum and multiplication does form a complete (idempotent) dioid; in particular, the operation $*$, given as $$J^* = \sum_{i\in \mathbb{N}} J^i$$ is well-defined for an ideal $J$. Such structures have a partial ordering "$\le$" given by $$J \le K \iff J + K = K.$$ Some preliminary results are that for fixed ideals $A$ and $B$, the affine equation $X = AX + B$ has a least (with respect to $\le$) solution $A^*B$, and any solution $X$ satisfies $X = A^*X$.

It seems like these dioids are well studied in other contexts, but I have never actually seen anything (in the web-literature anyhow) about dioids of ideals. There is also something called a cost dioid that is well studied, again for motivating reasons totally unrelated to sets of ideals, which requires the additional condition of the existence of roots of elements. Although this condition fails for general ideals, it may be true for some rings, or for some subsets of the collection of ideals of a ring. Maybe we can steal some of this dioid theory and apply it to the ideal context? Comments welcome!


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