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.
http://mathforum.org/kb/thread.jspa?messageID=6599151
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 $R$?
 A: 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 $M$.
A: 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://alpha.math.uga.edu/~pete/factorization2010.pdf
for more on this point of view.
A: 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.
A: 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).
A: 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 $R$.
A: 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!

