Defining the Product of Ideals 
Where does the "naive definition" of the product of two ideals $I$ and $J$, $IJ = \{ ij \mid i \in I, j \in J \}$ fall apart?
(The product of two ideals $I$ and $J$ is defined to be $IJ := \sum_i a_ib_i$, where each $a_i$ is in $I$ and each $b_i$ is in $J$.)

Note: This is a question from a problem set I'm working on. I am not expecting a full solution for the answers, but I'd like some help knowing where to look. It seems to work fine in $\mathbb{Z}$ and $\mathbb{C}[x]$.
 A: I think you'll find of interest my old sci.math post below.

From: Bill Dubuque 
Date: 30 Jul 2003 23:54:07 -0400
Message-ID: 
Bill Dubuque  wrote:

Rasmus Villemoes  wrote:
  >

In my algebra textbook, the product of two ideals I,J is defined as
{ sum_{i=1..n} a_i b_i | n >= 1 , a_i in I and b_i in J }
Now it is rather easy to prove that IJ is an ideal in R. The last
    question of the exercise is:
Is A = { ab | a in I, b in J } an ideal of R.
Now the preceding questions strongly suggest that the answer in
    general is no, but I can't find a counterexample. Clearly, (since it
    is understood that R is commutative), if one of I or J is a principal
    ideal, the set A is an ideal, so a counterexample has to consist of a
    non-PID and two ideals generated by at least two elements each  [...]

HINT: Find proper ideals whose product contains an irreducible element,
e.g.  p in (p,a)(p,b)  if  (a,b) = (1)
Examples abound.

Domains where ideals multiply simply as  IJ = { ij : i in I, j in J },
are called condensed domains. Below are reviews of related papers.

84a:13019  13F99 
Anderson, David F.; Dobbs, David E.
On the product of ideals.
Canad. Math. Bull. 26 (1983), no. 1, 106-114.  

In this paper the authors define an integral domain R to be a condensed
domain provided  IJ = {ij: i in I, j in J}  for all ideals I and J of R.
Bezout domains are condensed domains. The main results of the paper
characterize condensed domains within some large class of domains. For
example, it is shown that a GCD-domain is condensed if and only if it is 
a Bezout domain, and a Krull domain is condensed if and only if it is a
principal ideal domain. For a Noetherian domain  R  to be condensed it 
is necessary that  dim R <= 1  and that the integral closure of  R  be 
a principal ideal.
       Reviewed by J. T. Arnold

86h:13017  13F05 (13B20 13G05)
Anderson, David F.(1-TN); Arnold, Jimmy T.(1-VAPI); Dobbs, David E.(1-TN)
Integrally closed condensed domains are Bezout.
Canad. Math. Bull. 28 (1985), no. 1, 98-102.

An integral domain  R  is termed quasicondensed if  I^n = {i_1 i_2...i_n :
i_j in I  for  1 <= j <= n}  for each positive integer  n  and each 
two-generated ideal  I = (a,b)  of  R.  R  is said to be condensed if
IJ = {ij: i in I, j in J} for all ideals  I  and  J  of  R. The main theorem
shows that an integral domain is a Bezout domain if and only if it is 
integrally closed and condensed. An example (a  D+M  construction) is given
of an integrally closed quasicondensed domain which is not a Bezout domain.
       Reviewed by Anne Grams

90e:13019  13F30 (13B20 13G05)
Gottlieb, Christian (S-STOC)
On condensed Noetherian domains whose integral closures are discrete
valuation rings.
Canad. Math. Bull. 32 (1989), no. 2, 166-168.

Following D. F. Anderson and the reviewer [same journal 26 (1983), no. 1, 
106-114; MR 84a:13019] an integral domain  R  is said to be condensed in case
IJ = {ij : i in I, j in J}  for all ideals  I,J  of  R. The author defines an
integral domain  R  to be strongly condensed if for every pair  I,J  of ideals
of R, either IJ = aJ for some  a in I  or IJ = Ib for some  b in J. Suppose
henceforth that  R  is a Noetherian integral domain whose integral closure  R'
is a discrete valuation ring. It is proved that if  R  is condensed, then  R
contains an element of value 2 (in the associated discrete rank 1 valuation).
It is not known whether the converse holds, nor whether all condensed domains
are strongly condensed. As a partial converse, it is proved that R is strongly
condensed under the following conditions: (R',M')  is a finitely generated
R-module,  R'/M' is isomorphic to  R/M  and R contains an element of value 2.
       Reviewed by David E. Dobbs

1 955 608  13A15 (13Bxx)
Anderson, D. D.; Dumitrescu, Tiberiu
Condensed domains.
Canad. Math. Bull. 46 (2003), no. 1, 3-13.
http://journals.cms.math.ca/cgi-bin/vault/view/anderson8107

Abstract: 
An integral domain D with identity is condensed (resp., strongly condensed) if
for each pair of ideals I,J of D, IJ = {ij : i in I, j in J} (resp., IJ = iJ
for some i in I or IJ = Ij for some j in J). We show that for a Noetherian
domain D, D is condensed if and only if Pic(D) = 0 and D is locally condensed,
while a local domain is strongly condensed if and only if it has the 
two-generator property. An integrally closed domain D is strongly condensed 
if and only if D is a Bezout generalized Dedekind domain with at most one
maximal ideal of height greater than one. We give a number of equivalencies
for a local domain with finite integral closure to be strongly condensed.
Finally, we show that for a field extension k < K, the domain D = k + XK[[X]]
is condensed if and only if [K:k] <= 2 or [K:k] = 3 and each degree-two
polynomial in  k[X] splits over  k, while D is strongly condensed if and 
only if [K:k] <= 2.
A: Suppose $I=J=(x,y)\subseteq \mathbb R[x,y]$. Then $x^2$ and $y^2$ are in your naive definition, but their sum is not.
It is an enlightening exercise to try to see what exactly $\mathbb Z$ and $\mathbb C[x]$ have that make the naive definition work...
A: If you read Remark 2.7 of my paper "On a property of pre-Schreier domains"
[Comm. Algebra 15 (1987) 1895-1920] you would find that the question of
discrepancy between products of two ideals $A,B$ $(AB=\{\Sigma $ $%
a_{i}b_{i}|a_{i}\in A,b_{i}\in B\}$ in a ring and the product of two ideals $%
I,J$ $(IJ=\{a_{i}b_{i}|a_{i}\in I$ and $b_{i}\in I\}$ in a semigroup was
raised in the above mentioned paper of mine. I was naive and thought
everyone was my friend. So I would send preprints of even half written
papers around. I have a feeling that the Anderson-Dobbs paper, that is
mentioned by Bill Dubuque, was an effort to lift the idea from a pre-print
of the above paper with very little credit (they do mention the pre-print,
but without giving a clue to why they mention a result from Zafrullah.) At
the end of that remark I do mention the Anderson-Dobbs' paper. I think I
handled it beautifully, mentioning the heist and at the same time praising
it. But now I am bitter, as I don't see any light at the end of the tunnel.
For as I look back I see this kind of activity around most of my solo
papers. In any case, the property $\ast $ came right out of this discussion
of the discrepancy of the definitions of products of ideals. Recall that a
domain $D$ is a $\ast $-domain if $((\cap (a_{i}))(\cap (b_{j}))=\cap
_{ij}(a_{i}b_{j})$ for all $a_{1},...,a_{n},b_{1},....b_{m}\in D$
equivalently, $\in K=qf(D).$ I address an effort to lift the $\ast $%
-property in https://lohar.com/mithelpdesk/hd2006.pdf
Anyone interested in products of ideals or pre-Schreier domains may like to
study the $\ast $-property, as a Prufer domain has the $\ast $-property and
a so called Prufer $v$-multiplication domain (PVMD) that is a $\ast $-domain
is a locally GCD domain.
