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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 $\mathbb{C}[x]$.

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Since you say it is an assignment, I've taken the liberty of adding the [homework] tag. –  Arturo Magidin Feb 10 '11 at 21:45
    
@Arturo: Thanks, I didn't think to add that tag. –  Michael Chen Feb 10 '11 at 21:49
2  
Since it seems that the question from your problem set is precisely to explain where the naive definition goes awry, I've made it into a block-quote so it is clear that this is the homework speaking, and not you (it is not entirely clear to me if your assigment might instead be to work with the definition $IJ=\sum a_ib_i$, but you, on your own initiative, are wondering why the "naive definition" does not work). –  Arturo Magidin Feb 10 '11 at 22:03
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2 Answers

up vote 11 down vote accepted

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...

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For the $n$th time, I had to solve a captcha to edit this... –  Mariano Suárez-Alvarez Feb 10 '11 at 21:53
2  
Is it because they are principal ideal domains? That seems to be the common thread I see. –  Michael Chen Feb 10 '11 at 22:06
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@Michael: if you can prove that the naive definition does work in a principal ideal domain, then yes. If not, no :=) –  Mariano Suárez-Alvarez Feb 10 '11 at 22:08
1  
Also, finding an example which is not a principal ideal domain but where nonetheless the naive definition works is a good exercise. –  Mariano Suárez-Alvarez Feb 10 '11 at 22:13
2  
@Michael: Principal ideal rings will suffice; they don't have to be domains. –  Arturo Magidin Feb 10 '11 at 22:14
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I think you'll find of interest my old sci.math post below.

From: Bill Dubuque <w...@nestle.ai.mit.edu>
Date: 30 Jul 2003 23:54:07 -0400
Message-ID: <y8zhe538i74.fsf@nestle.ai.mit.edu>

Bill Dubuque <w...@nestle.ai.mit.edu> wrote:
>Rasmus Villemoes <burner+use...@imf.au.dk> 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.
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