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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 $\mathbb{C}[x]$.

Community
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Michael Chen
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    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

3 Answers3

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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...

Mariano Suárez-Álvarez
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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

<p>{ sum_{i=1..n} a_i b_i | n >= 1 , a_i in I and b_i in J }</p>

<p>Now it is rather easy to prove that IJ is an ideal in R. The last
question of the exercise is:</p>

<p>Is A = { ab | a in I, b in J } an ideal of R.</p>

<p>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  [...]</p>

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.

Bill Dubuque
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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.

user26857
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mzafrullah
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