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Let $R$ be a commutative ring with identity and $I,J,K$ be ideals of $R$. If $I\supseteq J$ or $I\supseteq K$, we have the following modular law $$ I\cap (J+K)=I\cap J + I\cap K$$

I was wondering if there are situations in which the modular law holds in which the hypothesis that $I$ contains at least one of $J,K$ is relaxed.

One example is when $R$ is a polynomial ring or power series ring and $I,J,K$ are monomial ideals.

Of course one containment always holds $I\cap (J+K)\supseteq I\cap J +I\cap K$. In what other situations does the other containment hold?

1 Answers 1

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Such domains are known as Prüfer domains. They are non-Noetherian generalizations of Dedekind domains. Their ubiquity stems from a remarkable confluence of interesting characterizations. For example, they are those domains satisfy either the Chinese Remainder Theorem for ideals, or Gauss's Lemma for polynomial content ideals, or for ideals: $\rm\ A\cap (B + C) = A\cap B + A\cap C\:,\ $ or $\rm\ (A + B)\ (A \cap B) = A\ B\:,\ $ or $\rm\ A\supset B\ \Rightarrow\ A\:|\:B\ $ for fin. gen. $\rm\:A\:$ etc. It's been estimated that there are close to 100 such characterizations known, e.g. see my sci.math post for 30 odd characterizations. Below is an excerpt:

THEOREM $\ \ $ Let $\rm\:D\:$ be a domain. The following are equivalent:

(1) $\rm\:D\:$ is a Prüfer domain, i.e. every nonzero f.g. (finitely generated) ideal is invertible.
(2) Every nonzero two-generated ideal of $\rm\:D\:$ is invertible.
(3) $\rm\:D_P\:$ is a Prufer domain for every prime ideal $\rm\:P\:$ of $\rm\:D.\:$
(4) $\rm\:D_P\:$ is a valuation domain for every prime ideal $\rm\:P\:$ of $\rm\:D.\:$
(5) $\rm\:D_P\:$ is a valuation domain for every maximal ideal $\rm\:P\:$ of $\rm\:D.\:$
(6) Every nonzero f.g. ideal $\rm\:I\:$ of $\rm\:D\:$ is cancellable, i.e. $\rm\:I\:J = I\:K\ \Rightarrow\ J = K\:$
(7) $\: $ (6) restricted to f.g. $\rm\:J,K.$
(8) $\rm\:D\:$ is integrally closed and there is an $\rm\:n > 1\:$ such that for all $\rm\: a,b \in D,\ (a,b)^n = (a^n,b^n).$
(9) $\rm\:D\:$ is integrally closed and there is an $\rm\: n > 1\:$ such that for all $\rm\:a,b \in D,\ a^{n-1} b \ \in\ (a^n, b^n).$
(10) Each ideal $\rm\:I\:$ of $\rm\:D\:$ is complete, i.e. $\rm\:I = \cap\ I\: V_j\:$ as $\rm\:V_j\:$ run over all the valuation overrings of $\rm\:D.\:$
(11) Each f.g. ideal of $\rm\:D\:$ is an intersection of valuation ideals.
(12) If $\rm\:I,J,K\:$ are nonzero ideals of $\rm\:D,\:$ then $\rm\:I \cap (J + K) = I\cap J + I\cap K.$
(13) If $\rm\:I,J,K\:$ are nonzero ideals of $\rm\:D,\:$ then $\rm\:I\ (J \cap K) = I\:J\cap I\:K.$
(14) If $\rm\:I,J\:$ are nonzero ideals of $\rm\:D,\:$ then $\rm\:(I + J)\ (I \cap J) = I\:J.\ $ ($\rm LCM\times GCD$ law)
(15) If $\rm\:I,J,K\:$ are nonzero ideals of $\rm\:D,\:$ with $\rm\:K\:$ f.g. then $\rm\:(I + J):K = I:K + J:K.$
(16) For any two elements $\rm\:a,b \in D,\ (a:b) + (b:a) = D.$
(17) If $\rm\:I,J,K\:$ are nonzero ideals of $\rm\:D\:$ with $\rm\:I,J\:$ f.g. then $\rm\:K:(I \cap J) = K:I + K:J.$
(18) $\rm\:D\:$ is integrally closed and each overring of $\rm\:D\:$ is the intersection of localizations of $\rm\:D.\:$
(19) $\rm\:D\:$ is integrally closed and each overring of $\rm\:D\:$ is the intersection of quotient rings of $\rm\:D.\:$
(20) Each overring of $\rm\:D\:$ is integrally closed.
(21) Each overring of $\rm\:D\:$ is flat over $\rm\:D.\:$
(22) $\rm\:D\:$ is integrally closed and prime ideals of overrings of are extensions of prime ideals of $\rm\:D.$
(23) $\rm\:D\:$ is integrally closed and for each prime ideal $\rm\:P\:$ of $\rm\:D,\:$ and each overring $\rm\:S\:$ of $\rm\:D,\:$ there is at most one prime ideal of $\rm\:S\:$ lying over $\rm\:P.\:$
(24) For polynomials $\rm\:f,g \in D[x],\ c(fg) = c(f)\: c(g)\:$ where for a polynomial $\rm\:h \in D[x],\ c(h)\:$ denotes the "content" ideal of $\rm\:D\:$ generated by the coefficients of $\rm\:h.\:$ (Gauss' Lemma)
(25) Ideals in $\rm\:D\:$ are integrally closed.
(26) If $\rm\:I,J\:$ are ideals with $\rm\:I\:$ f.g. then $\rm\: I\supset J\ \Rightarrow\ I|J.$ (contains $\:\Rightarrow\:$ divides)
(27) the Chinese Remainder Theorem $\rm(CRT)$ holds true in $\rm\:D\:,\:$ i.e. a system of congruences $\rm\:x\equiv x_j\ (mod\ I_j)\:$ is solvable iff $\rm\:x_j\equiv x_k\ (mod\ I_j + I_k).$
(28) Each finitely generated torsion-free $\rm\,D$-module is projective.

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    This is absolutely great. Thanks a lot. I will have to spend some time to study all these characterizations, but this is exactly what I wanted.2010-12-13
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    @Timothy: Enjoy - they are fascinating to study.2010-12-13
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    Would you recommend any survey papers that discuss the equivalent characterizations?2010-12-13
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    @Timothy: The references listed in the [Wikipedia article](http://en.wikipedia.org/wiki/Pr%C3%BCfer_domain) are a good root to start from. Many of the original research articles are online and can easily be located by keyword searches. I don't recall any comprehensive survey articles. Please let me know if you locate such.2010-12-13
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    @Bill: Yeah I started looking through the wikipedia article as well, but thought I can ask for your inputs as well. I found a book on Prufer domains. On page 2 the authors discuss several characterizations and refer to a book by Gilmer for more. http://books.google.com/books?id=W7YR0e7OKB4C&printsec=frontcover&dq=prufer+domains#v=onepage&q=prufer%20domains&f=false2010-12-13
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    @Timothy: Yes, Gilmer's book is one of the standard references and contains a wealth of useful material. However it's not easy to find (e.g. it's not on any ebook site).2010-12-13
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    @Timothy: For a mere $130 you can purchase the book (look through www.bookfinder.com).2010-12-13
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    @Timothy: I managed to recall the paper where I found many of the above characterizations: S. Bazzoni and S. Glaz, [Prüfer rings](http://www.math.uconn.edu/~glaz/My_Articles/PruferRings.Springer06.pdf), Multiplicative Ideal Theory in Commutative Algebra, Springer-Verlag (2006), pp. 55–72.2010-12-13
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    @Bill: Incidentally I was just perusing through the same paper. Thanks again.2010-12-13
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    @BillDubuque I directed a user to your answer here who asked the same type of question. I saw the comment thread about the difficulty of obtaining Gilmer's book and just thought I would post a link for how to get it at a more recent edition at a reasonable price (50 dollars) I had a tough time finding it myself, so this might be useful. http://www.campusbookstore.com/GeneralBooks/Details/QPPAM12-Multiplicative-Ideal-Theory-by-Gilmer2015-06-04
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    @CPM Iiirc someone mentioned that it has since appeared on various ebook sites.2015-06-04
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    @BillDubuque Okay, that is great to hear. It is a very useful book so I am glad to see it is more readily available. Commutative Rings by Kaplansky was also very difficult to find for a book cited so much.2015-06-04
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    I want to point out that rings (not necessarily domains) satisfying the condition stated by the OP are called *arithmetical*.2016-04-26