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Both in abstract algebra and measure theory is there term ring/algebra, but their definition are different and we can not deduce one from the other, the only requirement in definition they share is that they be closed under two operations: addition and multiplication in abstract algebra while difference and union in measure theory. Why we use the same term for these two different notions in two different branches of mathematics? Is there any connection between them? Thanks!

To be more specific, my intention is to compare "ring in measure theory" with "ring in abstract algebra", not "ring" with "algebra". I use notation "ring/algebra" in the post because the term algebra has the same situation, that is, compare "algebra in measure theory" with "algebra in abstract algebra".

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Why not use the same term? –  Mariano Suárez-Alvarez Apr 27 '10 at 4:29
    
It's been a while since I read it, but doesn' Halmos' Measure Theory book have some brief discussion of this? –  Yemon Choi Apr 27 '10 at 5:25
    
I found my question in Ex.(4), page 22 of Halmos' Measure Theory. –  zzzhhh Apr 29 '10 at 4:39
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3 Answers

up vote 8 down vote accepted

A ring of sets is a ring(usual definition) with the operations intersection(multiplication) and symmetric difference(addition). A sigma ring is a special kind of a ring of sets. Now a sigma algebra (which would possibly more appropriately be called sigma field) is a sigma ring where every element has a complement (multiplicative inverse) is a sigma ring with unity (equivalent to every element has a complement),thus it is a Boolean algebra with respect to intersection and union, and a Boolean ring with respect to intersection and symmetric difference. Every Boolean ring is an algebra over $\mathbb F_2$(thank you Mark Meckes for the correction). I hope this connection is enough to justify the use of these terms.

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Thank you but, to be more specific, my intention is to compare "ring in measure theory" with "ring in abstract algebra", not "ring" with "algebra". I use notation "ring/algebra" in the post because the term algebra has the same situation, that is, compare "algebra in measure theory" with "algebra in abstract algebra". –  zzzhhh Apr 27 '10 at 4:59
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zzzhhh, I think you should read both Gjergji and Pete's answers more closely. –  Qiaochu Yuan Apr 27 '10 at 5:02
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The complement of a set is not a multiplicative inverse; the product is the empty set, which is the additive identity. So a field/algebra of sets is not a field in the sense of abstract algebra. On the other hand, an algebra of sets is an algebra in the sense of abstract algebra: it is a ring as you explained, and also a vector space over the field $\mathbb{F}_2$ of 2 elements, which makes it an $\mathbb{F}_2$-algebra. This is why I personally prefer the terminology "sigma algebra" over "sigma field". –  Mark Meckes Apr 27 '10 at 13:10
    
Thank you for the deeper exposition, Prof. Meckes. –  zzzhhh Apr 27 '10 at 21:09
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There is a mathematical relationship between the two concepts: given an algebra of sets, one can define a commutative ring by taking addition to be symmetric difference and multiplication to be intersection. This ring is a Boolean ring, i.e., $x^2 = x$ for all elements $x$. Conversely, the Stone Representation Theorem says that every Boolean ring can be viewed as the ring associated to some algebra of sets. See e.g. Section 4.5

http://math.uga.edu/~pete/integral.pdf

for more detailed statements and proofs.

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Thank you but, to be more specific, my intention is to compare "ring in measure theory" with "ring in abstract algebra", not "ring" with "algebra". I use notation "ring/algebra" in the post because the term algebra has the same situation, that is, compare "algebra in measure theory" with "algebra in abstract algebra". –  zzzhhh Apr 27 '10 at 4:59
add comment

For what it's worth, a ring is an algebra over the integers.

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