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Quotient of the Mutlplicative Monoid of a Ring

  1. Oct 14, 2012 #1
    In abstract algebra (ring theory specifically), when we are dealing with factorization, UFD's, and so on, we are often only interested in the multiplicative structure of the ring, not the additive structure. So here is the basic situation we face: we a start with an integral domain (R,+,*) (i.e. a commutative ring with no zero divisors). Then from that we take the multiplicative monoid (M,*) where M = R - {0}. What structure does M have? It's a commutative monoid with the cancellation property: ab=ac implies b=c. (A monoid is a set endowed with an operation that is associative and has an identity.)

    But we're still not done! In the context of factorization and associated topics, we also don't care about multiplication by units (invertible elements). We call two elements a and b of M "associates" if a=bu for some unit u. So my question is, can we usefully construct out of M yet another structure which consists of equivalence classes under the associate equivalence relation?

    You can take a quotient of a group with a subset if the subset is a normal subgroup. You can take a quotient of a ring with a subset if the subset is an ideal. So what is the condition that a subset of a monoid has to satisfy in order to be able to construct a quotient monoid? And whatever the condition is, does the group G of units of M satisfy this condition, so that M/G would be our quotient monoid, and would be the quotient of M under the associate equivalence relation? And assuming that we are allowed to form the quotient monoid M/G, what is the structure of this monoid?

    Any help would be greatly appreciated.

    Thank You in Advance.
     
    Last edited: Oct 14, 2012
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  3. Oct 15, 2012 #2

    micromass

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    Group theory and ring theory are very peculiar theories if you want to look at quotient spaces. The idea is that that you quotient by a normal subgroup/ideal. However, with other algebraic object, things are not that simple. Sure, there exists a notion of quotient spaces, but it is not just quotienting by a "subset" of the space.

    The idea is to use so called congruence relations. With a monoid M, a congruence relation is an equivalence relation [itex]\sim[/itex] on M, such that if [itex]x\sim y[/itex] and [itex]a\sim b[/itex], then [itex]ax\sim by[/itex].
    We can then look at [itex]M/\sim[/itex], the set of quotient classes. This turns out to be a monoid under the operation [itex][a]*=[ab][/itex].

    Notice, that with groups and rings, the notion of congruence relation also makes sense. For example, a congruence relation on a group is an equivalence relation [itex]\sim[/itex] such that

    • If [itex]x\sim y[/itex] and [itex]a\sim b[/itex], then [itex]ax\sim by[/itex]
    • If [itex]x\sim y[/itex], then [itex]x^{-1}\sim y^{-1}[/itex]

    So given such an equivalence relation, we can form a quotient group [itex]G/\sim[/itex].
    However, it turns out that there is a bijection between the congruence relations and normal subgroups. Indeed, given a normal subgroup N, we can say that [itex]x\sim y[/itex] if and only if [itex]xy^{-1}\in N[/itex].
    Conversely, given a congruence relation [itex]\sim[/itex], we can form a normal subgroup [itex]N=\{x\in G~\vert~x\sim e\}[/itex].
    So the quotient groups with the congruence relations are exactly the quotient groups by normal subgroups. The same thing holds for rings.

    However, the theory of congruence relations has the benefit that it is applicable to other algebraic structures such as monoids, semigroups, lattices, etc. Furthermore, what is also nice is that versions of the isomorphism theorems hold in general for congruence relations as well. All of this is studied in "universal algebra".
     
  4. Oct 15, 2012 #3
    Thanks micromass! Clearly the associate equivalence relation is a congruence relation on M (a commutative monoid with cancellation property). So now my only question is, what is the structure of quotient monoid M/~ where a~b means a and b are associates, i.e. a=bu for some unit u? I don't really know monoid theory (if there is such a field), so I don't know how I would go about doing it.
     
  5. Oct 15, 2012 #4

    micromass

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    I don't think this is such an easy question. A somewhat easier question we can ask ourself: if F is a field, then what is the structure of the group [itex]F\setminus \{0\}[/itex]. For finite fields, this is easy: the groups are exactly the cyclic groups of order [itex]p^n-1[/itex] with p prime.
    For infinite fields, things are somewhat harder. This paper classifies such groups: http://www.math.uga.edu/~pete/Dicker1966.pdf [Broken] but the classification does not seem to be very nice.
     
    Last edited by a moderator: May 6, 2017
  6. Oct 15, 2012 #5
    micromass, I didn't really mean how would you classify the structure of M/~ upto isomorphism. I was just asking informally, what are the properties of M/~? In other words, what are the general properties of a commutative monoid with cancellation property and no nontrivial units?
     
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