Abstract Nonsense

Crushing one theorem at a time

Maximal Ideals (Pt. I)


Point of Post: In this post we discuss the notion of maximal ideals, give several characterizations for commutative unital rings, etc.

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Motivation

Just as is the case for prime ideals there are two natural ways to motivate the introduction to maximal ideals. That said, unlike the case for prime ideals where any kid off the street can relate to their importance by pointing the value of primes in \mathbb{Z} the value of maximal ideals is a slightly more delicate issue, but one which plays a pivotal role in much of advanced ring theory. In fact, a very common technique in more advanced ring theory goes as follows:

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\begin{array}{c}\text{Prove theorem for}\\ \text{local rings}\end{array}\implies \begin{array}{c}\text{Prove result holds}\\ \text{for subrings of}\\ \text{local rings}\end{array}\implies\begin{array}{c}\text{Embed arbitrary ring}\\ \text{into local ring}\end{array}

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where a local ring is a ring R with a unique maximal ideal \mathfrak{m}, often denoted (R,\mathfrak{m}). The reason for this (analogous to the embedding of integral domains for fields) is that, as we shall (maybe not so soon) see, local rings are particularly simple.

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So what are these ‘maximal’ ideals that are so important? As stated there are two natural ways to motivate why anyone would ever even consider such a kind of ideal. As in the case with prime ideals we start with a simple minded, but important motivating factor. In fact, this motivation is in direct parallel with the first motivating factor for prime ideals. Namely, for a commutative unital ring R we decided that prime ideals were the answer to “what kind of ideal \mathfrak{a} is characterized by having R/\mathfrak{a} be an integral domain. Well, while I can’t speak for the whole of the mathematical community, I feel that it’s not natural to ask when R/\mathfrak{a} is an integral domain. No, I believe that most people would want the whole shebang and ask when R/\mathfrak{a} is a field. Well, the answer to this question is “when \mathfrak{a} is a maximal ideal”. Well, what precisely are ‘maximal ideals’? Well, let’s see if we can take them, by definition, to be ideals whose attendant quotient ring is a field and see if we can work backwards towards their ‘real’ definition. So, we begin by picking a characterization of fields that is particularly amenable to our current situations. Considering we are dealing strictly with ideals it makes sense that this would be the characterization of fields in terms of ideals. In particular, one can recall that if R is commutative and unital then R is a field if and only if all it’s ideals are trivial or non-propert. In simpler terms, R will be a field precisely when the only ideals of R are (0)=\{0\} and (1)=R. Thus, an ideal \mathfrak{a} of R will be ‘maximal’ if and only if ideal lattice of R/\mathfrak{a} looks like

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\begin{array}{c}R/\mathfrak{a}\\ \vert\\ (\mathfrak{a})\end{array}

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But, now we’re cooking with fire since the fourth isomorphism theorem tells us this is the case only when the ideal structure of R looks like

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\begin{array}{ccc} & R &\\ \swarrow & \downarrow & \searrow \\ & \mathfrak{a} &\end{array}

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Or, in other words, R/\mathfrak{a} will have the desired ideal lattice if and only if \mathfrak{a} is maximal (in terms of containment) amongst the proper ideals of R. This is the definition of maximal that we shall take. Namely, for us an ideal \mathfrak{m} (appropriately dubbed) will be called maximal when \mathfrak{m}\subseteq\mathfrak{a} implies \mathfrak{a}=R and \mathfrak{m}\ne R.

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The second motivation is one which, while equally important, doesn’t quite have the same verve as the previous motivation. Namely, a heuristic statement with some real truth behind it, is that a ring with a particularly simple lattice structure has a particularly simple overall ring structure. Thus, to find examples of rings which are easy to deal with, a sensible thing to do is to single out what is simple in lattice land and then transfer this back over to ring land. In particular, it’s not hard to see (if one draws a few examples of Hasse diagrams) that rings with few maximal ideals are ‘simple’.

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Maximal Ideals

Let R be a ring. We define a proper ideal \mathfrak{m} of R to maximal if whenever \mathfrak{a} is an ideal of R such that \mathfrak{m}\subseteq\mathfrak{a} then \mathfrak{a}. In other words, \mathfrak{m} is a maximal ideal if it’s maximal in the poset of all proper ideals of R. We denote the set of all maximal ideals of a ring R by \text{MaxSpec}(R). As noted above in the introduction, we have by the fourth isomorphism theorem that:

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Theorem: Let R be a commutative unital ring. Then, \mathfrak{m}\in\text{MaxSpec}(R) if and only if R/\mathfrak{m} is a field.

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To see how this could fail for non-unital rings one can easily see that 4\mathbb{Z} is maximal in 2\mathbb{Z} yet 2\mathbb{Z}/4\mathbb{Z} is not a field. That said, it obviously fails for a stupid reason, the idea is still there. In particular, if one defines a simple ring to be a ring with only trivial ideals ((0) and R) then it’s evident that a more general version of the above says:

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Theorem: An ideal \mathfrak{m} in an ideal R is maximal if and only if R/\mathfrak{m} is simple.

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References: 

1. Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.

2. Bhattacharya, P. B., S. K. Jain, and S. R. Nagpaul. Basic Abstract Algebra. Cambridge [Cambridgeshire: Cambridge UP, 1986. Print.

3.  Reid, Miles. Undergraduate Commutative Algebra. Cambridge: Cambridge UP, 1995. Print.

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August 31, 2011 - Posted by | Algebra, Ring Theory | , , , , ,

1 Comment »

  1. […] be a field and let be irreducible. We know then that is a maximal ideal and thus is a field, which can be thought of as an extension of via the map sending . In fact, […]

    Pingback by Field Extensions « Abstract Nonsense | March 7, 2012 | Reply


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