Abstract Nonsense

Crushing one theorem at a time

Second Ring Isomorphism Theorem


Point of Post: In this post we state and prove the second ring isomorphism theorem which takes the form (S+\mathfrak{a})/\mathfrak{a}\cong S/(S\cap\mathfrak{a}).

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Motivation

Proving the second ring isomorphism theorem is literally almost immediate from the first ring isomorphism theorem. But, what is not apparent is why anyone would, based just on our knowledge so far, even consider the question “Is (S+\mathfrak{a})/\mathfrak{a}\cong S/(S\cap\mathfrak{a})?” Here’s perhaps one of the thought processes which would produce this question. So, suppose we are given a ring R, a subring S, and an ideal \mathfrak{a}\in\mathfrak{L}(R). We know , of course, we can form the quotient ring R/\mathfrak{a}. Well, what if we (for some reason, I’m sure there’s a few you could think of) wanted to form the ‘quotient ring’ S/\mathfrak{a}. Of course the reason why this doesn’t make sense, the reason why quotient ring is in scare quotes is that S/\mathfrak{a} could be utterly meaningless since there is no reason why \mathfrak{a}\subseteq S needs to be true. Perhaps though we don’t want to completely give up on creating a quotient ring using S and \mathfrak{a} as the main ingredients–perhaps we’d like to slightly modify the situation in such a way that we get something close to the notion of what S/\mathfrak{a} should be, but stay within the realm of usual quotient rings.  There are two natural ways one could go about doing this, either we could ‘enlarge’ S to a bigger subring which contains \mathfrak{a} or we could ‘shrink’ \mathfrak{a} so that it sits inside S. Of course, since our main goal was to take S/\mathfrak{a}and approximate it, we would like to, in both cases, change S and \mathfrak{a} minimally. In other words, we would like to take the minimal enlargement of S which contains \mathfrak{a} and the minimal reduction of \mathfrak{a} for which S contains it. Practically, this means considering S+\mathfrak{a} and S\cap\mathfrak{a} respectively. So, we’ve convinced ourselves that if we can’t really consider S/\mathfrak{a} our best bet (for closest approximation) is to consider (S+\mathfrak{a})/\mathfrak{a} or S/(S\cap\mathfrak{a}). So, the obvious question now is “which do we choose?” Luckily, what the second ring isomorphism theorem says is that we have no difficult choice at hand because the two rings are, in fact, isomorphic. Indeed, the second ring isomorphism theorem says that, using our notation we’ve been using, \mathfrak{a} is an ideal of S+\mathfrak{a}, S\cap\mathfrak{a} is an ideal of S, and (S+\mathfrak{a})/\mathfrak{a}\cong S/(S\cap\mathfrak{a}). The second ring isomorphism theorem is often called the ‘Diamond Isomorphism Theorem’ considering the following Hasse diagram of the sets in question in the poset \left(\mathfrak{L}(R),\subseteq\right) (in fact, it really applies to the poset \left(2^R,\subseteq\right) but usually the theorem is stated for when S\in\mathfrak{L}(R)):

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\displaystyle \begin{array}{ccccc} & & R & & \\ & & \vert & & \\ & & S+\mathfrak{a} & &\\ & \swarrow & & \searrow & \\ S & & & & \mathfrak{a}\\ & \searrow & & \swarrow & \\ & & S\cap\mathfrak{a} & & \end{array}

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Second Ring Isomorphism Theorem

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We now proceed to prove the above discussed theorem. Namely:

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Theorem (Second Ring Isomorphism Theorem): Let R be a ring, S a subring of R, and \mathfrak{a}\in\mathfrak{L}(R). Then, S+\mathfrak{a} is a subring of R, \mathfrak{a}\in\mathfrak{L}(S+\mathfrak{a}), S\cap \mathfrak{a}\in\mathfrak{L}(S), and

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\displaystyle \frac{S+\mathfrak{a}}{\mathfrak{a}}\cong\frac{S}{S\cap\mathfrak{a}}

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Proof: Consider the series of maps S\overset{\iota}{\hookrightarrow} S+\mathfrak{a}\overset{\pi}{\twoheadrightarrow} (S+\mathfrak{a})/\mathfrak{a} (where \iota is the inclusion map) evidently [by definition] this is a surjection S\twoheadrightarrow (S+\mathfrak{a})/\mathfrak{a} and evidently s+\mathfrak{a}=\mathfrak{a} if and only if s\in\mathfrak{a}, in other words s\in S\cap\mathfrak{a}. Thus, \ker\left(S\twoheadrightarrow (S+\mathfrak{a})/\mathfrak{a}\right)=S\cap\mathfrak{a}. Thus, by the first ring isomorphism theorem we may conclude that S/(S\cap\mathfrak{a})\cong (S+\mathfrak{a})/\mathfrak{a}. \blacksquare

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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.

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June 28, 2011 - Posted by | Algebra, Ring Theory | , , , , , , ,

2 Comments »

  1. […] to the second ring isomorphism theorem the third isomorphism theorem has an almost immediate proof from the first ring isomorphism […]

    Pingback by Third Ring Isomorphism Theorem « Abstract Nonsense | July 1, 2011 | Reply

  2. […] four isomorphism theorems we are going to discuss have the same feel and motivation as the first, second, third, and fourth ring isomorphism theorems with “ring” replaced by “left […]

    Pingback by The Module Isomorphism Theorems « Abstract Nonsense | November 11, 2011 | Reply


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