Abstract Nonsense

Crushing one theorem at a time

Generated Ideals and the Lattice of Ideals (Pt. II)

Point of Post: This post is a continuation of this one.

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But, when R is unital we have a different description of \displaystyle \left(\bigcup_{\beta\in\mathcal{B}}\mathfrak{a}_\beta\right). Namely, we know for a unital ring that \displaystyle \left(\bigcup_{\beta\in\mathcal{B}}\mathfrak{a}_\beta\right) is the set of all finite sums of the form r_1a_1s_1+\cdots+r_na_ns_n with \displaystyle a_j\in\bigcup_{\beta\in\mathcal{B}}\mathfrak{a}_\beta, r_j,s_j\in R, and n\in\mathbb{N}. But, since each \mathfrak{a}_\beta is an ideal it’s clear that left and right multiplying by R-elements doesn’t affect anything and so for unital rings \displaystyle \left(\bigcup_{\beta\in\mathcal{B}}\mathfrak{a}_\beta\right) is just the set of all sums of the form a_1+\cdots+a_n such that \displaystyle a_j\in\bigcup_{\beta\in\mathcal{B}}\mathfrak{a}_\beta. In other words, \displaystyle \left(\bigcup_{\beta\in\mathcal{B}}\mathfrak{a}_\beta\right) is the set of all finite sums where the summands lie in \displaystyle \bigcup_{\beta\in\mathcal{B}}\mathfrak{a}_\beta. This motivates the following definition: let R be a ring (not necessarily unital) and \left\{J\right\}_{\beta\in\mathcal{B}} be a collection of subrings we define then the sum of \left\{J_\beta\right\}_{\beta\in\mathcal{B}} to be the set

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\displaystyle \sum_{\beta\in\mathcal{B}}J_\beta=\left\{\sum_{j=1}^{n}a_j:a_j\in\bigcup_{\beta\in\mathcal{B}}J_\beta\text{ and }n\in\mathbb{N}\right\}

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in other words the sum of a collection of subrings is the set of all finite sums where the summands belong to the union of the subrings. What we’ve seen above is that if R is unital ring and \left\{\mathfrak{a}_\beta\right\}_{\beta\in\mathcal{B}} is a collection of ideals of R then \displaystyle \left(\bigcup_{\beta\in\mathcal{B}}\mathfrak{a}_\beta\right)=\sup_{\beta\in\mathcal{B}}\mathfrak{a}_\beta is equal to \displaystyle \sum_{\beta\in\mathcal{B}}\mathfrak{a}_\beta. One would hope that this holds true for not only two-sided ideals but left and right as well (this should be clear since the same argument applies) but also for non-unital rings. Indeed, this is true. To prove this we start with the following theorem:

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Theorem: Let R be a ring (not necessarily unital) and \left\{J_\beta\right\}_{\beta\in\mathcal{B}} be a collection of  left,right, or two-sided ldeals of R. Then, \displaystyle \sum_{\beta\in\mathcal{B}}J_\beta is a left, right, or two-sided ideal of R respectively.

Proof: For notational convenience let \displaystyle J denoted \displaystyle \sum_{\beta\in\mathcal{B}}J_\beta. Let then x,y\in J be arbitrary. Then, x=s_1+\cdots+s_n and y=s'_1+\cdots+s'_m with s_j,s'_j\in J_\beta for some \beta\in\mathcal{B}. We see then that x-y=t_1+\cdots+t_{n+m} where t_j=s_j if j\leqslant n and t_j=-s'_j for j>n, and since clearly each t_j belongs to some J_\beta we may conclude that x-y\in J. Thus, J is a subgroup of R. We now suppose that each J_\beta is an ideal since the other two cases are the same. To prove that J is an ideal we let r\in R be arbitrary and note that

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\displaystyle rx=rs_1+\cdots+rs_n\text{ and }xr=s_1r+\cdots+s_nr

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and since each J_\beta is a two-sided ideal we see that each rs_j and s_jr belongs to whichever J_\beta s_j did. Thus, rx and xr are (finite) sums of elements from \displaystyle \bigcup_{\beta\in\mathcal{B}}J_\beta and thus in J. Since x and r were arbitrary the conclusion follows. \blacksquare

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With this in mind we prove the following:

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Theorem: Let R be a ring and \left\{J_\beta\right\} be a collection of left, right, or two-sided ideals then \sup \left\{J_\beta\right\}_{\beta\in\mathcal{B}} is equal to \displaystyle \sum_{\beta\in\mathcal{B}}J_\beta (where, while the result remains the same, the meaning of supremum depends on whether the ideals are left, right, or two-sided in the sense that we are dealing with the lattices \mathfrak{L}_L(R), \mathfrak{L}_R(R), and \mathfrak{L}(R) respectively–we won’t make explicit note of this in the proof since all the manipulations apply equally well to all three, but keep this in mind).

Proof: Evidently since \sup \left\{J_\beta\right\}_{\beta\in\mathcal{B}} is a (left, right or two-sided) ideal containing \displaystyle \bigcup_{\beta\in\mathcal{B}}J_\beta is must contain all finite sums of elements from \displaystyle \bigcup_{\beta\in\mathcal{B}}J_\beta and thus it must contain \displaystyle \sum_{\beta\in\mathcal{B}}J_\beta. Conversely, we have just proven that \displaystyle \sum_{\beta\in\mathcal{B}}J_\beta is a (left, right, or two sided) ideal which clearly contains \displaystyle \bigcup_{\beta\in\mathcal{B}}J_\beta and thus by definition it must contain \sup\left\{J_\beta\right\}_{\beta\in\mathcal{B}}. The conclusion follows. \blacksquare

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Modularity and the Product of Ideals

You’ll now recall though that I claimed that \mathfrak{L}(R) is a modular lattice. A modular lattice is a lattice \left(L,\subseteq,\wedge,\vee\right) that has the property that if a\geqslant c a\vee (b\wedge c)=(a\vee b)\wedge c. For us this shall mean that if \mathfrak{a,b,c}\in\mathfrak{L}(R) with \mathfrak{a}\subseteq \mathfrak{b} then \mathfrak{a}\cap(\mathfrak{b}+\mathfrak{c})=\mathfrak{a}+(\mathfrak{a}\cap\mathfrak{c}). Indeed:

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Theorem: Let R be a ring and \mathfrak{a,b,c}\in\mathfrak{L}(R). Then,

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\mathfrak{a}\cap\left(\mathfrak{b}+\mathfrak{c}\right)\supseteq (\mathfrak{a}\cap\mathfrak{b})+(\mathfrak{a}\cap\mathfrak{c})

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if \mathfrak{a}\supseteq\mathfrak{b} then equality holds.

Proof: Let x\in (\mathfrak{a}\cap\mathfrak{b})+(\mathfrak{a}\cap\mathfrak{c}) then x=y+z where y\in\mathfrak{a}\cap\mathfrak{b} and z\in\mathfrak{a}\cap\mathfrak{c}. We note firstly that since y,z\in\mathfrak{a} that x=y+z\in\mathfrak{a}. But, since y\in\mathfrak{b} and z\in\mathfrak{c} we also see that x=y+z\in\mathfrak{b}+\mathfrak{c} and thus x=y+z\in\mathfrak{a}\cap(\mathfrak{b}+\mathfrak{c}).

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Suppose now that \mathfrak{a}\subseteq\mathfrak{b}, we aim to show that \mathfrak{a}\cap(\mathfrak{b}+\mathfrak{c})=\mathfrak{a}+(\mathfrak{a}\cap\mathfrak{c}). We know from the previous paragraph that

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Conversely, if x\in\mathfrak{a}\cap(\mathfrak{b}+\mathfrak{c}) we then know that x\in\mathfrak{a} and x=y+z where y\in\mathfrak{b} and z\in\mathfrak{c}. But, since y\in \mathfrak{b}\subseteq\mathfrak{a} as well by assumption we know that x-y=z\in\mathfrak{a} and so z\in\mathfrak{a}\cap\mathfrak{c}, thus x=y+z\in\mathfrak{a}+(\mathfrak{a}\cap\mathfrak{c}). \blacksquare

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Of course, a similar result holds true for elements of \mathfrak{L}_L(R) and \mathfrak{L}_L(R). Thus, collecting all the facts from this post we get:

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Theorem: Let R be a ring, then \left(\mathfrak{L}(R),\subseteq,\cap,+\right) forms a complete modular lattice.

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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.


June 23, 2011 - Posted by | Algebra, Ring Theory | , , , , , , , , ,


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