# Abstract Nonsense

## 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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$\mathfrak{a}\cap(\mathfrak{b}+\mathfrak{c})\supseteq(\mathfrak{a}\cap\mathfrak{b})+(\mathfrak{a}\cap\mathfrak{c})=\mathfrak{a}+(\mathfrak{a}\cap\mathfrak{c})$

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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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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.

June 23, 2011 -