Abstract Nonsense

Fourth Ring Isomorphism Theorem (Lattice Isomorphism Theorem) (Pt. I)

Point of Post: In this post we discuss the notion of the fourth isomorphism theorem, but phrase it more in terms of lattices and Galois connections than is often done in most texts.

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Motivation

The fourth isomorphism theorem is much more involved than the other three isomorphism theorems (in content), but is perhaps one of the most useful results in basic ring theory. The fourth isomorphism theorem answers a fundamental question surely anyone trying to do actual ring theory (in the sense of dealing with real live rings, i.e. examples) will ask: “Is there a way I can find all the ideals of $R/\mathfrak{a}$ if I know all (or even better, just some of) the ideals of $R$?” Since quotient rings will often popup and ideals are of integral importance it makes sense that this is an important question. The beauty of the fourth isomorphism theorem is not only do we get a resounding “Yes!” to our question, but the connection between ideals turns out to be much, much more than we ever could have anticipated. Indeed, it turns out that there is a natural and astoundingly simple connection between subrings of $R$ containing $\mathfrak{a}$ and subrings of $R/\mathfrak{a}$ which besides being simple has many added properties. Moreover, we shall see that the map between these two sets of subrings  will reduce nicely to a map between ideals of $R$ containing $\mathfrak{a}$ and ideals of $R/\mathfrak{b}$.

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A nice way to picture precisely what the fourth ring isomorphism theorem says involves Hasse diagrams. Namely, what we shall see is that there is a natural correspondence between the lattice (we will show it is a lattice) of ideals of $R$ containing $\mathfrak{a}$ and the ideals of $R/\mathfrak{a}$ which is such that if the Hasse diagram for the first lattice is of the form

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$\begin{array}{ccccccc}\nwarrow & & \nearrow & & & \uparrow &\\ & \mathfrak{b} & & & & \mathfrak{c} & \\ & & \nwarrow & & \nearrow & & \\ & & & \mathfrak{a} & & & \end{array}$

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then the Hasse diagram for the second lattice will be

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$\begin{array}{ccccccc}\nwarrow & & \nearrow & & & \uparrow &\\ & \mathfrak{b}/\mathfrak{a} & & & & \mathfrak{c}/\mathfrak{a} & \\ & & \nwarrow & & \nearrow & & \\ & & & \{\mathfrak{a}\} & & & \end{array}$

$\text{ }$ Order Theoretic Prerequisites

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Let $\left(L,\wedge,\vee\right)$ be a lattice. We call $K\subseteq L$ a sublattice of $L$ if $\left(K,\wedge,\vee\right)$ is itself a lattice. We denote this by $K\leqslant L$ when convenient. It’s important to note that even if $L$ is a complete lattice and $K\leqslant L$ we need not have that $K$ is a complete lattice. An example of this can be seen in topology where one considers the topology $\mathscr{T}\subseteq 2^X$ of some space $X$ and note that while $\left(2^X,\cap,\cup\right)$ is a complete lattice and $\mathscr{T}\leqslant 2^X$ yet $\mathscr{T}$ need not be closed under arbitrary infima (i.e. while $\displaystyle \left\{\left(\frac{-1}{n},\frac{1}{n}\right)\right\}_{n\in\mathbb{N}}$ is in the usual topology on $\mathbb{R}$ the infimum of this collection of sets is $\{0\}$ which is not in the topology). Thus, we call $K\leqslant L$ a complete sublattice of it’s complete as a lattice. It’s obvious that such properties of lattices such as modularity are preserved under sublattices. The only thing we prove about sublattices is the following fact:

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Theorem: Let $\left(L,\wedge,\vee\right)$ be a lattice. Define $L^{\geqslant a}$ for some $a\in L$ to be the set of all elements $b\in L$ with $b\geqslant a$. Then, $L^\geqslant b$ is a sublattice of $L$. Furthermore, if $L$ is complete then $L^{\geqslant a}$ is a complete sublattice and $\displaystyle \sup_L\Omega=\sup_{L^{\geqslant}}\Omega$ for all $\Omega\subseteq L^{\geqslant}$.

Proof: Let $x,y\in L^{\geqslant a}$ be arbitrary. Note then that since $a\leqslant x,y$ we have by definition that $a\leqslant x\wedge y$ and evidently since $x\vee y\geqslant x$ we have that $x\vee y\geqslant a$ and so $x\wedge y,x\vee y\in L$. Since $x,y$ were arbitrary the conclusion follows.

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Suppose now that $L$ is complete. To show that $L^{\geqslant a}$ is complete it evidently suffices to show that $\displaystyle \sup_{L}\Omega\in L^{\geqslant a}$ for every $\Omega\subseteq L^{\geqslant}$ since then evidently each $\Omega$ then has a supremum and $\displaystyle \sup_{L^{\geqslant a}}\Omega=\sup_L \Omega$. But, this is fairly evidently follows immediately from the same logic as the reason why $x\wedge y,x\vee y\in L^{\geqslant a}$. The conclusion follows. $\blacksquare$

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Next, let $\left(P,\leqslant\right)$ and $\left(Q,\preceq\right)$ be posets. A mapping $F:P\to Q$ is called monotone if $a\leqslant b$ implies $F(a)\preceq F(b)$ (i.e. that it preserves order). A monotone Galois connection between $P$ and $Q$ is a pair of monotone maps $F:P\to Q$ and $G:Q\to P$ with the property that $q\leqslant F(p)$ if and only if $p\leqslant G(q)$ where $p\in P$ and $q\in Q$. In this context $F$ is called the lower adjoint and $G$ the upper adjoint.

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Finally, let $\left(L,\wedge,\vee\right)$ and $\left(K,\wedge,\vee\right)$ two lattices (where $\wedge$ and $\vee$ may be different operations) we call a map $F:L\to K$ a lattice homomorphism if $F\left(a\wedge b\right)=F(a)\wedge F(b)$ and $F(a\vee b)=F(a)\vee F(b)$ for all $a,b\in L$. If $F$ is bijective we call $F$ a lattice isomorphism and it’s easy to see that $F^{-1}$ is a lattice homomorphism $K\to L$. A complete lattice homomorphism is a map $F:L\to K$, where $L,K$ are complete lattices, which satisfies $F(\sup\Omega)=\sup F(\Omega)$ for every $\Omega\subseteq L$ and a complete lattice isomorphism is a bijective complete lattice homomorphism–evidently then $F^{-1}$ is a complete lattice isomorphism. It’s easy to see that every lattice or complete lattice homomorphism is monotone.

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The Fourth Isomorphism Theorem

Before we actually state the fourth ring isomorphism theorem we make note of a quick fact about the lattice $\mathfrak{L}(R)$ for some ring $R$. Indeed, let $\mathfrak{a}\in\mathfrak{L}(R)$ and define $\mathfrak{L}^{\supseteq \mathfrak{a}}\left(R\right)$ to be the set of all ideals of $R$ containing $\mathfrak{a}$. From the above theorems and comments we may conclude that $\mathfrak{L}^{\supseteq \mathfrak{a}}\left(R\right)$ is a modular complete sublattice of $\mathfrak{L}(R)$. With this in mind we see that:

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Theorem(Fourth Isomorphism Theorem/Lattice Isomorphism Theorem): Let $R$ be a ring and $\mathfrak{a}$ be an ideal of $R$. The map

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$F:\left\{\begin{array}{c}\text{Subrings of }R\\ \text{that contain }\mathfrak{a}\end{array}\right\}\to\left\{\begin{array}{c}\text{Subrings of the}\\ \text{quotient ring }R/\mathfrak{a}\end{array}\right\}:\mathfrak{b}\mapsto \pi(\mathfrak{b})=\mathfrak{b}/\mathfrak{a}$

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(where $\pi:R\twoheadrightarrow R/\mathfrak{a}$ is the canonical projection) is a bijection which reduces to a complete lattice isomorphism

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$F:\mathfrak{L}^{\supseteq\mathfrak{a}}(R)\to\mathfrak{L}(R/\mathfrak{a})$

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Moreover $F$ and $F^{-1}$ form a montone Galois connection between $\mathfrak{L}^{\supseteq \mathfrak{a}}$ and $\mathfrak{L}(R/\mathfrak{a})$.

Proof: We first show that $F$ is a bijection. This turns out to be relatively simple. Suppose that $A,B$ are subrings of $R$ are such that $F(A)=F(B)$. Let $a\in A$ then $a+\mathfrak{a}\in F(A)=F(B)$ and so there exists some $b\in B$ such that $a+\mathfrak{a}=b+\mathfrak{b}$ or that $a-b\in \mathfrak{a}\subseteq B$ and thus $(a-b)+b=a\in B$ and thus $A\subseteq B$. The opposite inclusion follows similarly and thus we may conclude that $F$ is injective. To prove that $F$ is surjective we merely note that if $S$ is a subring of $R/\mathfrak{a}$ then $\pi^{-1}(S)$ is a subring of $R$ and since $\mathfrak{a}=\pi^{-1}(\{0\})\subseteq \pi^{-1}(S)$ we may conclude that $\pi^{-1}(S)$ is a subring of $R$ containing $\mathfrak{a}$, surjectivity then follows since $\pi(\pi^{-1}(S))=S$ (since $\pi$ is surjective).

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What we next prove  that $F\left(\mathfrak{L}^{\geqslant \mathfrak{a}}(R)\right)=\mathfrak{L}(R/\mathfrak{a})$. We first note that $F\left(\mathfrak{L}^{\geqslant \mathfrak{a}}(R)\right)\subseteq F\left(\mathfrak{L}(R/\mathfrak{a}\right)$–this is immediate from the fact that for any $\mathfrak{b}\in \mathfrak{L}^{\geqslant \mathfrak{a}}(R)$ we have that $\pi(\mathfrak{b})=F(\mathfrak{b})$ is an ideal of $\mathfrak{L}(R/\mathfrak{a})$ since $\pi$ is epic. The proof that the opposite inclusion is true we merely note that if $J\in\mathfrak{L}(R/\mathfrak{a})$ then $\pi^{-1}(J)$ is an ideal of $R$ since we know the preimage of ideals under morphisms are ideals. Since $\mathfrak{a}=\pi^{-1}(\{0\})\subseteq\pi^{-1}(J)$ we may thus conclude that $\pi^{-1}(J)\in\mathfrak{L}^{\supseteq\mathfrak{a}}(R)$ and since $F(\pi^{-1}(J))=\pi(\pi^{-1}(J))=J$ (once again because $\pi$ is epic) surjectivity follows.

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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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