Abstract Nonsense

Crushing one theorem at a time

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

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

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To prove that F is a complete lattice isomorphism we begin by letting \Omega\subseteq \mathfrak{L}^{\supseteq\mathfrak{a}}(R) be arbitrary. We recall that \displaystyle \sup\Omega is equal to \displaystyle \sum_{\mathfrak{b}\in\Omega}\mathfrak{b}. But, it’s trivial to see that since \pi is surjective that we have

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\displaystyle F\left(\sup\Omega\right)=\pi\left(\sum_{\mathfrak{b}\in\Omega}\mathfrak{b}\right)=\sum_{\mathfrak{b}\in\Omega}\pi(\mathfrak{b})=\sup F(\Omega)

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Now, let \Omega\subseteq\mathfrak{L}^{\supseteq\mathfrak{a}}(R). We claim that

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\displaystyle F\left(\inf\Omega\right)=\pi\left(\bigcap_{\mathfrak{b}\in\Omega}\mathfrak{b}\right)=\bigcap_{\mathfrak{b}\in\Omega}\pi(\mathfrak{b})=\inf F(\Omega)

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The inclusion

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\displaystyle \pi\left(\bigcap_{\mathfrak{b}\in\Omega}\mathfrak{b}\right)\subseteq\bigcap_{\mathfrak{b}\in\Omega}\pi(\mathfrak{b})

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we get for free as a common set-theoretic fact. To see the reverse inclusion let y be in \displaystyle \bigcap_{\mathfrak{b}\in\Omega}\pi(\mathfrak{b}). Fix some \mathfrak{b}_0\in\Omega, since y\in\pi(\mathfrak{b}_0) clearly we have that y=\pi(x_0) for some x_0\in\mathfrak{b}_0. I claim that x_0\in\mathfrak{b} for all \mathfrak{b}\in\Omega from where the desired inclusion follows. To see this let \mathfrak{b}\in\Omega be arbitrary, since y\in \pi(\mathfrak{b}) there exists some x\in\mathfrak{b} for which \pi(x)=y=\pi(x_0). Thus, since \pi is a morphism we have that x_0-x\in\ker\pi=\mathfrak{a}\subseteq\mathfrak{b} and so (x_0-x)+x=x_0\in\mathfrak{b}. Since \mathfrak{b} was arbitrary we have that x is in \displaystyle \bigcap_{\mathfrak{b}\in\Omega}\mathfrak{b}. Thus, \displaystyle y=\pi(x)\in\pi\left(\bigcap_{\mathfrak{b}\in\Omega}\mathfrak{b}\right). Since y was arbitrary the equality follows.

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Putting these together with the previous result that F is a bijection shows that F is a complete lattice isomorphism as claimed.

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It remains to show that F and F^{-1} form a monotone Galois connection between \mathfrak{L}^{\supseteq\mathfrak{a}}(R) and \mathfrak{L}(R/\mathfrak{a}). For this theorem that particularly means that \mathfrak{b}/\mathfrak{a}\subseteq\mathfrak{c}/\mathfrak{a} if and only if \mathfrak{b}\subseteq\mathfrak{c}. But, this follows immediately from the fact that F is monotone (since F is a complete lattice isomorphism) and the obvious fact that the inverse of a monotone function is monotone.

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Putting this all together the theorem follows. \blacksquare

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Remark: It’s also fairly clear to see that F is a semiring  homomorphism. Moreover, if I would have defined the lattice of subrings we would see that the above could be extended to that case in the same manner.

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A Generalization

Although not used quite as often the above lattice isomorphism theorem enables us to prove the following ‘generalization’:

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Generalized Fourth Isomorphism Theorem: Let R and S be rings and f:R\twoheadrightarrow S an epimorphism. Then the map F:\mathfrak{L}^{\supseteq\ker f}(R)\to\mathfrak{L}(S) given by \mathfrak{b}\mapsto f(\mathfrak{b}) is a lattice ismorphism. Moreover, F and F^{-1} form a monotone Galois connection between \mathfrak{L}^{\supseteq\ker f}(R) and \mathfrak{L}(S).

Proof: This follows immediately from the fourth ring isomorphism theorem and the first ring isomorphism theorem since the map R/\ker f\to S:r+\ker f\mapsto f(r) is an isomorphism. \blacksquare

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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.


July 9, 2011 - Posted by | Algebra, Ring Theory | , , , , , , ,


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