# Abstract Nonsense

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

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

$\text{ }$

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

$\text{ }$

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

$\text{ }$

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

$\text{ }$

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

$\text{ }$

The inclusion

$\text{ }$

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

$\text{ }$

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.

$\text{ }$

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

$\text{ }$

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.

$\text{ }$

Putting this all together the theorem follows. $\blacksquare$

$\text{ }$

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.

$\text{ }$

$\text{ }$

A Generalization

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

$\text{ }$

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. $\blacksquare$

$\text{ }$

$\text{ }$

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.