There are various possible ways to define a lattice.
One way is to define a partial order $\leq$ on a set $L$. We say, for subset $S \subseteq L$, that an element $u \in L$ that $u$ is an upper bound for $S$ if $s \leq u$ for all $s \in S$.
If there exists an element $u$ such that if for any $x \in L$ such that $s \leq x$ for all $s \in S$, we have $u \leq x$, then $u$ is said to be the least upper bound, or join of $S$. A subset of a partially-ordered set need not have any upper bounds, nor any least upper bounds, but if it has a least upper bound, it is unique.
If every two-element subset $\{s_1,s_2\}$ has a join, $L$ is said to be a join-semilatiice.
Dually, putting the bound on the left of the $\leq$ sign, and the elements of $S$ on the right, we obtain a similar definition of lower bound, and greatest lower bound, also called meet, and of a meet-semilattice.
A lattice is a partially-ordered set that is both a join-semilattice, and a meet-semilattice.
Note that this endows $L$ with two binary operations: join and meet. These are usually denoted by $\vee$ and $\wedge$, respectively.
The $R$-submodules of a given $R$-module $M$ can be given a partial order, by set inclusion.
We have, for two such $R$-submodules $N_1$ and $N_2$:
$N_1 \vee N_2 = \langle N_1,N_2\rangle = N_1 + N_2$
(This is a consequence of the closure condition of an abelian group: any $R$-submodule which contains both $N_1$ and $N_2$ surely contains the sum-including the smallest such submodule, and just as clearly the sum is contained in any submodule of $M$ that contains each summand).
$N_1 \wedge N_2 = N_1 \cap N_2$
(Again clearly the intersection is maximal as a SET, and since it is a submodule itself, it is the maximal submodule contained in both $N_1$ and $N_2$).
The lattice-isomorphism theorem states that if we have an $R$-module homomorphism $\phi: M \to M'$ with $\text{ker }\phi = N$:
$L =$ the sublattice of the lattice of submodules of $M$ over $N$ (i.e. the submodules $A$ for which $N \leq A$, that is: $N \subseteq A$, for our submodule $N$ of $M$),
$L' = $ the lattice of submodules of $\phi(M)$.
then we have a lattice isomorphism:
$\psi: L \to L'$, given by: $\psi(A) = \phi(A)$.
It is not uncommon to state this in term of the binary operations, instead of the partial order:
$\psi(N_1 + N_2) = \psi(N_1) + \psi(N_2)$
$\psi(N_1 \cap N_2) = \psi(N_1) \cap \psi(N_2)$
that is: $\psi$ respects join and meet. It is not uncommon to use the "same letter" for $\psi$ and $\phi$ even though one is defined on $M$ and one is defined on $L$.
Given the FIRST isomorphism theorem, one can "suppress $\phi$", using instead the isomorphic modules $A/N$. Then the statements about $\psi$ above become the same statements about "sums and intersections commuting" found in Dummit and Foote.
Moreover, in any category where we have an analogue of the first homomorphism theorem (such a category necessarily possesses kernels), we obtain a similar lattice-isomorphism theorem:
Groups (kernels are normal subgroups)
Abelian groups (kernels are subgroups)
Rings (kernels are ideals)
Vector spaces (kernels are null spaces)
So, normally, in algebra, you have some "big ugly thing". To get a handle on this big ugly thing, you start to look at its pieces. There may be too many pieces to be able to sort out "what's what". So we filter out "the dull bits", and get a smaller thing that preserves "the interesting bits", but which is of a more manageable size.
Since we have an isomorphism (of some sort) we can "pull-back" what we learn to the big ugly thing. It's sleight-of-hand, in some sense, but it often helps.
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Euge said:
It's almost right. In the definition of $\mathcal{B}$, the $A/N$ should just be $A$
Perhaps less confusingly:
$\mathcal{B} = \{B\ |\ B \text{ is a submodule of }M/N\}$
to avoid confusion with the $A$ used in the definition of $\mathcal{A}$.
It is clear that $A/N \in \mathcal{B}$, using $A/N$ in the definition of $\mathcal{B}$ presupposes we already know the map $A \to A/N$ is surjective, which actually needs to be shown.
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This is, in fact, the same theorem as in:
http://mathhelpboards.com/linear-abstract-algebra-14/first-isomorphism-theorem-vector-spaces-knapp-theorem-2-27-a-11639.html
but in the more general setting for modules. When you can see it as "obviously true" in several settings simultaneously, you have begun to make real progress (letting our ring be $\Bbb Z$ we obtain the same theorem for abelian groups as a corollary. If our module is, in fact $M = R$, and use the natural $R$ action on an $R/I$-module, we can also obtain the corresponding theorem for rings as a corollary. If $R$ is a field, we obtain the theorem for vector spaces as a corollary. Groups (and monoids, and semigroups) are sort of "the odd man out here", they aren't as "nice". So modules are just *that* cool.).
