3.2. Submodules, Quotient Modules and Isomorphism Theorems.
Let \(M\) be a \(R\)-module. Then a set \(N \subset M\) is said to be a submodule of \(M\) if \(N\) is also a \(R\)-module.
\textcolor{MidnightBlue}{What do we need for \(N \subset M\) to be a submodule?} For \(N\) to be a submodule,
- \(N\) needs to be an nonnempty abelian group
- Axioms (a) - (d) in Definition 1.1 must be satisfied
- \(N\) needs to be closed under multiplication of \(R\). That is, \(\cdot : R \times M|_{N} \to N\), where \(M|_N\) is \(M\) restricted to \(N\) (namely, just \(N\)).
However, since \(N \subset M\), axioms (a) - (d) are already satisfied for \(N\). In addition, if \(N\) is a nonempty subgroup of \(M\) then it is automatically abelian. Thus \(N\) is a \(R\)-submodule of \(M\) if \(N\) is a subgroup of \(M\) and \(N\) is closed under multiplication of elements from \(R\). This leads us to the following submodule test.
[(Submodule Test.)] Let \(M\) be an \(R\)-module and \(N \subset M\) be nonempty. Then \(N\) is an \(R\)-submodule of \(M\) if and only if \(an_1 + bn_2 \in N\) for all \(n_1, n_2 \in N\) and \(a, b \in R\).
(\(\implies\)) If \(N\) is an \(R\)-submodule of \(N\) then obviously \(an_1 + bn_2 \in N\) for all \(n_1, n_2 \in N\) and \(a, b \in R\).
(\(\impliedby\)) Suppose \(an_1 + bn_2 \in N\) for all \(n_1, n_2 \in N\) and \(a, b \in R\). First observe that \(N\) is nonempty. Now setting \(a = 1\) and \(b = -1\) we see that \(n_1 - n_2 \in N\) for all \(n_1, n_2 \in N\), and thus by the subgroup test we see that \(N\) is a subgroup of \(N\).
Since \(an_1 + bn_2 \in N\) for all \(a, b \in R\) we see that \(N\) is closed under multiplication of elements of \(R\).
Since \(N\) is an abelian subgroup of \(M\) and is closed under multiplication of elements of \(R\), we see that \(N\) is an \(R\)-submodule as desired.
\noindent Example.\ An immediate example we can create from our previous discussions the fact that if \(f: M \to N\) is an \(R\)-module homomorphism then
- 1. \(\ker(f)\) is an \(R\)-submodule of \(M\).
- 2. \(\im(f)\) is an \(R\)-submodule of \(N\).
As we saw in group and ring theory, arbitrary intersections of subgroups or subrings resulted in subgroups and subrings. Thus the following theorem should be of no surprise.
Let \(R\) be a ring and \(M\) an \(R\)-module. If \(\{N_\alpha\}_{\alpha \in \lambda}\) be a set of \(R\)-submodules of \(M\), then \(N = \bigcap_{\alpha \in \lambda}N_{\alpha}\) is a submodule of \(M\).
First observe that \(N = \bigcap_{\alpha \in \lambda}N_{\alpha}\) is nonempty, since \(0 \in N_\alpha\) (the identity) for all \(\alpha \in \lambda\). Thus for any \(n_1, n_2 \in N\) we know that \(n_1, n_2 \in N_\alpha\) for all \(\alpha \in \lambda\). Since each such \(N_\alpha\) is an \(R\)-submodule, we know that \(an_1 + bn_2 \in N_\alpha\) for all \(\alpha \in \lambda\) for any \(a, b \in R\). Hence, \(an_1 + bn_2 \in N\) for all \(a, b \in R\), proving that \(N\) is an \(R\)-submodule as desired.
Note that what ring \(R\) is under discussion, we will just state a \(R\)-submodule as simply a submodule. \ \ Quotient Modules. \ \ As we discovered quotient groups in group theory and quotient rings in ring theory, it should again be no surprise that we can formalize the concept of quotient modules.
In group theory, a quotient group \(G/H\) only made sense if the group \(H\) being quotiened out was normal to \(G\). This guaranteed that our desired group operation in the quotient group worked and made sense as desired. In ring theory, a quotient ring \(R/I\) only made sense if the ring \(I\) being quotiened out was an ideal of \(R\). Since we wanted \(R/I\) to be a ring, we needed not only addition but multiplication to be well-defined, but well-definedness only worked when \(I\) was an ideal.
In both cases, we couldn't quotient out just any subgroup or a subring to get a quotient group or quotient ring. They had to be special subsets (e.g. normal groups, ideals). However, in module theory, it does happen to be the case that we can just quotient out a submodule to get a quotient module.
\textcolor{purple}{ To define a quotient module, we first consider an \(R\)-module \(M\) and a submodule \(N\) of \(M\). To turn \(R/N\) into an \(R\)-module, we first turn this into an abelian group, which we can perfectly do since \(N\) is a subgroup of \(M\), an abelian group, so \(M/N\) makes sense. A result from group theory tells us that if \(M\) is abelian then \(M/N\) is abelain. \ \indent Next, to turn this into an \(R\)-module we define scalar multiplication as
where \(r \in R\), and multiplication of elements as
As always, when defining a quotient object we're worried about the ability of our multiplication to preserve equivalence of elements. This is usually where we run into trouble in group theory or ring theory, in which case we modify the set \(N\) which we're quotienting out. In group theory, we'd turn \(N\) into normal group, and in ring theory we'd turn \(N\) into an ideal. Here we'll leave \(N\) alone, since it works out in the end. \ \indent Thus suppose that
that is, \(m = m' + n\) for some \(n \in N\). Then to check if our multiplicaton is well-defined, we observe that for \(a \in R\)
and since \(N\) is a submodule, it is closed under scalar multiplication of elements of \(R\). Hence, \(an \in N\), so that
Thus we see that \(am + N = am' + N\), so that our scalar multiplication is well-defined. } This leads to the following definition.
Let \(R\) be a ring and \(M\) an \(R\)-module. If \(N\) is a submodule of \(M\), then we defined \(M/N\) to be the quotient \(R\)-module of \(M\) with respect to \(N\). As we showed earlier, this is in fact an \(R\)-module.
As before, it should be no surprise that the Noether Isomorphism Theorems apply to modules as well. In fact, the Noether Isomorphism Theorems were first introduced by Emmy Noether for modules; not through groups or for rings. The Isomorphism Theorems hold for groups and rings since abelian groups and rings are special cases of modules.
First, we introduce two homomorphisms which seem as if they are so stupidly simple that they don't even deserve a definition; yet, they do.
Let \(R\) be a ring and \(M\) and \(N\) be \(R\)-module homomorphisms. Then we define the following \(R\)-module homomorphisms.
- 1. The map \(\pi: M \to M/N\) given by
is said to be the projection map. Note that \(\pi\) is surjective, and that \(\ker(\pi) = N\) (since \(m + N = N\) if and only if \(m \in N\).) * 2. The map \(i: M/N \to M\) given by
is known as the inclusion map. More generally, if \(M' \subset M\), the inclusion map can also be defined as \(i: M' \to M\) where
for all \(m' \in M'\). Note that \(i\) is injective, and in the first case \(\im(i) = M/N\cup \{0\}\) and in the second case \(\im(i) = M'\).
[(First Isomorphism Theorem)] Let \(R\) be a ring and \(M\) and \(N\) be \(R\)-modules. If \(f: M \to N\) is an \(R\)-module homomorphism, then
\vspace{-0.8cm}
The proof is the same as before. Define the map \(\phi: M/\ker(f) \to N\) as
\textcolor{NavyBlue}{We quickly show that this is well-defined.} If \(m + \ker(f) = m' + \ker(f)\) for some \(m, m' \in M\), then \(m = m' + k\) for some \(k \in K\). Therefore,
\textcolor{NavyBlue}{Next, we show this is in fact an \(R\)-module homomorphism.} Linearity is obvious, so we check the second criterion. Now for any \(a \in R\) we see that
where we pulled the \(a\) outside from \(f(am)\) to make \(af(m)\) from the fact that \(f\) is an \(R\)-module homomorphism.
\textcolor{NavyBlue}{Now we make two observations.} First, we see that there is a one-to-one correspondence between \(M/\ker(f) \to \im(f)\). Second, this implies that \(\phi\) is an isomorphism between the two modules, so that
as desired.
[(Second Isomorphism Theorem.)] Let \(R\) be a ring and \(M\) and \(N\) and \(P\) be submodules of \(M\). Then
\vspace{-0.8cm}
\begin{minipage}{0.35 \textwidth} \begin{figure}[H]
\end{figure} \end{minipage} \hfill \begin{minipage}{0.6\textwidth} The diagram on the left is the same one we used in group theory and ring theory. That is, the second isomorphism theorem can still be described using the diamond diagram. \end{minipage}
Construct the projection map \(\pi : M \to M/P\) and let \(\pi'\) be the restriction of \(\pi\) to \(N\). Then we see that \(\ker(\pi') = N \cap P\), while
Thus by the First Isomorphism Theorem we have that
as desired.
[(Third Isomorphism Theorem)] Let \(R\) be a ring and \(M\) an \(R\)-module. Suppose \(N\) and \(P\) submodules such that \(P \subset N\). Then
\vspace{-0.8cm}
Construct the map \(f: M/P \to M/N\) by defining \(f(m + P) = m + N\) where \(m+P \in M/P\) and \(m + N \in M/N\). First observe that this is a surjective mapping since \(P \subset M\), so the correspondence \(m + P \to m + N\) will cover all of \(M/N\).
Now observe that
Therefore, by the First Isomorphism Theorem
as desired.
[(Fourth Isomorphism Theorem)] Let \(R\) be a ring and \(M\) an \(R\)-module. Suppose \(N\) is a submodule of \(M\). Then every submodule of \(M/N\) is of the form \(P/N\) where \(N \subset P \subset M\). Another way to understand this statement is to realize there is a one to one correspondence between the submodules of \(M\) containing \(N\) and the submodules of \(M/N\).