We call the operations \(M\times M\xrightarrow{+}M\) and \(R\times M\xrightarrow{\cdot} M\) module addition and scalar multiplication by \(R\), respectively.
If \(I\) is a left ideal of \(R\text{,}\) then \(I\) is a left \(R\)-module, where module addition is ring addition, and scalar multiplication is (left) multiplication by elements of \(R\text{.}\) Note that this scalar multiplication is well-defined precisely because \(I\) is closed under multiplication on the left by arbitrary elements of \(R\text{.}\)
Let \(I\) be a left ideal of \(R\text{.}\) The quotient group \(R/I\) is an \(R\)-module, where module addition is the usual coset addition and scalar multiplication is defined as
for all \(r\in R\) and \(s+I\in R/I\text{.}\) Once again, the fact that this operation is well-defined is a result of \(I\) being closed under multiplication on the left by elements of \(R\text{.}\) Indeed, assume \(s+I=s'+I\text{,}\) so that \(s'=s+I\text{.}\) We then have
If \(R\) and \(S\) are rings and \(\phi\colon R\rightarrow S\text{,}\) then \(S\) has the structure of an \(R\)-module. Module addition is just ring addition in \(S\text{,}\) and scalar multiplication by elements of \(R\) is defined as follows:
That this satisfies the scalar multiplication axioms follows in a straightforward manner from (a) the fact that \(\phi\) is a ring homomorphism, (b) the fact that \(R\) is a ring, and (c) the fact that \(S\) is a ring. You are invited to verify this yourself.
Let \(F\) be a field. An \(F\)-module \(M\) is the same thing as a vector space over \(F\text{.}\) Indeed, modules should be considered as a straightforward generalization of a vector space, where the base field of the vector space is replaced with an arbitrary ring.
A \(\Z\)-module is the same thing as an abelian group. Indeed, if \(M\) is a \(\Z\)-module, then in particular \(M\) is a an abelian group with respect to its module addition. Conversely, given an abelian group \((G,+)\text{,}\) it is easy to see (using an induction argument) that the operation
\begin{align*}
\Z\times G\amp\rightarrow G \\
(n,g) \amp \mapsto ng
\end{align*}
is the unique scalar multiplication by \(\Z\) that can be defined on \(G\text{.}\) Recall that for an abelian group \((G,+)\text{,}\) given \(n\in \Z\) and \(g\in G\text{,}\) we define \(ng\) is defined to be the \(n\)-fold addition of \(g\) with itself if \(n\) is positive, the \(\abs{n}\)-fold addition of \(-g\) with itself if \(n\) is negative, and \(0_G\) if \(n=0\text{.}\)
In practice, to verify a subset of a module \(M\) is a submodule, instead of using the definition above directly, we use a procedure that is pretty much an analogue of the procedure for showing something is a subspace in the context of vector spaces.
Show that for all \(m_1,m_2\in M\) and \(r,s\in R\text{,}\) if \(m_1,m_2\in N\text{,}\) then \(rm_1+sm_2\in N\text{.}\) Using logical shorthand, prove the following implication:
\begin{align*}
m_1,m_2\in N \amp \implies rm_1+sm_2\in N\text{.}
\end{align*}
Taking \(r=0\) and \(s=-1\) and using the fact that \((-1)m=-m\) in any \(R\)-module (prove this yourself!), we see that \(N\) is closed under additive inverses.
Let \(R\) be a ring. An \(R\)-module homormorphism between two \(R\)-modules \(M\) and \(N\) is a group homomorphism \(\phi\colon M\rightarrow N
\text{,}\) satisfying \(\phi(rm)=r\phi(m)\) for all \(r\in R\text{.}\)
The set of all \(R\)-module homomorphisms from \(M\) to \(N\) is denoted \(\Hom_R(M,N)\text{.}\) An \(R\)-module homomorphism from \(M\) to itself is called an \(R\)-module endomorphism. We write \(\End_R(M)=\Hom_R(M,M)\text{:}\) i.e., \(\End_R(M)\) is the set of all endormorphisms of \(M\text{.}\)
Let \(M\) and \(N\) be \(R\)-modules. Given \(\phi,\psi\in \Hom_R(M,N)\) and \(r\in R\text{,}\) we define \(R\)-module homomorphisms \(\phi+\psi\) and \(r\phi\) as follows:
Additionally, \(\End_R(M)\) is a ring with respect to addition and composition, and the map \(\phi\colon R\rightarrow \End_R(M)\) is a ring homomorphism. If \(R\) is commutative, the homomorphism \(\phi\) gives \(\End_R(M)\) the structure of an \(R\)-algebra.
If \((M_i)_{i\in I}\) is a family of submodules of \(M\) indexed by the nonempty set \(I\text{,}\) then \(\bigcap_{i\in I}M_i\) is a submodule of \(M\text{.}\)
Given a subset \(A\) of \(M\text{,}\) let \(\mathcal{M}=\{M'\subseteq M\mid M' \text{ a submodule containing} A\}\text{,}\) and define \((A)=\bigcap_{M'\in \mathcal{M}}M'\text{,}\) the intersection of all submodules of \(M\) containing \(A\text{.}\)
\((A)\) is the smallest submodule of \(M\) (with respect to inclusion) containing \(A\text{.}\)
Let \(M\) be an \(R\) module, and let \(A\) a subset of \(M\text{.}\) An \(R\)-linear combination of the elements of \(A\) is an element of \(M\) of the form \(\sum_{i=1}^{n}r_im_i\text{,}\) where \(n\) is a positive integer, \(m_1,m_2,\dots, m_n\in A\text{,}\) and \(r_1,r_2,\dots, r_n\in R\text{.}\)
Given an \(R\)-module \(M\) and a subset \(A\subseteq M\text{,}\) the set \((A)\) defined as the intersection of all submodules of \(M\) containing \(A\) is called the submodule of \(M\) generated by \(A\).
The concrete description of \((A)\) as the set of \(R\)-linear combinations of elements of \(A\) is the \(R\)-module analogue of the span of a collection of vectors. Later we will see an analogue of the the notion of linear independence.
Let \(N\) be a submodule of the \(R\)-module \(M\text{.}\) We define a scalar multiplication on the quotient group \(M/N\) as follows: given \(r\in R\text{,}\)\(m+N\in M/N\text{,}\)\(r(m+N)=rm+N\text{.}\) This is a well-defined operation, and \(M/N\text{,}\) together with coset addition and this scalar multiplication, is an \(R\)-module called the quotient module of \(M\) by \(N\text{.}\)
Theorem2.17.12.The isomorphism theorems for modules.
Let \(M\) and \(N\) be \(R\)-modules.
Let \(M'\) be a submodule of the \(R\)-module \(M\text{,}\) and let \(\pi\colon M\rightarrow M/M'\) be the quotient map. Given any \(\phi\in \Hom_R(M,N)\) satisfying \(M'\subseteq \ker \phi\text{,}\) there is a unique \(\overline{\phi}\in \Hom_R(M/M', N)\) satisfying \(\overline{\phi}=\overline{\phi}\circ \pi\text{:}\) namely, the map \(\overline{\phi}\colon M/M'\rightarrow N\) defined as \(\overline{\phi}(m+M')=\phi(m)\text{.}\)
If \(\phi\colon M\rightarrow N\) is an \(R\)-module homomorphism, then \(M/\ker\phi\cong \im \phi\) as \(R\)-modules, and the map \(\overline{\phi}\colon M/\ker\phi\rightarrow \im N\) defined as \(\overline{\phi}(m+\ker\phi)=\phi(m)\) is an isomorphism.
In the special case where \(M'\) is a submodule of \(M\) and \(\pi\colon M\rightarrow M/M'\) is a quotient map, we write \(\pi(M'')=M''/M'\) for any submodule \(M''\) of \(M\text{.}\) Thus, in this case, if \(M''\) is a submodule of \(M\) containing \(M'\text{,}\) the isomorphism above is written
defines a bijection between set of submodules of \(M\) containing \(\ker\phi\) and the set of a submodules of \(N\text{,}\) with inverse given by the map
\begin{align*}
N' \amp \mapsto \phi^{-1}(N')\text{.}
\end{align*}
Moreover this bijection preserves inclusions, intersections, sums, and quotients of submodules.
