Definition: Let \(R\) be a ring (for now, not necessarily commutative and not necessarily having a unit). A left \(R\)-module is an abelian group \(M\) together with a map \(R\times M\to M\) (written \((r,m)\mapsto rm\)) such that:
If \(R\) has a unit element \(1\), we also require \(1m=m\) for all \(m\in M\).
A right module is defined by a map \(M\times R\to M\) and written \((m,r)\mapsto mr\) and satisfying the property \[ (m r_1)r_2 = m(r_1 r_2). \]
If \(R\) is not commutative, these really are different, since for a left module:
while for a right module
If \(R\) is commutative, and \(M\) is a left \(R\)-module, then we can define a right \(R\) module \(M'\) with the same underlying abelian group \(M\) and by defining \(m' r=(r m)'\). This works because
\[ (m'r_1) r_2 = (r_1 m)'r_2 = (r_2(r_1 m))'=((r_2 r_1)m)' =((r_1 r_2)m)' = m'(r_1 r_2) \]
If \(R\) is a field, then a left (or right) \(R\)-module is the same as a vector space.
If \(M\) is an abelian group, and \(R\) is a ring, then a left \(R\)-module structure on \(M\) is the same as a ring map
\[R\to \End (M).\]
If \(\phi_r\) is the endomorphism associated to \(r\in R\), then \(rm=\phi_{r}(m)\). The associativity comes from defining the ring structure on \[\End (M)\] as the usual composition of functions: \[ \phi_{r_1 r_2}=\phi_{r_1}\circ\phi_{r_2}. \]
Definition: If \(M\) is a left \(R\)-module, then a submodule \(N\) of \(M\) is a subgroup with the property that, if \(n\in N\), then \(rn\in N\) for all \(r\in R\).
Observation: A ring \(R\) is a left module over itself by ring multiplication. The (left) ideals of \(R\) are exactly the left submodules of \(R\).
If \(F\) is a field and \(n>1\), let \(R=M_{n}(F)\) be the \(n\times n\) matrix ring over \(F\). The matrices with arbitrary first column and zeros elsewhere form a left ideal \(J\) and therefore a left submodule of \(R\) as left \(R\)-module. But \(J\) is not a right \(R\)-submodule.
A field \(F\) is a one-dimensional vector space over itself, and a commutative ring \(R\) is a module (left and right) over itself with the ideals of \(R\) being the submodules.
Let \(R\) be a ring with unity and let \(n\ge 1\) be a positive integer. Then
\[ R^{n}=\lbrace (r_1,\ldots, r_n) : r_{i}\in R\mathrm{\ for\ } i=1,\ldots,n\rbrace \]
is an \(R\) module with componentwise addition and multiplication given by \(r(r_1,\ldots, r_n)=(rr_1,\ldots, rr_n)\).
This is called the free \(R\)-module of rank \(n\).
More generally, if \(R\subset S\) is a subring, and \(M\) is an \(S\)-module, then it is an \(R\)-module. This is called restriction of scalars.
Let \(M\) be an abelian group. Then it is automatically a \(\Z\)-module where we define
\[ nx=\overbrace{x+x+\cdots+x}^{n}. \]
Furthermore, given any \(\Z\)-module, it must be the case that
\[ nx = (\overbrace{1+1+\cdots+1}^{n})x = \overbrace{x+x+\cdots+x}^{n}. \]
(Note: this is why we require \(1x=x\) when \(R\) is a ring with unity in the module axioms).
Further, submodules of \(M\) (as \(\Z\)-module) are just the subgroups of \(M\) (as abelian group).
Suppose that \(M\) is a left \(R\) module and \(I\subset R\) is a two-sided ideal with the property that, for all \(y\in I\), and all \(x\in M\), we have \(yx=0\). In this case we say that \(I\) annihilates \(M\) or that \(IM=0\).
With this hypothesis, we may view \(M\) as an \(R/I\) module by defining \((r+I)m=rm\) for any coset representative \(r+I\in R/I\). This is well-defined since two different coset representatives \(r,r'\) satisfy \(r'=r+i\) for some \(i\in I\) and therefore \(r'm=(r+i)m=rm\) since \(im=0\).
If \(M\) is an abelian group and \(m\in Z\) is a positive integer such that \(mM=0\), then \(M\) can be viewed as a module over \(\Zn{m}\) by this process.
This operation is a special case of a general operation called base change or extension of scalars that we will study in more detail later.
Let \(F\) be a field, let \(V\) be a vector space over \(F\), and let \(T:V\to V\) be an \(F\)-linear transformation. Define a homomorphism
\[ F[x]\to \End(V) \]
by sending
\[x^{k} \mapsto T^{k}=\overbrace{T\circ T\circ\cdots\circ T}^{n}.\]
This construction makes \(V\) into a module for \(F[x]\) which depends on the choice of the linear transformation \(T\).
Let \(V=F^{2}\) and let \(T\) be the linear transformation given by the matrix
\[ T = \begin{pmatrix} 0 & 1 \\ 1 & 1\end{pmatrix} \]
If \(e_0\) and \(e_1\) are the standard basis elements of \(F^{2}\) then
\[\begin{aligned} Te_0&=&e_1 \\ T^2e_0=Te_1=e_0+e_1=e_0+Te_0&=&(1+T)e_0\\ \end{aligned} \]
Therefore \((T^2-T-1)e_0=0\) and
\[(T^2-T-1)e_1=(T^2-T-1)Te_0=T(T^2-T-1)e_0=0\]
so the polynomial \(x^2-x-1\) is in the kernel of the map from \(F[x]\to \End(V)\).
By the base change construction above this means that \(V\) can be viewed as a module over \(F[x]/(x^2-x-1)\).
We saw above that, given an \(F\)-vector space \(V\) with a linear transformation \(T\), we get an \(F[x]\) module where \(x\) acts on \(V\) through \(T\).
Conversely, suppose that \(M\) is an module over \(F[x]\). Then \(M\) is an \(F\) vector space (via the restriction of scalars from \(F[x]\) to \(F\)). Furthermore, the element \(x\in F[x]\) acts on \(M\) as an \(F\)-linear transformation because that’s what the module axioms amount to.
Therefore there is an equivalence between
\[ \begin{xy}\xymatrix{ \lbrace F[x]-\mathrm{modules}\rbrace\ar@2{<->}[d]\\ \lbrace \mathrm{vector\ spaces\ }V\mathrm{\ over\ }F\mathrm{\ with\ a\ given\ linear\ map\ }T:V\to V\rbrace \\} \end{xy} \]
In the correspondence above, a submodule of an \(F[x]\) module \(M\) corresponds to a subspace \(W\subset V\) that is preserved by \(T\), meaning \(TW\subset W\).
Thus, not all subspaces of \(V\) correspond to submodules.
In the example given earlier, the only \(T\)-stable proper subspace of \(V\) is the zero subspace.
If we consider instead the linear map on \(F^{2}\) satisfying \(Ue_0=0\) and \(Ue_1=e_0\), then the one dimensional subspace spanned by \(e_0\) is \(U\)-stable and \(F^{2}\) viewed as an \(F[x]\) module via \(U\) has a submodule corresponding to that subspace.
Proposition: A subset \(N\) of a left \(R\)-module \(M\) is a submodule if it is nonempty and, for all \(x,y\in N\) and \(r\in R\), we have \(x+ry\in N\). Alternatively, if \(N\) is a subgroup of the abelian group \(M\) and \(rN\subset N\) for all \(r\in R\) then \(N\) is a submodule.
Definition: Let \(R\) be a commutative ring with unity. An \(R\)-algebra is a (not necessarily commutative) ring \(S\) with a ring homomorphism \(f:R\to S\) carrying \(1_R\) to \(1_S\) such that \(f(R)\) is in the center of \(S\).
The polynomial ring \(F[x]\) is an \(F\)-algebra, as is the matrix ring \(M_{n}(F)\) where the homomorphism \(f:F\to M_{n}(F)\) embeds \(F\) as the diagonal matrices. More generally, any \(F\)-algebra \(A\), where \(F\) is a field, contains \(F\) in its center and the identites of \(A\) and \(F\) are the same.
The ring \(\Zn{p}\) is a \(\Z\)-algebra. In fact any ring \(S\) with \(1\) is a \(\Z\) algebra by the map sending \(n\in\Z\) to \(n 1_S\).
The ring \(\Q[x]\) is a \(\Z[x]\) algebra.
We typically omit the explicit map \(f\) and just think of \(R\) as “contained in” \(A\); this can be misleading since \(f\) doesn’t need to be injective, but it works in practice.
Definition: A map of \(R\)-algebras \(f:A\to B\) is a ring homomorphism that is \(R\)-linear in the sense that \(f(ra)=rf(a)\) for all \(r\in R\) and \(a\in A\).
Any homomorphism of rings with unity is a \(\Z\)-algebra morphism.
Definition: Let \(R\) be a ring and let \(M\) and \(N\) be (left) \(R\)-modules. A function \(f:M\to N\) is an \(R\)-module homomorphism if:
Note that, if \(R\) is a field, then \(M\) and \(N\) are vector spaces and an \(R\)-module homomorphism is just a linear map.
A module isomorphism is a bijective homomorphism.
We let \(\Hom_{R}(M,N)\) denote the set of \(R\)-module homomorphisms from \(M\) to \(N\).
Let \(R\) be a ring and let \(M\) and \(N\) be \(R\)-modules. Let \(f:M\to N\) be a homomorphism.
Let \(M\) be an \(R\) module and let \(N\subset M\) be a submodule.
Definition: Let \(M/N\) be the quotient abelian group. Then \(M/N\) is an \(R\)-module where \(R\) acts on cosets by
\[ r(x+N)=rx+N. \]
This is called the quotient module of \(M\) by \(N\).
The \(R\)-module structure is well defined because if \(x+N=y+N\), then \(x=y+n\) for some \(n\in N\), and \(rx = ry+rn\). Since \(N\) is a submodule, \(rn\in N\) so \(rx+N=ry+N\).
Notice that \(N\) can be any submodule, there is no “normality” condition like for groups.
There is always a “projection” homomorphism \(\pi:M\to M/N\) defined by \(\pi(m)=m+N\) which has kernel \(N\).
If \(A\) and \(B\) are submodules of a module \(M\), then \(A+B\) is the smallest submodule of \(M\) containing both \(A\) and \(B\). Alternatively it is:
\[ A+B=\lbrace a+b : a\in A, b\in B\rbrace \]
Let \(M\), \(N\), and \(K\) be \(R\) modules, and let \(f:M\to K\) be a homomorphism with \(N\subset\ker(f)\). Then there is a unique homomorphism \(\overline{f}:M/N\to K\) making this diagram commutative:
\[ \begin{xy} \xymatrix { M\ar[rd]^{f}\ar[d]^{\pi} & \\ M/N\ar[r]_{\overline{f}} & K\\ } \end{xy} \]
The isomorphism theorems for abelian groups give isomorphism theorems for modules.
The proofs of all of these facts are found by checking that the group isomorphisms respect the action of the ring \(R\).
The set \(\Hom_{R}(M,N)\) is an abelian group: \((f+g)(m)=f(m)+g(m)\) and the zero map is the identity.
If \(R\) is commutative then \(\Hom_{R}(M,N)\) is an \(R\)-module if we set \((rf)\) to be the function \((rf)(m)=r(f(m))=f(rm)\). We need \(rf\) to be a module homomorphism, which means we need:
\[ (rf)(sm)=s(rf)(m). \]
This works out ok if \(R\) is commutative since
\[ (rf)(sm)=f(rsm)=f(srm)=s(f(rm))=s((rf)(m)) \]
but it fails if \(R\) is not commutative.
The set \(\Hom_{R}(M,M)\) is a ring with multiplication given by composition. The identity map gives an identity for this ring.
If \(R\) is commutative then, given \(r\in R\), we have an element \(\phi_r\in\Hom_{R}(M,M)\) given by \(\phi_r(m)=rm\). This is a homomorphism because
\[ \phi_r(sm)=rsm=srm=s\phi_r(m) \]
but this fails in general if \(R\) is not commutative. Thus, if \(R\) is commutative, \(\Hom_{R}(M,M)\) is an \(R\)-algebra.
If \(M=R^{n}\), then \(\Hom_{R}(M,M)\) is the ring of \(n\times n\) matrices with entries from \(R\).