Suppose that \(R\) is a ring and \(M\) is an \(R\)-module. Let \(N_1,\ldots, N_{k}\) be submodules of \(M\). Then the sum \(N_1+\ldots+N_{k}\) is the collection
\[ N_1+\ldots+N_{k}=\lbrace n_1+\cdots+n_{k} : n_{i}\in N_{i}\rbrace \]
It is a submodule of \(M\) and the smallest submodule containing all the \(N_{i}\).
One can also consider infinite collections of submodules:
\[ \sum_{i\in I} N_{i} = \lbrace \sum_{j\in J} n_{j} : n_{j}\in N_{j},\ J\subset I\mathrm{\ finite\ }\rbrace \]
Suppose \(A\subset M\). Then the submodule \(RA\) of \(M\) generated by \(A\) is the smallest submodule of \(M\) containing \(A\). In practice it is the collection
\[ RA = \lbrace r_1 a_1 + \cdots + r_k a_k : r_1,\ldots, r_k\in R, a_1,\ldots, a_k\in A, k\in\Z, k\ge 0\rbrace \]
In linear algebra, we would say that \(RA\) is the submodule of \(M\) that is spanned by \(A\) and this terminology can be used here as well.
We can also say that \(RA\) is the set of (finite) \(R\)-linear combinations of elements of \(A\).
Suppose that \(V\) is a \(\Q\)-vector space of dimension \(n\) and \(w_1,\ldots, w_k\) are a set of vectors in \(V\).
Since \(V\) is also a \(\Z\) module (by “restriction of scalars”) we can consider the sub-\(\Z\)-module of \(V\) generated by the \(w_{i}\). This is all \(\Z\)-linear combinations of the \(w_i\).
For example if \(V=\Q^{2}\) and \(A=\lbrace w_1, w_2\rbrace\) are the standard basis elements then \(\Z A\) is the subset of \(V\) of vectors with integer coefficients in the standard basis.
Definition: An \(R\)-module \(M\) is finitely generated if there is a finite subset \(A\subset M\) such that \(RA=M\).
Note that \(\Q\) is finitely generated as a \(\Q\)-module (in fact it’s generated by one element) but not as a \(\Z\)-module.
For vector spaces, finitely generated means finite dimensional. A generating set is the same as a spanning set.
A set \(m_1,\ldots, m_k\) in an \(R\)-module \(M\) is linearly independent if, whenever \(\sum r_i m_i=0\), all \(r_i=0\).
For vector spaces, a maximal linearly independent set (meaning a linearly independent set which becomes dependent when any nonzero element is added to it) automatically spans the vector space, and we call this a basis.
For modules, this fails. Consider \(\Z^{2}\) and let \(e_1=[2,0]\) and \(e_2=[0,2]\). If \(e=[a,b]\) then \[ 2e-ae_1-be_2=0 \] so \(e_1, e_2\) is a maximal linearly independent set. But they don’t generate all of \(\Z^{2}\).
Definition: An \(R\) module \(M\) is cyclic if it is generated by one element: \(M=Ra\) for some \(a\in M\).
If \(R=\Z[i]\), then \((1+i)R\) is a cyclic module for \(R\) generated by \((1+i)\). But if we view \((1+i)R\) as a \(\Z\)-module inside the \(\Z\)-module \(R=\Z+\Z i\) then \((1+i)R\) is generated over \(\Z\) by \(1+i\) and \((1+i)i=i-1\); it is not cyclic as a \(\Z\)-module.
Proposition: Let \(M\) be a cyclic \(R\)-module. Then \(M\) is isomorphic to \(R/I\) where \(I\) is a left ideal of \(R\).
Proof: Let \(m\in M\) generate \(M\). Consider the map \(f: R\to M\) defined by \(f(r)=rm\). This is a module homomorphism since
\[ f(r_1 r_2)=r_1 r_2 m = r_1 (r_2 m) = r_1 f(r_2c). \]
(Remember that we are thinking of \(R\) here as an \(R\)-module, not a ring.)
The kernel of the map \(f(r)=rm\) is the set \(I=\lbrace r\in R : rm=0\rbrace\).
This is a left ideal since if \(rm=0\) then \(srm=0\) for all \(s\in R\).
Since \(M\) is cyclic, the map \(f\) is surjective.
Therefore by the isomorphism theorem \(M\) is isomorphic to \(R/I\).
Recall that a module \(M\) for \(F[x]\) is the same as an \(F\)-vector space \(V\) together with a linear map \(T:V\to V\).
If \(M\) is cyclic then there is an \(m\in M\) so that every \(m'\in M\) is given by \(p(x)m\) for some \(p(x)\in F[x].\)
This means that that there is a vector \(v\in V\) so that every vector \(v'\in V\) is of the form \(p(T)v\). In other words, the set \(v, Tv, T^2v,\ldots, T^{n}v,\ldots\) spans \(V\).
If \(V=F^{2}\) and \(T\) satisfies \(Te_{1}=0\) and \(Te_{2}=e_{2}\) then \(V\) is not cyclic.
If \(Te_{1}=0\) and \(Te_{2}=e_{1}\) then \(V\) is cyclic and generated by \(e_{2}\). Also \(T^2e_2=0\) and so as an \(R\)-module \(V\) is isomorphic to \(F[x]/(x^2)\).
Suppose that \(M_{1},\ldots, M_{k}\) are \(R\) modules. The direct product \(M_1\times\cdots\times M_{k}\) of the \(M_{i}\) is the set of “vectors” \((m_1,\ldots, m_k)\) with \(m_{i}\in M_{i}\). Addition and multiplication by \(R\) are done componentwise.
Suppose that \(M\) is an \(R\)-module and \(N_1,\ldots, N_{k}\) are submodules of \(M\). There is a module homomorphism
\[ N_1\times\cdots\times N_{k}\to N_1+\cdots N_{k}\subset M \]
defined by sending \((n_1,\ldots, n_k)\to n_1+\cdots n_k\).
Definition: The sum map above is an isomorphism if and only if either of the following two conditions are satisfied:
If \(M\) is isomorphic to \(N_1\times \cdots \times N_{k}\) via the sum map, we say that
\[ M = N_1\oplus N_2\oplus \cdots \oplus N_{k} \]
and say that \(M\) is the internal direct sum of the \(N_{i}\).
Suppose that \(I\) is a set and \(M_{i}\) is an \(R\)-module for each \(i\in I\).
The direct product \(\prod_{I} M_{i}\) is the collection of all functions \(f:I\to \cup_{i\in I} M_{i}\) such that \(f(i)\in M_{i}\). It is an \(R\)-module: \((f+g)(i)=f(i)+g(i)\) and \((rf)(i)=r(f(i))\).
The direct sum \(\oplus_{I} M_{i}\) is the submodule of \(\prod_{I} M_{i}\) consistsing of functions \(f\) with the additional property that there is a finite subset \(J\subset I\) such that \(f(i)=0\) unless \(i\in J\).
Notice that if \(I\) is finite then these two things are the same.
Suppose that \(I=\N\), the natural numbers, and \(M_{i}\) is a family of \(R\)-modules indexed by \(I\). Then:
Notice that, if each \(M_{i}\) is countable, then so is \(\oplus_{i\in I}M_{i}\), but \(\prod_{i\in I}M_{i}\) is not.
Definition: A module \(M\) is free on a set \(A\) of generators if, for every nonzero element \(m\) of \(M\), there are unique nonzero \(r_1,\ldots, r_k\) in \(R\) and elements \(a_1,\ldots, a_k\) in \(A\) such that
=======Definition: A module \(M\) is free on a set \(A\) of generators if, for every nonzero element \(m\) of \(M\), there are unique nonzero \(r_1,\ldots, r_k\) in \(R\) and elements \(a_1,\ldots, a_k\) in \(A\) such that
>>>>>>> 1d46efb62a5e048ffe0959473099170f82f5a9fd\[ m = r_1 a_1 + \cdots + r_k a_k. \]
Such a set \(A\) is called a basis of \(M\), so a module \(M\) is free if it has a basis.
If \(A=\lbrace a_1,\ldots,a_n\rbrace\) is finite, then \(M\) is free on \(A\) if the map
\[ \oplus_{i=1}^{n} R \to M \]
<<<<<<< HEADdefined by \((r_1,\ldots, r_n)\mapsto r_1 a_1+\cdots+r_n a_n\) is an isomorphism. So basically \(M\) is free on a set \(A\) with \(n\) elements if and only if it is isomorphic to \(R^{n}\).
If \(R=\Z\), then \(M=\Z/m\Z\oplus \Z/m\Z\) is not free on \((1,0)\) and \((0,1)\). Every \(m\in M\) is a linear combination \(r_1(1,0)+r_2(0,1)\) for \(r_1,r_2\in\Z\), but \(r_1\) and \(r_2\) are not uniquely determined. In fact \(M\) is not free on any set of generators.
Any vector space over \(F\) is a free \(F\)-module.
A principal ideal in a (commutative) ring is a free module, but a non-principal ideal is not. Consider \(I=(2,1+\sqrt{-5})\subset R=\Z[\sqrt{-5}]\). Choose any two elements of this ideal, say \(x\) and \(y\). Then \(-y\cdot x + x\cdot y=0\) which shows that the map \(R\oplus R\to I\) is not injective. On the other hand we know that the ideal is not principal.
defined by \((r_1,\ldots, r_n)\mapsto r_1 a_1+\cdots+r_n a_n\) is an isomorphism. So basically \(M\) is free on a set \(A\) with \(n\) elements if and only if it is isomorphic to \(R^{n}\).
If \(R=\Z\), then \(M=\Z/m\Z\oplus \Z/m\Z\) is not free on \((1,0)\) and \((0,1)\). Every \(m\in M\) is a linear combination \(r_1(1,0)+r_2(0,1)\) for \(r_1,r_2\in\Z\), but \(r_1\) and \(r_2\) are not uniquely determined. In fact \(M\) is not free on any set of generators.
Any vector space over \(F\) is a free \(F\)-module.
A principal ideal in a (commutative) ring is a free module, but a non-principal ideal is not. Consider \(I=(2,1+\sqrt{-5})\subset R=\Z[\sqrt{-5}]\). Choose any two elements of this ideal, say \(x\) and \(y\). Then \(-y\cdot x + x\cdot y=0\) which shows that the map \(R\oplus R\to I\) is not injective. On the other hand we know that the ideal is not principal.
Let \(A\) be a set. There exists a
module \(F(A)\), called the free
module on \(A\), which contains
\(A\) as a subset.
It satisfies the following property.
Let \(M\) be any module and let \(f:A\to M\) be any map of sets. Then there is a unique module homomorphism \(\Phi:F(A)\to M\) such that the following diagram commutes:
\[ \begin{xy} \xymatrix { A\ar[r]^{\subset}\ar[rd]^{f} & F(A)\ar[d]^{\Phi}\\ & M\\ } \end{xy} \]
If \(V\) is a vector space and \(B\) is a basis, then \(V\) is free on \(B\). A linear map from \(V\to W\) is determined by where you send \(B\). In this situation, \(f:B\to W\) is the map of sets sending the basis of \(V\) to a subset of \(W\), and \(\Phi\) is the resulting linear map.
If \(A\) is any set, then \(F(A)\) is the \(R\)-module of “formal linear combinations of elements of \(A\)”: the set of sums \(\sum r_{i}a_{i}\) over finite collections \(\lbrace a_1,\ldots,a_n\rbrace\) of elements of \(A\).
Alternatively it is the set of functions \(f:A\to R\) that are zero for all but a finite subset of \(A\) with pointwise addition and scalar multiplication.
Any two free modules on the same set are isomorphic via the module map induced by the identity map on \(A\).
Let \(R\) be an integral domain.
Definition: The rank of an \(R\)-module is the maximum number of \(R\)-linear independent elements of \(M\).
Proposition: Let \(M\) be a free \(R\) module of rank \(n\). Then any \(n+1\) elements of \(M\) are linearly dependent. Thus any submodule of \(M\) has rank at most \(n\).
Proof: Let \(m_1,\ldots, m_{n+1}\) be elements of \(M\) and let \(e_1,\ldots, e_n\) be a basis of \(M\). Each \(m_{i}\) is an \(R\)-linear combination of the \(e_i\). We can view the \(m_i\) as vectors in \(F^{n}\) where \(F\) is the fraction field of \(R\). They are linearly dependent in \(F^{n}\), meaning there is a relation \[ \sum f_i m_{i}=0 \] where the \(f_i\) are in \(F\). Clearing denominators gives a relation over \(R\). ## Torsion
Suppose that \(R\) is a ring with unity.
Definition: Let \(M\) be an \(R\)-module. An element \(m\in M\) is a torsion element if \(rm=0\) for some nonzero \(r\in R\). The set of torsion elements in \(M\) is called \(\Tor(M)\).
Lemma: If \(R\) is an integral domain and \(M\) is an \(R\)-module, then the set of torsion elements is a submodule.
Proof: If \(m_1\) and \(m_2\) are torsion, \(r_1 m_1=0\) and \(r_2 m_2=0\), with both \(r_1\) and \(r_2\) nonzero, then \(r_1 r_2 (m_1+m2)=0\) and \(r_1 r_2 (m_1 m_2)=0\), and \(r_1 r_2\) is nonzero since \(R\) is an integral domain.
If \(R\) is an integral domain, an \(R\)-module \(M\) is called torsion-free if \(\Tor(M)=0\).
Any free module is torsion-free, but the converse is false. For example, non-principal ideals in integral domains are not free. This follows from the following lemma.
Lemma: An ideal of \(R\) is free if and only if it is principal.
Proof: \(R\) is a free module of rank \(1\), so a submodule has rank at most \(1\); if it has rank \(1\), it is a principal ideal.