Glossary of Common Terms
Operation/action on a set/object/module
Let \(G\) be a group and let \(S\) be a set. An operation or an action of \(G\) on \(S\) is a homomorphism \[\pi: G \to \text{Perm}(S)\] of \(G\) into the group of permutations of \(S\). We then call \(S\) a \(G\)-set. We denote the permutation associated with an element \(x \in G\) by \(\pi_x\). Thus the homomorphism is denoted by \(x \mapsto \pi_x\). That is, \[ \pi_{g_1 g_2} = \pi_{g_1} \circ \pi_{g_2} \in \text{Perm}(S). \] Given \(s \in S\), the image of \(s\) under the permutation \(\pi_x\) is \(\pi_x(s)\). From such an operation we obtain a mapping \[G \times S \to S\] \[G \times S \ni (x,s) \mapsto \pi_x(s) =: xs \in S.\]
Algebra
p. 121 Lang (2002).
Let \(A\) be a commutative ring. Let \(E\), \(F\) be modules. By a bilinear map \[g: E \times E \to F\] we mean a map such that given \(x \in E\), the map \(g(x, \cdot)\) is \(A\)-linear: \[ g(x,y_1+y_2) = g(x,y_1) + g(x, y_2) \] for all \(y_1, y_2 \in E\) (additive group homomorphism) and \[ g(x,a y_1) = a g(x, y_1), \] for all \(a \in A\) and \(y_1 \in E\), and given \(y \in E\), the map \(g(\cdot, y)\) is also \(A\)-linear.
By an \(A\)-algebra, we mean a module \(E\) together with a bilinear map \(g: E \times E \to E\). We view such a map as a law of composition on \(E\).
We note that the group ring \(A[G]\) (or monoid ring when \(G\) is a monoid) is an \(A\)-algebra, also called the group (or monoid) algebra. Actually the group algebra can be viewed as a special case of the following situation.
Let \(f: A \to B\) be a ring homomorphism such that \(f(A)\) is contained in the center of \(B\), i.e., \(f(a)\) commutes with every element of \(B\) for every \(a \in A\). Then we may view \(B\) as an \(A\)-module (how?), defining the operation of \(A\) on \(B\) by the map \[(a,b) \mapsto f(a)b\] for all \(a \in A\) and \(b \in B\). The axioms for a module are trivially satisfied (why?), and the multiplicative law of composition \(B \times B \to B\) (what is this? \(B\) is a-priori a ring, so it is just the multiplicative law) is clearly bilinear (i.e. \(A\)-bilinear). In this book, unless otherwise specified, by an algebra over \(A\), we shall always mean a ring homomorphism as above. We say that the algebra is finitely-generated if \(B\) is finitely generated as a ring over \(f(A)\).
A non-commutative, non-associative algebra is one for which the law of composition \(g\) is non-commutative: \(g(e_1, e_1) \ne g(e_2,e_1)\) for some \(e_1,e_2 \in E\), and for which the law of composition \(g\) is also not associative: \(g(g(e_1,e_2),e_3) \ne g(e_1, g(e_2,e_3))\) for some \(e_1, e_2, e_3 \in E\).
Module, Submodule
Let \(A\) be a ring. A left-module over \(A\), denoted by \(M\), is an abelian group, usually written additively, together with an operation of \(A\) on \(M\) (viewing \(A\) as a multiplicative monoid by R2), such that, for all \(a,b \in A\) and \(x,y \in M\), we have \((a+b)x = ax + bx\) and \(a(x+y) = ax + ay\).
One can prove that \(a(-x) = -(ax)\) and that \(0x = 0\). By definition of an operation, we have \(1x = x\).
Let \(M\) be a module. By a submodule \(N\) of \(M\) we mean an additive subgroup such that \(AN \subset N\).
For example, consider the subset of \(M( 2 \times 2, \mathbb C)\), the 2 by 2 complex matrices, given by \[V = \{\alpha_1 B_1 + \alpha_2 B_2 + \alpha_3 B_3 \mid \alpha \in \mathbb R^3\},\] where \(B_1 = \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix}\), \(B_2 = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}\), and \(B_3 = \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix}\). Note that \(V\) is an abelian group under matrix addition. Furthermore, there is an operation of the ring \(\mathbb R\) on \(V\), given by the usual multiplication of a matrix by a scalar, and note that this is indeed a well-defined operation because multiplying by a real scalar is an operation which stays inside the group \(V\) (this would not be the case for complex scalars.) Thus, we can view \(V\) as an \(\mathbb R\)-module. If we view \(M( 2 \times 2, \mathbb C)\) as an \(\mathbb R-\)module, then we can say that \(V\) is an \(\mathbb R\)-submodule of \(M( 2 \times 2, \mathbb C)\).
Vector Space
A vector space \(V\) is a module \(V\) over a field \(\mathbb F\).
Free Module
A free module \(M\) is a module which admits a basis or the zero module.
A vector space is always free.
Ring
Basic Definition
p.83 Lang (2002).
A Ring \(A\) is a set, together with two laws of composition called multiplication and addition respectively, and written as a product and as a sum respectively, satisfying the following conditions
(R1) With respect to addition, \(A\) is a commutative group.
(R2) The multiplication is associative, and has a unit element.
(R3) For all \(x,y,z \in A\), we have \((x+y)z = xz + yz\) and \(z(x+y) = zx + zy\). This is called distributivity.
We denote the unit element for addition by \(0\) and the unit element for multiplication by \(1\).
We say \(A\) is commutative if \(xy = yx\) for all \(x,y \in A\).
Ring Homomorphism
p.88 Lang (2002)
By a ring-homomorphism, one means a mapping \(f: A \to B\) where \(A\), \(B\) are rings, and such that \(f\) is a monoid-homomorphism for the multiplicative structures on \(A\) and \(B\) and also a monoid-homomorphism for the additive structure.
Convolution Product
Page 85 Lang Algebra.
Universal Objects
Let \(\mathfrak{C}\) be a category. An object \(P\) of \(\mathfrak{C}\) is called universally attracting if there exists a unique morphism of each object of \(\mathfrak{C}\) into \(P\), and is called universally repelling if for every object of \(\mathfrak{C}\) there exists a unique morphism of \(P\) into this object.
When the context is clear, we call objects \(P\) as above universal. Since a universal object \(P\) admits the identity morphism into itself, it is clear that if \(P\), \(P'\) are two universal objects in \(\mathfrak{C}\), then there exists a unique isomorphism between them.
Example (Lang (2002) Section : Tensor Product.) Let \(R\) be a commutative ring, and let \(E_1, \cdots, E_n\) be \(R\)-modules. Let \(\mathfrak{C}\) be the category of multilinear maps with domain \(E_1 \times \cdots \times E_n\). Given \(f,g \in \text{Ob}(\mathfrak C)\), that is, \[ f: E_1 \times \cdots \times E_n \to F \text{ and } g: E_1 \times \cdots \times E_n \to G, \] for \(R\)-modules \(F\) and \(G\), we define a morphism \(f \to g\) to be an \(R\)-linear map \(h: F \to G\) such that we get a “commutative diagram”: \[ f \circ h = g. \] Let \(\phi: E_1 \times \cdots \times E_n \to M/N\), be the usual multilinear map where \(M\) is the free module generated by \(E_1 \times \cdots \times E_n\) and \(N\) is the submodule generated by elements of a specific form. The definition of a tensor product is that it is a universal object in the category \(\mathfrak{C}.\) Then \(\phi\) is a tensor product. Why? Observe that \(\phi\) is universally repelling. Indeed, if \(g: E_1 \times \cdots \times E_n \to G\) is given, then one can show that there is a unique morphism \(\phi \to g\).
Free abelian group generated by \(S\)
Page 38 Lang.
Let \(S\) be a set. We shall define the abelian group generated by \(S\) as follows. Let \(\mathbb Z\langle S \rangle\) be the set of maps \(\phi: S \to \mathbb Z\) such that \(\phi(x) = 0\) for almost all \(x \in S\). Then \(\mathbb Z \langle S \rangle\) is an abelian group (addition being the usual addition of maps). If \(k\) is an integer and \(x\) is an element of \(S\), we denote by \(k \cdot x\) the map \(\phi\) such that \(\phi(x) = k\) and \(\phi(y) = 0\) if \(y \ne x\). Then it is obvious that every element \(\phi\) of \(\mathbb Z \langle S \rangle\) can be written in the form \[ \phi = k_1 \cdot x_1 + \cdots + k_n \cdot x_n \] for some integers \(k_i\) and elements \(x_i \in S\) for \(i \in [n]\), all the \(x_i\) being distinct. Furthermore, \(\phi\) admits a unique such expression, because if we have \[ \phi = \sum_{x \in S}k_x \cdot x = \sum_{x \in S} k_x' \cdot x \] then \(0 = \sum_{x \in S} (k_x - k_x') \cdot x\), whence \(k_x' = k_x\) for all \(x \in S\). We map \(S\) into \(\mathbb Z \langle S \rangle\) by the map \(f_S = f\) such that \(f(x) = 1 \cdot x\). It is then clear that \(f\) is injective, and that \(f(S)\) generates \(\mathbb Z \langle S \rangle\). If \(g: S \to B\) is a mapping of \(S\) into some abelian group \(B\), then we can define a map \[ g_*: \mathbb Z \langle S \rangle \to B \] such that \[ g_* \left( \sum_{x \in S} k_x \cdot x \right) = \sum_{x \in S} k_x g(x). \] This map is a homomorphism (trivial) and we have \(g_* \circ f = g\) (also trivial). It is the only homomorphism which has this property, for any such homomorphism \(g_*\) must be such that \(g_*(1 \cdot x) = g(x)\).
It is customary to identify \(S\) in \(\mathbb Z(S)\), and we sometimes omit the dot when we write \(k_x x\) or a sum \(\sum k_x x\).
If \(\lambda: S \to S'\) is a mapping of sets, there is a unique homomorphism \(\overline{\lambda}\) making a certain diagram commutative. The diagram implements the relation \[ \overline{\lambda} \circ f_S = f_{S'} \circ \lambda, \] where \[f_S: S \to \mathbb Z \langle S \rangle,\] \[f_{S'}: S' \to \mathbb Z \langle S' \rangle,\] \[\lambda: S \to S',\] \[\overline{\lambda}: \mathbb Z \langle S \rangle \to \mathbb Z \langle S \rangle.\]
In fact \(\overline{\lambda}\) is none other than \((f_{S'} \circ \lambda)_*\), with the notation of the preceding paragraph. The proof of this statement is a trivial exercise.
We shall denote \(\mathbb Z \langle S \rangle\) also by \(F_{ab}(S)\), and call \(F_{ab}(S)\) the free abelian group generated by \(S\). We call elements of \(S\) its free generators.
As an exercise, show that every abelian group \(A\) is a factor group of a free abelian group \(F\). If \(A\) is finitely generated, show that one can select \(F\) to be finitely generated also.
If the set \(S\) consists of \(n\) elements, then we say that the free abelian group \(F_{ab}(S)\) is the free abelian group on \(n\) generators. If \(S\) is the set of \(n\) letters \(x_1, \cdots, x_n\), we say that \(F_{ab}(S)\) is the free abelian group with free generators \(x_1, \cdots, x_n\).
An abelian group is free if and only if it is isomorphic to a free abelian group \(F_{ab}(S)\) for some set \(S\). Let \(A\) be an abelian group, and let \(S\) be a basis for \(A\). Then it is clear that \(A\) is isomorphic to the free abelian group \(F_{ab}(S)\).
Group \(G\) generated by a set \(S\)
Let \(G\) be a group and \(S\) a subset of \(G\). We say that \(S\) generates \(G\), or that \(S\) is a set of generators of \(G\), if every element of \(G\) can be expressed as a product of elements of \(S\) or inverses of elements of \(S\), i.e., as a product \(x_1 \cdots x_n\) where each \(x_i\) or \(x_i^{-1}\) is in \(S\). The set of all such products is a subgroup of \(G\) (the empty product is the unit element), and is the smallest subgroup of \(G\) containing \(S\). Thus \(S\) generates \(G\) if and only if the smallest subgroup of \(G\) containing \(S\) is \(G\) itself. If \(G\) is generated by \(S\), we write \(G = \langle S \rangle\).
Monoid, submonoid
A monoid is a set \(G\), with a law of composition which is associative, and having a unit element (so that in particular \(G\) is non-empty). A law of composition is just a map \(G \times G \ni (x,y) \mapsto xy \in G\) which is denoted multiplicatively as above, or additively, as in \(G \times G \ni (x,y) \mapsto x+y \in G\). An element \(e \in G\) such that \(ex = xe = x\) for all \(x \in G\) is called a unit element. A unit element is unique. When the law of composition is additive, then we write \(0 \in G\) for the unit element. Given \(x,y,z \in G\), the product of these elements can be formed in two ways \((xy)z\) and \(x(yz)\). \(G\) being associative means these two ways produce the same element in \(G\).
A submonoid \(H\) of \(G\) is a subset of \(G\) containing the unit element and which is closed under the law of composition. That is, \(x,y \in H\) implies that \(xy \in H\). \(H\) is itself a monoid under the law of composition induced by \(G\), i.e. the restriction to \(H \times H \to H\).
Group, subgroup
A group \(G\) is a monoid (set, law of composition is associative, and there is a unit element) such that every element \(x \in G\) has an inverse \(x^{-1} \in G\). That is, if \(x \in G\), then \(xx^{-1} = x^{-1}x = e\).
A subgroup \(H\) of \(G\) is a subset of \(G\) which is closed under taking inverses and the law of composition, and contains the unit element.
Group Homomorphism, Isomorphism, Automorphism, Endomorphism
p.10 Lang (2002).
Let \(G, G'\) be monoids. A homomorphism is a mapping \(f: G \to G'\) such that \(f(xy) = f(x)f(y)\) for all \(x,y \in G\) and the identity of \(G\) is mapped to the identity of \(G'\). If in addition \(G, G'\) are groups, then \(f\) is called a group homomorphism.
A homomophism \(f: G \to G'\) is called an isomorphism if there exists a homomorphism \(g: G' \to G\) such that \(f \circ g\) and \(g \circ f\) are the identity mappings (in \(G'\) and \(G\) respectively).
An isomorphism \(f: G \to G\) is called an automorphism.
A homomorphism of \(G\) into itself is called an endomorphism.
Left/right coset
p. 12 Lang
Let \(G\) be a group and let \(H\) be a subgroup. A left coset of \(H\) in \(G\) is a subset of \(G\) of type \(aH\), for some element \(a\) of \(G\). An element of \(aH\) is called a coset representative of \(aH\). The map \(x \mapsto ax\) induces a bijection of \(H\) onto \(aH\). Hence any two left cosets have the same cardinality.
Observe that if \(a,b\) are elements in \(G\) and \(aH\), \(bH\) are cosets having one element in common, then they are equal. Indeed, let \(ax = by\) with \(x,y \in H\). Then \(a = byx^{-1}\). But \(yx^{-1} \in H\). Hence, \(aH = b(yx^{-1})H = bH\), because for any \(z \in H\) we have \(zH = H\).
We conclude that \(G\) is the disjoint union of the left cosets of \(H\). A similar remark applies to the right cosets (subsets of \(G\) of type \(Ha\)).
Factor Ring
p.89 Lang
Let \(\mathfrak{a}\) be an ideal of the ring \(A\). We can construct the factor ring \(A/\mathfrak{a}\) as follows. Viewing \(A\) and \(\mathfrak{a}\) as additive groups, let \(A/\mathfrak{a}\) be the factor group. We define a multiplicative law of composition on \(A/\mathfrak{a}\): If \(x + \mathfrak{a}\) and \(y + \mathfrak{a}\) are two cosets of \(\mathfrak{a}\), we define \((x + \mathfrak{a})(y+\mathfrak{a})\) to be the coset \((xy+ \mathfrak{a})\). This coset is well-defined, for if \(x_1, y_1\) are in the same coset as \(x,y\) respectively, then one verifies at once that \(x_1y_1\) is in the same coset as \(xy\).
p.91
The factor ring \(A/\mathfrak{a}\) is also called a residue class ring. Cosets of \(\mathfrak{a}\) in \(A\) are called residue classes modulo \(\mathfrak{a}\), and if \(x \in A\), then the coset \(x + \mathfrak{a}\) is called the residue class of \(x\) modulo \(\mathfrak{a}\).
Factor module
p. 119 Lang
Let \(M\) be an \(A\)-module, and \(N\) a submodule. We shall define a module structure on the factor group \(M/N\) (for the additive group structure). Let \(x + N\) be a coset of \(N\) in \(M\), and let \(a \in A\). We define \(a(x+N)\) to be the coset \(ax + N\). It is trivial to verify that this is well-defined (i.e. if \(y\) is in the same coset as \(x\), then \(ay\) is in the same coset as \(ax\)), and that this is an operation of \(A\) on \(M/N\) satisfying the required condition, making \(M/N\) into a module, called the factor module of \(M\) by \(N\).
Factor group
p. 14 Lang
Let \(f: G \to G'\) be a group homomorphism, and let \(H\) be its kernel. If \(x\) is an element of \(G\), then \(xH = Hx\) because both are equal to \(f^{-1}(f(x))\). We can also rewrite this relation as \(xHx^{-1} = H\).
Conversely, let \(G\) be a group, and let \(H\) be a subgroup. Assume that for all elements \(x\) of \(G\) we have \(xH \subset Hx\) (or equivalently \(xHx^{-1} \subset H\)). If we write \(x^{-1}\) instead of \(x\), we get \(H \subset xHx^{-1}\), whence \(xHX^{-1} = H\). Thus our condition is equivalent to the condition \(xHx^{-1} = H\) for all \(x \in G\). A subgroup \(H\) satisfying this condition will be called normal. We shall now see that a normal subgroup is the kernel of a homomorphism.
Let \(G'\) be the set of cosets of \(H\). (By assumption, a left coset is equal to a right coset, so we need not distingush between them.) If \(xH\) and \(yH\) are cosets, then their product \((xH)(yH)\) is also a coset, because \[xHyH = xyHH = xyH.\]
By means of this product, we have defined a law of composition on \(G'\) which is associative. It is clear that the coset \(H\) itself is a unit element for this law of composition, and that \(x^{-1}H\) is an inverse for the coset \(xH\). Hence, \(G'\) is a group.
Let \(f: G \to G'\) be the mapping such that \(f(x)\) is the coset \(xH\). Then \(f\) is clearly a homomorphism, and (the subgroup) \(H\) is contained in its kernel. If \(f(x) = H\), then \(xH = H\). Since \(H\) contains the unit element, it follows that \(x \in H\). Thus \(H\) is equal to the kernel, and we have obtained our desired homomorphism.
The group of cosets of a normal subgroup \(H\) is denoted by \(G/H\) (which we read \(G\) modulo \(H\), or \(G \mod H)\). The map \(f\) of \(G\) onto \(G/H\) constructed above is called the canonical map, and \(G/H\) is called the factor group of \(G\) by \(H\).
Free Module over a ring \(A\) generated by a non-empty set \(S\)
p. 137 Lang
We let \(A\langle S \rangle\) be the set of functions \(\phi: S \to A\) such that \(\phi(x) = 0\) for almost all \(x \in S\). If \(a \in A\) and \(x \in S\), we denote by \(ax\) the map \(\phi\) such that \(\phi(x) = a\) and \(\phi(y) = 0\) for \(y \ne x\). Then as for abelian groups, given \(\phi \in A \langle S \rangle\), there exists elements \(a_i \in A\) and \(x_i \in S\) such that \[\phi = a_1x_1 + \cdots + a_n x_n.\] It is immediately verified that the family of functions \(\delta_x\) \((x \in S)\) such that \(\delta_x(x) = 1\) and \(\delta_x(y) = 0\) for all \(y \ne x\) form a basis for \(A \langle S \rangle\). In other words, the expression of \(\phi\) as \(\sum a_i x_i\) above is unique. This construction can be applied when \(S\) is a group or a monoid \(G\), and gives rise to the group algebra.
So \(A \langle S \rangle\) is the free module over the ring \(A\) generated by \(S\).
Category
p. 53 Lang
A category \(\mathfrak C\) consists of a collection of objects \(\text{Ob}(\mathfrak C)\); and for two objects \(A,B \in \text{Ob}( \mathfrak C)\) a set \(\text{Mor}(A,B)\) called the set of morphisms of \(A\) into \(B\); and for three objects \(A,B,C \in \text{Ob}(\mathfrak C)\) a law of composition (i.e. a map) \[\text{Mor}(B,C) \times \text{Mor}(A,B) \to \text{Mor}(A,C)\] satisfying the following axioms
(Cat 1) Two sets \(\text{Mor}(A,B)\) and \(\text{Mor}(A',B')\) are disjoint unless \(A = A'\) and \(B = B'\), in which case they are equal.
(Cat 2) For each object \(A\) of \(\mathfrak C\) there is a morphism \(\text{id}_A \in \text{Mor}(A,A)\) which acts as left and right identity for the elements \(\text{Mor}(A,B)\) and \(\text{Mor}(B,A)\) respectively, for all objects \(B \in \text{Ob}(\mathfrak C)\).
(Cat 3) The law of composition is associative (when defined), i.e. given \(f \in \text{Mor}(A,B)\), \(g \in \text{Mor}(B,C)\) and \(h \in \text{Mor}(C,D)\) then \[(h \circ g) \circ f = h \circ (g \circ f),\] for all objects \(A,B,C,D\) of \(\mathfrak C\).
Isomorphism, automorphism, endomorphism in Categories
p. 54 Lang
An element \(f \in \text{Mor}(A,B)\) is also written \(f: A \to B\) or \[A \xrightarrow{f} B.\] A morphism \(f\) is called an isomorphism if there exists a morphism \(g: B \to A\) such that \(g \circ f\) and \(f \circ g\) are the identities in \(\text{Mor}(A,A)\) and \(\text{Mor}(B,B)\) respectively. If \(A = B\), then we also say that the isomorphism is an automorphism.
A morphism of an object \(A\) into itself is called an endomorphism. The set of endomorphisms of \(A\) is denoted by \(\text{End}(A)\). It follows at once from our axioms that \(\text{End}(A)\) is a monoid.
Let \(A\) be an object of a category \(\mathfrak C\). We denote by \(\text{Aut}(A)\) the set of automorphisms of \(A\).
Direct Product
p. 9 Lang.
Let \(G_1, G_2\) be groups. Let \(G_1 \times G_2\) be the direct product as sets, so \(G_1 \times G_2\) is the set of all pairs \((x_1, x_2)\) with \(x_i \in G_i\). We define the product componentwise by \[(x_1, x_2)(y_1,y_2) = (x_1y_1, x_2y_2).\] Then \(G_1 \times G_2\) is a group, whose unit element is \((e_1, e_2)\) (where \(e_i\) is the unit element of \(G_i\).) Similarly, for \(n\) groups we define \(G_1 \times \cdots G_n\) to be the set of \(n\)-tuples with \(x_i \in G_i\) (\(i=1,\cdots,n\)), and componentwise multiplication. Even more generally, let \(I\) be a set, and for each \(i \in I\), let \(G_i\) be a group. Let \(G = \prod G_i\) be the set theoretic product of the sets \(G_i\). Then \(G\) is the set of all families \((x_i)_{i \in I}\) with \(x_i \in G_i\). We can define a group structure on \(G\) by componentwise multiplication, namely, if \((x_i)_{i \in I}\) and \((y_i)_{i \in I}\) are two elements of \(G\), we define their product to be \((x_iy_i)_{i \in I}\). We define the inverse of \((x_i)_{i \in I}\) to be \((x_i^{-1})_{i \in I}\). It is then obvious that \(G\) is a group called the direct product of the family.
Direct Sum
p. 36 Lang.
Let \((A_i)_{i \in I}\) be a family of abelian groups. We define their direct sum \[A = \bigoplus_{i \in I} A_i\] to be the subset of the direct product \(\Prod A_i\) consisting of all the families \((x_i)_{i \in I}\) with \(x_i \in A_i\) such that \(x_i = 0\) for all but a finite number of indices \(i\). Then it is clear that \(A\) is a subgroup of the product. For each index \(j \in I\), we map \[\lamba_j: A_j \to A\] by letting \(\lambda_j(x)\) be the element whose \(j\)-th component is \(x\), and having all other components equal to \(0\). Then \(\lambda_j\) is an injective homomorphism.
Proposition 7.1. Let \(\{f_i: A_i \to B\}\) be a family of homomorphisms into an abelian group \(B\). Let \(A = \bigoplus A_i\). There exists a unique homomorphism \[f: A \to B\] such that \(f \circ \lambda_j = f_j\) for all \(j\).
Proof. Define \(f: A \to B\) by \[ f((x_i)_{i \in I}) = \sum_{i \in I} f_i(x_i), \] and argue accordingly. q.e.d.
The property in Prop 7.1 is called the universal property of the direct sum.
p.37 Lang
Let \(A\) be an abelian group and \(B,C\) subgroups. If \(B+C = A\) and \(B \cap C = \{0\}\) then the map \[B \times C \to A\] given by \((x,y) \mapsto x+y\) is an isomorphism (as we already noted in the non-commutative case). Instead of writing \(A = B \times C\), we shall write \[A = B \oplus C\] and say that \(A\) is the direct sum of \(B\) and \(C\). We use a similar notation for the direct sum of a finite number of subgroups \(B_1, \cdots, B_n\) such that \[B_1 + \cdots + B_n = A\] and \[B_{i+1} \cap (B_1 + \cdots + B_i) = 0.\] In that case we write \(A = B_1 \oplus \cdots \oplus B_n.\)
Graded Algebra
p. 172 Lang.
Let \(A\) be an algebra over a field \(k\). By a filtration of \(A\) we mean a sequence of \(k\)-vector spaces \(A_i\) \((i=0,1,\cdots)\) such that \[ A_0 \subset A_1 \subset A_2 \subset \cdots \] and \[\cup A_i = A,\] and \(A_iA_j \subset A_{i+j}\) for all \(i,j \ge 0\). In particular, \(A\) is an \(A_0\)-algebra (why?). We then call \(A\) a filtered algebra. Let \(R\) be an algebra. We say that \(R\) is graded if \(R\) is a direct sum \(R = \bigoplus R_i\) of subspaces such that \(R_iR_j \subset R_{i+j}\) for all \(i,j \ge 0\).
(Note that \(A_iA_j\) uses the \(k\)-bilinear form in the definition of the \(k\)-algebra \(A\).)Basis
p.38 Lang.
Let \(A\) be an abelian group. Let \(\{e_i\}_{i \in I}\) be a family of elements of \(A\). We say that this family is a basis for \(A\) if the family is not empty, and if every element of \(A\) has a unique expression as a linear combination \[x = \sum x_ie_i\] with \(x_i \in \mathbb Z\), and almost all \(x_i = 0\). Thus the sum is actually a finite sum. An abelian group is said to be free if it has a basis. If that is the case, it is immediate that if we let \(Z_i = \mathbb Z\) for all \(i \in I\), then \(A\) is isomorphic to the direct sum \[A \cong \bigoplus_{i \in I} A_i.\]
p. 135 Lang.
Let \(M\) be a module over a ring \(A\) and let \(S\) be a subset of \(M\). We shall say that \(S\) is a basis of \(M\) if \(S\) is not empty, if \(S\) generates \(M\), and if \(S\) is linearly independent. If \(S\) is a basis of \(M\), then in particular \(M \ne \{0\}\) if \(A \ne \{0\}\) and every element of \(M\) has a unique expression as a linear combination of elements of \(S\). Similarly, let \(\{x_i\}_{i \in I}\) be a non-empty family of elements of \(M\). We say that it is a basis of \(M\) if it is linearly independent and generates \(M\).
Dual Space and Dual Module
p. 142 Lang
Let \(E\) be a free module over a commutative ring \(A\). We view \(A\) as a free module of rank 1 over itself. By the dual module \(E^\vee\) of \(E\) we shall mean the module \(\text{Hom}(E,A)\). Its elements will be called functionals. Thus a functional on \(E\) is an \(A\)-linear map \(f: E \to A\). If \(x \in E\) and \(f \in E^\vee\), we sometimes denote \(f(x)\) by \(\langle x, f \rangle\). Keeping \(x\) fixed, we see that the symbol \(\langle x, f \rangle\) as a function of \(f \in E^\vee\) is \(A\)-linear in its second argument, and hence that \(x\) induces a linear map on \(E^\vee\), which is \(0\) if and only if \(x = 0\). Hence we get an injection \(E \to E^{\vee \vee}\) which is not always an injection.
Let \((x_i)_{i \in I}\) be a basis of \(E\). For each \(i \in I\), let \(f_i\) be the unique functional such that \(f(x_i) = \delta_{ij}\). Such a linear map exists by general properties of bases.
Theorem 6.1. Let \(E\) be a finite free module over the commutative ring \(A\), of finite dimension \(n\). Then \(E^\vee\) is also free, and \(\dim E^\vee = n\). If \(\{x_1, \cdots, x_n\}\) is a basis for \(E\), and \(f_i\) is the functional such that \(f_i(x_j) = \delta_{ij}\), then \(\{f_1, \cdots, f_n\}\) is a basis for \(E^\vee\).
Proof. Show that the \(f_i\) generate the dual space and are linearly independent in the dual space. Be careful, \(f_i\) is only the indicator function on the basis \(x_i\), but scales appropriately. q.e.d.
Ideal
p. 86 Lang.
A left ideal \(\mathfrak{a}\) in a ring \(A\) is a subset of \(A\) which is a subgroup of the additive group of \(A\), such that \(A \mathfrak{a} \subset \mathfrak{a}\) and hence \(A \mathfrak{a} = A\), since \(A\) contains \(1\)). To define a right ideal, we require \(\mathfrak{a}A = \mathfrak{a},\) and a two-sided ideal is a subset which is both a left and a right ideal. A two-sided ideal is called simply an ideal in this section.
A ring \(A\) is said to be commutative if \(yx = xy\) for all \(x,y \in A\). In that case, every left or right ideal is two-sided.