Algebras

Now we define the notion of an \(A\)-algebra. An algebra is a structure obtained by adding to an \(A\)-module \(M\) a multiplication structure compatible with the addition and scalar multiplication of \(M\), which is the same process as defining on an abelian group \(G\) a multiplication compatible with the addition of \(G\). However, there are two problems with using §Definition of a Ring, ⁋Definition 1 directly.

First, the first problem is that for a general ring \(A\), \(\lMod{A}\) or \(\rMod{A}\) is not a monoidal category. In the definition above, we defined the multiplication of a ring as \(A\otimes A\rightarrow A\); similarly, when attempting to define multiplication on a module, \(M\otimes_AM\) may not be defined, so this approach cannot be used. Since \((\lMod{A},\otimes_A, A)\) is a symmetric monoidal category if \(A\) is a commutative ring, this is resolved by assuming that \(A\) is a commutative ring. (§Direct Products, Direct Sums, and Tensor Products of Modules, §§Tensor Products of Modules over Commutative Rings)

The second problem is a more practical one: while this structure is widely used in many fields, the properties required in each case are all different. That is, if we repeat §Definition of a Ring, ⁋Definition 1 as is and call a monoid object \(E\) of the symmetric monoidal category \((\lMod{A},\otimes_A,A)\) an \(A\)-algebra, then due to the associativity axiom and unit axiom of a monoid object, the multiplication of \(E\) must always satisfy associativity and have an identity. On the other hand, important examples such as [Lie Theory] §Lie Groups, ⁋Definition 8 may not satisfy these conditions.

For these reasons, from now on we always assume \(A\) to be a commutative ring, and we define an \(A\)-algebra in its most general form as follows.

Definition 1 If an \(A\)-module \(E\) is equipped with an \(A\)-bilinear map \(\mu:E\times E \rightarrow E\), we call \(E\) an \(A\)-algebra. In addition, depending on the properties satisfied by the multiplication \(\mu\) of \(E\), we define the following:

• If the multiplication of \(E\) is commutative, we call it a commutative \(A\)-algebra.
• If the multiplication of \(E\) is associative, we call it an associative \(A\)-algebra.
• If the multiplication of \(E\) has an identity element, we call it a unital \(A\)-algebra.

Then a monoid object of \((\lMod{A},\otimes_A,A)\) is precisely an associative unital \(A\)-algebra.

Meanwhile, an associative unital \(A\)-algebra can be thought of as an \(A\)-module \(E\) equipped with a multiplication, but it can also be regarded as a ring \(E\) equipped with a scalar multiplication. Fix an arbitrary ring homomorphism \(\rho:A \rightarrow E\), and assume that \(\rho(A)\) is contained in the center \(Z(E)\) of \(E\). Then, viewing \(E\) as a module over itself, we can consider the \(A\)-module \(\rho^\ast E\) defined via the restriction of scalars

\[\rho^\ast:\lMod{E} \rightarrow \lMod{A}\]

Then, on \(\rho^\ast E\), the multiplication of \(E\) defines \(\mu:\rho^\ast E\times \rho^\ast E \rightarrow \rho^\ast E\). Since

\[\mu(\alpha\cdot x, y)=(\rho(\alpha)x)y=\rho(\alpha)xy\]

and \(\rho(\alpha)\in Z(E)\), from this we can verify the following relation:

\[\alpha\cdot \mu(x,y)=\mu(\alpha\cdot x,y)=\mu(x,\alpha\cdot y)\]

Conversely, given any associative unital \(A\)-algebra \(E\),

\[\rho_E:A\rightarrow E;\qquad \alpha\mapsto \alpha\cdot 1_E\]

is a ring homomorphism, and these constructions are inverses of each other. That is, these data are precisely the same, and a ring homomorphism \(\rho:A \rightarrow E\) endowing \(E\) with an \(A\)-algebra structure in this way is called a structure morphism.

Definition 2 Let two \(A\)-algebras \(E,E'\) be given. An \(A\)-module homomorphism \(u: E \rightarrow E'\) is an \(A\)-algebra homomorphism if the equation

\[u(xy)=u(x)u(y)\]

holds for all \(x,y\in E\). If \(E\) and \(E'\) are unital and \(u(1)=1\), we call it a unital \(A\)-algebra homomorphism.

Then the following holds.

Proposition 3 The composition of \(A\)-algebra homomorphisms is an \(A\)-algebra homomorphism. A bijective \(A\)-algebra homomorphism is always an isomorphism. Moreover, the composition of unital \(A\)-algebra homomorphisms is unital, and the inverse of a bijective unital \(A\)-algebra homomorphism is also unital.

Proof

That the composition is an \(A\)-algebra homomorphism follows directly by substituting into the definition. Meanwhile, for a bijective \(A\)-algebra homomorphism \(u:E\rightarrow E'\), the fact that \(u^{-1}\) is \(A\)-linear is a consequence of §Modules, ⁋Proposition 7, and since for any \(x,y\in E'\) we have \(u(u^{-1}(x)u^{-1}(y))=xy\), the injectivity of \(u\) yields \(u^{-1}(xy)=u^{-1}(x)u^{-1}(y)\). In the case of a unital \(A\)-algebra homomorphism, in addition to this calculation, it suffices to verify that the following two identities

\[(u\circ v)(1)=u(1)=1,\qquad u(1)=1\implies 1=u^{-1}(1)\]

hold.

Therefore, the category of \(A\)-algebras is well-defined, and the category of unital \(A\)-algebras and unital \(A\)-algebra homomorphisms is also well-defined. We denote the category of associative unital \(A\)-algebras and unital \(A\)-algebra homomorphisms by \(\Alg{A}\), and write \(\cAlg{A}\) for its full subcategory consisting of commutative ones.

Examples of Algebras

Just as a free ring is defined on an abelian group, on any \(A\)-module, a free associative unital \(A\)-algebra is defined. This process is completely identical to the construction in §Definition of a Ring, ⁋Proposition 4. On reflection, this is natural: the only additional thing we need to show is that the multiplication defined here is compatible with the scalar multiplication structure of \(A\), and this is because requiring \(A\) to be commutative already defines a scalar multiplication structure on \(F(M)=\bigoplus_{n\geq 0}M^{\otimes n}\).

That is, the following holds.

Proposition 4 For any \(A\)-module \(M\), if we define

\[F(M)=\bigoplus_{n\geq 0} M^{\otimes n}\]

then \(F\) is the left adjoint to the forgetful functor \(\Alg{A} \rightarrow \lMod{A}\). ([Category Theory] §Adjoint Functors, ⁋Definition 1)

The \(F(M)\) obtained from the above proposition uses the addition, which is the data defined on \(M\), as is, and is obtained by defining multiplication as a new operation. On the other hand, using the multiplication defined on an appropriate group \(G\), it is also possible to make it into an \(A\)-algebra.

Definition 5 Fix a ring \(A\), not necessarily commutative, and an arbitrary group \(G\). Then the group ring \(AG\) is, first, as a set, the collection of finitely supported functions from \(G\) to \(A\), where the operations for two functions \(\alpha:G \rightarrow A\) and \(\beta: G \rightarrow A\) are given by

\[\alpha+\beta: x\mapsto \alpha(x)+\beta(x),\qquad \alpha\beta:x\mapsto \sum_{uv=x}\alpha(u)\beta(v)\]

If \(A\) is a commutative ring, \(AG\) also has a natural \(A\)-module structure and thus becomes an associative unital \(A\)-algebra. In this case, \(AG\) is called a group algebra.

For each \(x\in G\), if we define \(\delta_x: G \rightarrow A\) by

\[\delta_x(y)=\begin{cases}1&\text{if $x=y$}\\0&\text{if $x\neq y$}\end{cases}\]

then any element of \(AG\) can be written, for a family indexed by elements of \(G\), namely \(\alpha_x\in A\), as

\[\sum_{x\in G} \alpha_x\delta_x,\qquad\text{$\alpha_x=0$ for all but finitely many $x$}\]

For convenience, if we denote the function \(\delta_x\) by \(x\), the multiplication in \(AG\) can be written as

\[\left(\sum_{x\in G} \alpha_xx\right)\left(\sum_{y\in G} \beta_yy\right)=\sum_{x,y\in G}\alpha_x\beta_yxy=\sum_{z\in G}\left(\sum_{x\in G}\alpha_x\beta_{x^{-1}z}\right)z\]

Then the following holds.

Proposition 6 Consider two functors

\[A-: \Grp \rightarrow \Alg{A},\qquad (-)^\times:\Alg{A} \rightarrow \Grp\]

Here, \((-)^\times\) is the functor that, given an arbitrary \(A\)-algebra \(E\), yields the group of units \(E^\times\). Then the adjunction

\[A{-}\dashv (-)^\times\]

holds.

Proof

That is, for any group \(G\) and \(A\)-algebra \(E\), we must show the following isomorphism:

\[\Hom_{\Alg{A}}(AG, E)\cong \Hom_\Grp(G, E^\times)\]

First, suppose a group homomorphism \(f:G \rightarrow E^\times\) is given. Then from this, we can define \(\tilde{f}:AG \rightarrow E\) by the formula

\[\tilde{f}:AG \rightarrow E;\quad \sum_{x\in G} \alpha_x\delta_x\mapsto \sum_{x\in G} \alpha_xf(x)\]

That \(\tilde{f}\) preserves addition and scalar multiplication follows directly from the definition, and for multiplication as well, since

\[\begin{aligned}\tilde{f}\left(\left(\sum_{x\in G} a_xx\right)\left(\sum_{y\in G} b_yy\right)\right)&=\tilde{f}\left(\sum_{x,y\in G}a_xb_yxy\right)=\tilde{f}\left(\sum_{z\in G}\left(\sum_{x\in G}a_xb_{x^{-1}z}\right)z\right)\\&=\sum_{z\in G}\left(\sum_{x\in G}a_xb_{x^{-1}z}\right)f(z)=\sum_{x,y\in G} a_xb_yf(xy)\\&=\left(\left(\sum_{x\in G} a_xf(x)\right)\left(\sum_{y\in G} b_yf(y)\right)\right)=\tilde{f}\left(\sum_{x\in G} a_xx\right)\tilde{f}\left(\sum_{y\in G} b_yy\right)\end{aligned}\]

the desired identity holds. Also, since

\[\tilde{f}(1_{AG})=\tilde{f}(\delta_e)=f(e)=1_E\]

\(\tilde{f}\) is unital. Conversely, suppose an arbitrary unital \(A\)-algebra homomorphism \(u: AG \rightarrow E\) is given. Then the function defined by

\[\bar{u}: G \rightarrow E;\qquad x\mapsto u(\delta_x)\]

preserves multiplication. This is because for any \(x,y\in G\),

\[\bar{u}(xy)=u(\delta_{xy})=u(\delta_x\delta_y)=u(\delta_x)u(\delta_y)=\bar{u}(x)\bar{u}(y)\]

holds. Also, since \(u\) sends \(1_{AG}=\delta_e\) to \(1_E\), for any \(x\in G\),

\[\bar{u}(x)\bar{u}(x^{-1})=u(\delta_x\delta_{x^{-1}})=u(\delta_e)=1_E\]

holds, and therefore the image of \(\bar{u}\) is contained in \(E^\times\). These two processes are inverses of each other because \(\tilde{f}\) sends \(\delta_x\) to \(f(x)\), so \(\bar{\tilde{f}}=f\), and conversely, because the \(\delta_x\) generate \(AG\) as an \(A\)-module, the two \(A\)-algebra homomorphisms \(\tilde{\bar{u}}\) and \(u\) that agree on the \(\delta_x\) are equal.

The final example is already somewhat familiar, revisiting as an \(A\)-algebra what we briefly examined as a ring in §Graded Rings, ⁋Example 5. Earlier, in Definition 5, we exceptionally considered the case where \(A\) is not commutative in order to define general group rings, but when dealing with \(A\)-algebras, our assumption that \(A\) is always a commutative ring still remains in effect.

Definition 7 For a commutative ring \(A\), the polynomial algebra \(A[\x]\) is, as a set,

\[A[\x]=\{p(\x)=a_n\x^n+\cdots+a_1\x+a_0\mid a_i\in A\}\]

where addition and multiplication are given by polynomial addition and multiplication, and scalar multiplication by \(A\) is likewise given by scalar multiplication of polynomials. More generally, for any set \(S\), we can define \(A[S]\) in a similar manner.

In the case of the multivariable polynomial algebra \(A[S]\), for convenience of notation, it is customary to choose an index set \(I\) satisfying \(\lvert S\rvert=\lvert I\rvert\) and write the elements of \(S\) as \(\x_i\). Then, for any

\[\alpha=(\alpha_i)_{i\in I}\in \mathbb{N}_{\geq0}^I\qquad\text{finitely supported}\]

we write

\[\x^\alpha=\prod_{i\in I}\x_i^{\alpha_i}\]

and the elements of \(A[\x_i]_{i\in I}\) can be written as

\[\sum_{d=0}^n \sum_{\lvert\alpha\rvert=d} a_\alpha\x^\alpha,\qquad\text{$a_\alpha=0$ for all but finitely many $\alpha$}\]

Here, \(\lvert\alpha\rvert=\sum_{i\in I}\alpha_i\).

Similarly, the polynomial algebra also satisfies the following universal property.

Proposition 8 The functor that takes a set \(S\) and, with each element of \(S\) as a variable, associates the polynomial algebra \(A[\x_i]_{i\in I}\), \(A[-]:\Set \rightarrow \cAlg{A}\), is a left adjoint to the forgetful functor \(U: \cAlg{A} \rightarrow \Set\).

Proof

That is, we must show that the isomorphism

\[\Hom_{\cAlg{A}}(A[\x_i]_{i\in I}, E)\cong\Hom_\Set(S, UE)\]

holds. First, suppose an arbitrary function \(f:S \rightarrow U(E)\) is given. Then the map \(\tilde{f}: A[\x_i]_{i\in I} \rightarrow E\) defined by

\[\tilde{f}: \sum_{d=0}^n \sum_{\lvert\alpha\rvert=d} a_\alpha\x^\alpha\mapsto \sum_{d=0}^n\sum_{\lvert\alpha\rvert=d}a_\alpha\prod_{i\in I}f(\x_i)^{\alpha_i}\]

is a map from \(A[\x_i]_{i\in I}\) to \(E\) that is an \(A\)-algebra homomorphism. Indeed, \(\tilde{f}\) is by definition \(A\)-linear, and since \(\tilde{f}(\x^\alpha\x^\beta)=\tilde{f}(\x^{\alpha+\beta})=\tilde{f}(\x^\alpha)\tilde{f}(\x^\beta)\) holds on monomials, it preserves multiplication by the distributive law, and from the term with \(\alpha=0\), we have \(\tilde{f}(1)=1\).

Conversely, given any unital \(A\)-algebra homomorphism \(u:A[\x_i]_{i\in I} \rightarrow E\), we may define \(\bar{u}:S \rightarrow UE\) by \(\x_i\mapsto u(\x_i)\). These two correspondences are inverses of each other: since \(\tilde{f}\) sends \(\x_i\) to \(f(\x_i)\), we have \(\bar{\tilde{f}}=f\), and conversely, since the \(\x_i\) generate \(A[\x_i]_{i\in I}\) as an \(A\)-algebra, and because they agree on the \(\x_i\), the two \(A\)-algebra homomorphisms \(\tilde{\bar{u}}\) and \(u\) are equal.

That is, we can think of a polynomial algebra as a kind of free commutative associative unital \(A\)-algebra. Comparing Definition 5 and Definition 7, there is a slight similarity: if we generalize Definition 5 to consider the monoid ring, the polynomial ring of Definition 7 becomes the monoid ring formed by the monoid of elements of the form \(\x^\alpha\).

Subalgebras, Ideals, and Quotient Algebras

Definition 9 For an \(A\)-algebra \(E\), if a submodule \(F\) is closed under multiplication, then \(F\) is called a subalgebra of \(E\). Also, if \(E\) is unital and \(1_E\in F\), then \(F\) is called a unital subalgebra.

Meanwhile, just as we defined in §Definition of a Ring, ⁋Definition 7, we also define ideals of an \(A\)-algebra.

Definition 10 For an \(A\)-algebra \(E\), a submodule \(\mathfrak{a}\) is a left ideal of \(E\) if for any \(x\in \mathfrak{a}\) and \(\alpha\in E\), we have \(\alpha x\in \mathfrak{a}\). Similarly, we define a right ideal. A submodule that is both a left ideal and a right ideal is called a two-sided ideal.

In particular, when \(E\) is associative and unital, an ideal of the \(A\)-algebra is the same as an ideal of \(E\) obtained by forgetting the \(A\)-action defined on it and viewing \(E\) as a ring. Indeed, for any \(\alpha\in A\) and \(x\in E\), we have \(\alpha\cdot x=(\alpha\cdot 1_E)x\), so an ideal as a ring is automatically a submodule. On the other hand, since a unital subalgebra in Definition 9 contains \(1_E\), the only one that is both an ideal and a unital subalgebra is \(E\) itself. As an example, the kernel of an \(A\)-algebra homomorphism \(u:E \rightarrow F\)

\[\ker u=\{x\in E\mid u(x)=0\}\]

is a two-sided ideal of \(E\). Then we can define the following.

Definition 11 For any \(A\)-algebra \(E\) and any two-sided ideal of \(E\), \(\mathfrak{a}\), we call \(E/\mathfrak{a}\) the quotient algebra by \(\mathfrak{a}\) of \(E\).

By the two-sided ideal condition,

\[(x+\mathfrak{a})(y+\mathfrak{a})=xy+\mathfrak{a}\]

is well-defined independently of the choice of representatives. Moreover, since \(\mathfrak{a}\) is a submodule, the addition of \(E\) and the \(A\)-action also descend to \(E/\mathfrak{a}\), from which we obtain that \(E/\mathfrak{a}\) is an \(A\)-algebra. If \(E\) is associative, commutative, or unital, then \(E/\mathfrak{a}\) also has the respective property. Furthermore, the following holds.

Proposition 12 For an \(A\)-algebra homomorphism \(u:E \rightarrow F\) and its kernel \(\ker u\), the following hold.

1. \(\ker u\) is a two-sided ideal of \(E\), and \(x+\ker u \mapsto u(x)\) defines a well-defined isomorphism \(E/\ker u \rightarrow \im u\).
2. For a subalgebra of \(E\), \(E'\), \(E'+\ker u=\{x'+y\mid x'\in E', y\in\ker u\}\) is a subalgebra of \(E\), \(E'\cap\ker u\) is a two-sided ideal of \(E'\), and there exists an isomorphism \((E'+\ker u)/\ker u\cong E'/(E'\cap \ker u)\).
3. If two two-sided ideals of \(E\), \(\mathfrak{a}, \mathfrak{b}\), satisfy \(\mathfrak{b}\subseteq \mathfrak{a}\), then \(\mathfrak{a}/\mathfrak{b}\) is a two-sided ideal of \(E/\mathfrak{b}\) and \((E/\mathfrak{b})/(\mathfrak{a}/\mathfrak{b})\cong E/\mathfrak{a}\) holds.
4. For a two-sided ideal of \(E\), \(\mathfrak{a}\), there exists an inclusion-preserving bijection between the set of two-sided ideals of \(E/\mathfrak{a}\) and the set of two-sided ideals containing \(\mathfrak{a}\) in \(E\).

The proof of each part is obtained by the same construction as in §Quotient Rings and Ring Isomorphisms, ⁋Theorem 3. Here, only addition, bilinear multiplication, and the two-sided ideal condition are used, and the fact that each map is \(A\)-linear follows from the corresponding subset of \(E\) being closed under the \(A\)-action. Therefore, the four assertions hold for \(A\)-algebras without assuming associativity, commutativity, or a unit. Even when the \(A\)-algebras involved are associative or commutative, the same construction preserves those properties, and when unital \(A\)-algebras, unital homomorphisms, and unital subalgebras are used, the isomorphisms above are also unital.

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References

[Bou] Bourbaki, N. Algebra I. Elements of Mathematics. Springer. 1998.