Algebras

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

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

First, for a general ring \(A\), the category \(\lMod{A}\) or \(\rMod{A}\) is not monoidal. If \(A\) is a commutative ring, then \((\lMod{A},\otimes_A, A)\) becomes a symmetric monoidal category, so this issue 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)

Second, there is a historical issue: if we fix a commutative ring \(A\) as before and call a monoid object \(E\) in the symmetric monoidal category \((\lMod{A},\otimes_A,A)\) an \(A\)-algebra, then the associativity and unit axioms of a monoid object force the multiplication of \(E\) to always be associative and to possess an identity element. Yet many important examples studied so far often fail to satisfy these properties. To include all of them, we must drop associativity and the existence of an identity element, which also means partially abandoning a definition like §Definition of a Ring, ⁋Definition 1.

For these reasons, we shall henceforth assume that \(A\) is always a commutative ring, and define an \(A\)-algebra as follows.

Definition 1 If an \(A\)-bilinear map \(\mu:E\times E \rightarrow E\) is given on an \(A\)-module \(E\), then \(E\) is called an \(A\)-algebra. Moreover, depending on the properties satisfied by the multiplication \(\mu\) of \(E\), we define the following.

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

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

On the other hand, an associative unital \(A\)-algebra can be thought of as an \(A\)-module \(E\) equipped with a multiplication, but also 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 may consider the \(A\)-module \(\rho^\ast E\) defined via the restriction of scalars

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

Then on \(\rho^\ast E\) a map \(\mu:\rho^\ast E\times \rho^\ast E \rightarrow \rho^\ast E\) is defined via the multiplication of \(E\), and in this case

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

and since \(\rho(\alpha)\in Z(E)\), through this we can verify the identity

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

The ring homomorphism \(\rho:A \rightarrow E\) giving the \(A\)-algebra structure on \(E\) is called the structure morphism.

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

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

holds for all \(x,y\in E\). If \(E\) and \(E'\) are unital, then an \(A\)-algebra homomorphism is also understood to be unital; that is, we implicitly assume \(u(1)=1\).

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.

Proof

That the composition is an \(A\)-algebra homomorphism follows immediately from substituting the definition. 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 for any \(x,y\in E'\) we have \(u(u^{-1}(x)u^{-1}(y))=xy\), so from the injectivity of \(u\) we obtain \(u^{-1}(xy)=u^{-1}(x)u^{-1}(y)\). Finally, from \(u(1)=1\) we get \(u^{-1}(1)=1\).

These data define the category of \(A\)-algebras. Unless otherwise stated, we shall always assume that an \(A\)-algebra is associative and unital. In particular, an \(A\)-algebra homomorphism is also unital. Under this convention, we write \(\Alg{A}\) for the category of associative unital \(A\)-algebras, and \(\cAlg{A}\) for the category of commutative \(A\)-algebras.

Examples of Algebras

Just as a free ring is defined on an abelian group, a free algebra is defined on any \(A\)-module. This process is completely identical to what was done in §Definition of a Ring, ⁋Proposition 4; indeed, this is natural, since 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 already ensured by requiring that \(A\) be commutative so that the scalar multiplication structure on \(F(M)=\bigoplus_{n\geq 0}M^{\otimes n}\) is already defined.

That is, the following holds.

Proposition 4 For any \(A\)-module \(M\), defining

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

the functor \(F\) is the left adjoint of the forgetful functor \(\Alg{A} \rightarrow \lMod{A}\).

From the above proposition, the resulting \(F(M)\) uses the addition already present as data on \(M\), and obtains a new operation, multiplication. Conversely, it is also possible to make an \(A\)-algebra from a suitable group \(G\) by using a multiplication defined on \(G\).

Definition 5 Fix a (not necessarily commutative) ring \(A\) and an arbitrary group \(G\). Then the group ring \(AG\) is, as a set, the collection of finitely supported functions from \(G\) to \(A\); for two functions \(\alpha:G \rightarrow A\), \(\beta: G \rightarrow A\) their operations 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, then \(AG\) also carries an \(A\)-module structure, so it becomes an \(A\)-algebra. In this case \(AG\) is called the group algebra.

For each \(x\in G\), defining \(\delta_x: G \rightarrow A\) by the formula

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

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

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

For convenience, writing the function \(\delta_x\) as \(x\), the multiplication in \(AG\) can be written as the formula

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

Then the following holds.

Proposition 6 Consider the two functors

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

where \((-)^\times\) is the functor sending an \(A\)-algebra \(E\) to the group of units \(E^\times\). Then \(A{-}\dashv (-)^\times\).

Proof

That is, for any group \(G\) and \(A\)-algebra \(E\) we must show the 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 immediately from the definition, and for multiplication as well

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

so the desired identity holds. Conversely, given any \(A\)-algebra homomorphism \(u: AG \rightarrow E\), the function defined by the formula

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

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 \(x\in G\) we have

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

and therefore the image of \(\bar{u}\) lies in \(E^\times\). That these two processes are inverses of each other follows 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, two \(A\)-algebra homomorphisms \(\tilde{\bar{u}}\) and \(u\) that agree on the \(\delta_x\) must be equal.

The last example is already somewhat familiar: we revisit what was briefly examined as a ring in §Graded Rings, ⁋Example 5, now viewed as an \(A\)-algebra.

Definition 7 For a 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 the scalar multiplication by \(A\) is likewise given by multiplication by constants. More generally, for any set \(S\) the algebra \(A[S]\) can be defined in a similar manner.

In the case of multivariate polynomials \(A[S]\), for notational convenience it is customary to choose an index set \(I\) with \(\lvert S\rvert=\lvert I\rvert\) and denote the elements of \(S\) by \(\x_i\). 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 elements of \(A[\x_i]_{i\in I}\) can be written as

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

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

Similarly, polynomial algebras also satisfy the following universal property.

Proposition 8 The functor \(A[-]:\Set \rightarrow \cAlg{A}\) sending a set \(S\) to the polynomial algebra \(A[\x_i]_{i\in I}\) whose variables are the elements of \(S\) is the left adjoint of the forgetful functor \(U: \cAlg{A} \rightarrow \Set\).

Proof

That is, we must show the isomorphism

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

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 the formula

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

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

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

Thus a polynomial algebra can be thought of as a kind of free commutative \(A\)-algebra. Comparing Definition 5 and Definition 7 reveals a slight similarity; if we generalize Definition 5 to the notion of a monoid ring, then the polynomial ring of Definition 7 becomes the monoid ring made from the monoid consisting of elements of the form \(\x^\alpha\).

Subalgebras, Ideals, and Quotient Algebras

Definition 9 A submodule \(F\) of an \(A\)-algebra \(E\) is called a subalgebra of \(E\) if \(F\) is closed under the multiplication of \(E\) and \(1\in F\).

Meanwhile, we also define the ideal of an \(A\)-algebra in the same way as was defined in §Definition of a Ring, ⁋Definition 7.

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

That is, an ideal of an \(A\)-algebra is nothing other than the ideal of \(E\) viewed simply as a ring, forgetting the \(A\)-action defined on \(E\). Indeed, since \(E\) is unital, for any \(\alpha\in A\) and \(x\in E\) we have \(\alpha\cdot x=(\alpha\cdot 1)x\), so an ideal as a ring automatically becomes a submodule. On the other hand, a subalgebra in Definition 9 contains \(1\), so the only thing that is both an ideal and a subalgebra is \(E\) itself. As an example, the kernel

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

of an \(A\)-algebra homomorphism \(u:E \rightarrow F\) 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 \(\mathfrak{a}\) of \(E\), we call \(E/\mathfrak{a}\) the quotient algebra of \(E\) by \(\mathfrak{a}\).

Of course, having defined it this way we must prove that \(E/\mathfrak{a}\) is an \(A\)-algebra; but the fact that the ring structure is well defined is the same as the calculation before §Quotient Rings and Ring Homomorphisms, ⁋Definition 1, and we only need to add that since \(\mathfrak{a}\) is a submodule, the \(A\)-action on \(E\) descends to \(E/\mathfrak{a}\). Moreover, 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 \(E'\) of \(E\), the set \(E'+\ker u=\{x'+y\mid x'\in E', y\in\ker u\}\) is a subalgebra of \(E\), and \(E'\cap\ker u\) is a two-sided ideal of \(E'\), and there is an isomorphism \((E'+\ker u)/\ker u\cong E'/(E'\cap \ker u)\).
3. If two two-sided ideals \(\mathfrak{a}, \mathfrak{b}\) of \(E\) 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 \(\mathfrak{a}\) of \(E\), there is an inclusion-preserving bijection between the set of two-sided ideals of \(E/\mathfrak{a}\) and the set of two-sided ideals of \(E\) containing \(\mathfrak{a}\).

This is the \(A\)-algebra version of §Quotient Rings and Ring Homomorphisms, ⁋Theorem 3, and its proof proceeds in the same way except that we additionally check that the ring homomorphism obtained there is \(A\)-linear. The submodule conditions appearing in all four items come from the corresponding subsets being closed under the \(A\)-action of \(E\).

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References

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