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 directly.

First, for a general ring \(A\), neither \(\lMod{A}\) nor \(\rMod{A}\) is a monoidal category. This is resolved if we assume that \(A\) is a commutative ring, since then \((\lMod{A},\otimes_A, A)\) becomes a symmetric monoidal category. (§Products, Coproducts, and Tensor Products of Rings, §§Tensor Products of Rings)

Second, there is a historical issue: if we fix a commutative ring \(A\) and call a monoid object \(E\) in the symmetric monoidal category \((\lMod{A},\otimes_A,A)\) an \(A\)-algebra, as was traditionally done, then the associativity and unit axioms for a monoid object force the multiplication on \(E\) to always be associative and to have an identity element. However, many examples that have been studied and play important roles do not satisfy these properties. To include all of them, we must drop associativity and the existence of an identity element, which also means giving up the definition in §Definition of a Ring, ⁋Definition 1 to some extent.

For these reasons, from now on we always assume that \(A\) is 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. Furthermore, depending on the properties satisfied by the multiplication \(\mu\) on \(E\), we define the following.

• If the multiplication on \(E\) satisfies the commutative law, it is called a commutative \(A\)-algebra.
• If the multiplication on \(E\) satisfies the associative law, it is called an associative \(A\)-algebra.
• If the multiplication on \(E\) has 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 either as an \(A\)-module \(E\) equipped with a multiplication, or 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 \(\rho^\ast E\), a map \(\mu:\rho^\ast E\times \rho^\ast E \rightarrow \rho^\ast E\) is defined via the multiplication of \(E\), and at this time

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

and since \(\rho(\alpha)\in Z(E)\), we can verify the following equation:

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

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 equation

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

holds for all \(x,y\in E\).

The following proposition is not difficult to show.

Proposition 3 The composition of \(A\)-algebra homomorphisms is an \(A\)-algebra homomorphism. A bijective \(A\)-algebra homomorphism is always an isomorphism.

These data define the category of \(A\)-algebras, \(\Alg{A}\). We also define the category of commutative \(A\)-algebras, \(\cAlg{A}\).

Examples of Algebras

Just as a free ring is defined over an abelian group, a free algebra is defined over any \(A\)-module. This process is exactly the same as the one constructed in §Definition of a Ring, ⁋Proposition 4; in fact, this is natural, since all we additionally need to show is that the multiplication defined here is compatible with the scalar multiplication structure of \(A\), and this is already satisfied because a scalar multiplication structure on \(F(M)=\bigoplus_{n\geq 0}M^{\otimes n}\) is already defined by requiring \(A\) to be commutative.

That is, the following holds.

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

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

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

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

Definition 5 Fix a ring \(A\) (not necessarily commutative) and an arbitrary group \(G\). Then the group ring \(AG\) is, first of all, as a set the collection of finitely supported functions from \(G\) to \(A\), and 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 has the structure of an \(A\)-module, 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

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

any element of \(AG\) can be written as

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

for a family \(\alpha_x\in A\) indexed by elements of \(G\). For convenience, writing the function \(\delta_x\) as \(x\), 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 the two functors

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

Here \((-)^\times\) is the functor that sends an arbitrary \(A\)-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

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

That \(\tilde{f}\) preserves addition and scalar multiplication is obvious, and for multiplication as well,

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

so the desired equation holds. Conversely, if an arbitrary \(A\)-algebra homomorphism \(u: AG \rightarrow E\) is given, the function \(\bar{u}: G \rightarrow E\) defined by

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

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

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

and therefore the image of \(\bar{u}\) lies in \(E^\times\). That these two processes are inverse to each other is obvious.

The last example is already somewhat familiar.

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

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

and addition and multiplication are given by polynomial addition and multiplication, and scalar multiplication by \(A\) is likewise given by multiplication by a constant polynomial. More generally, for any set \(S\), \(A[S]\) can also be defined in a similar way.

In the case of multivariate polynomials \(A[S]\), for notational convenience it is common 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

\[\sum_{d=0}^n \sum_{\lvert\alpha\rvert=d} a_\alpha\x^\alpha.\]

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

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

Proposition 8 The functor \(A[-]:\Set \rightarrow \cAlg{A}\) that takes a set \(S\) and assigns the polynomial algebra \(A[\x_i]_{i\in I}\) with each element of \(S\) as a variable 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

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

is an \(A\)-algebra homomorphism from \(A[\x_i]_{i\in I}\) to \(E\). Conversely, when an arbitrary \(A\)-algebra homomorphism \(u:A[\x_i]_{i\in I} \rightarrow E\) is given, we define \(\bar{u}:S \rightarrow UE\) by \(\x_i\mapsto u(\x_i)\). It is easy to check that these are inverse to each other.

That is, the polynomial algebra can be thought of as a kind of free commutative \(A\)-algebra. Comparing Definition 5 and Definition 7, there is some similarity; if we generalize Definition 5 to think of a monoid ring, then the polynomial ring in Definition 7 becomes the monoid ring made from the monoid 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 \(E\).

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

Definition 10 A subalgebra \(\mathfrak{a}\) 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 special; it is simply the ideal of \(E\) when we forget the \(A\)-action on \(E\) and view \(E\) as just a ring. For 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 \(\mathfrak{a}\) of \(E\), we call \(E/\mathfrak{a}\) the quotient algebra of \(E\) by \(\mathfrak{a}\).

Of course, having defined it this way, one should prove that \(E/\mathfrak{a}\) is an \(A\)-algebra, but this is the same as for rings so we omit it. Also, 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 \(E\), \(E'+\ker u=\{x'+y\mid x'\in E', y\in\ker u\}\) 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 \(\mathfrak{a}\).