Algebras

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

To begin with, the first problem is that for a general ring \(A\), \(\lMod{A}\) or \(\rMod{A}\) is not a monoidal category. This is because if \(A\) is a commutative ring, then \((\lMod{A},\otimes_A, A)\) becomes a symmetric monoidal category, so 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 historical one: 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 multiplication of \(E\) must always satisfy associativity and have an identity element, due to the associativity and unit axioms of a monoid object. However, many important examples that have been studied so far often do not satisfy these properties. Therefore, in order to include all of them, we must give up associativity and the existence of an identity element, which also means partially giving up a definition like §Rings, ⁋Definition 1.

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 we call \(E\) an \(A\)-algebra. Also, depending on the properties satisfied by the multiplication \(\mu\) of \(E\), we define the following.

• If the multiplication of \(E\) satisfies commutativity, we call it a commutative \(A\)-algebra.
• If the multiplication of \(E\) satisfies associativity, 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 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 a multiplication defined on an \(A\)-module \(E\), but it can also be thought of as a scalar multiplication defined on a ring \(E\). 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 through 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

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

through this. In this case, the ring homomorphism \(\rho:A \rightarrow E\) giving an \(A\)-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 \(x,y\in E\).

It is not difficult to show the following proposition.

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 \(\Alg{A}\) of \(A\)-algebras. Also, we define the category \(\cAlg{A}\) of commutative \(A\)-algebras.

Examples of Algebras

Just as free rings are defined over abelian groups, free algebras are defined over any \(A\)-module. This process is exactly the same as the one constructed in §Rings, ⁋Proposition 4; thinking about it, this is natural, since all we need to additionally show is that the multiplication defined here is compatible with the scalar multiplication structure of \(A\), which is already guaranteed because requiring \(A\) to be commutative means that the scalar multiplication structure is already defined 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 of the forgetful functor \(\Alg{A} \rightarrow \lMod{A}\).

The \(F(M)\) obtained from the above proposition uses the addition, which is the data already defined on \(M\), as it is, and is obtained by defining a new operation, multiplication. On the other hand, it is also possible to make an \(A\)-algebra using the 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 \(A\); 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 \(x\in G\), if we define \(\delta_x: G \rightarrow A\) by the formula

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

then 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 the elements of \(G\). For convenience, if we write the function \(\delta_x\) as \(x\), then 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

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

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 any \(A\)-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

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

Then it is obvious that \(\tilde{f}\) preserves addition and scalar multiplication, and also in the case of multiplication,

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

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

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

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

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

and therefore the image of \(\bar{u}\) lies in \(E^\times\). That these two processes are inverses of 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 here addition and multiplication are given by polynomial addition and multiplication, and the 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 \(I\) satisfying \(\lvert S\rvert=\lvert I\rvert\) and denote the elements of \(S\) by \(\x_i\). Then for arbitrary

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

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 the formula

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

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

Likewise, 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 \(A[\x_i]_{i\in I}\) having 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 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 \(E\). Conversely, given an arbitrary \(A\)-algebra homomorphism \(u:A[\x_i]_{i\in I} \rightarrow E\), we define \(\bar{u}:S \rightarrow UE\) by \(\x_i\mapsto u(\x_i)\). It is easy to check that these are inverses of 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 a slight similarity: if we generalize Definition 5 to think 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 \(E\).

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

Definition 10 A subalgebra \(\mathfrak{a}\) of an \(A\)-algebra \(E\) is called a left ideal of \(E\) if \(\alpha x\in \mathfrak{a}\) holds for arbitrary \(x\in \mathfrak{a}\) and \(\alpha\in E\). Similarly, a right ideal is also defined. An ideal that is both a left ideal and a right ideal is called a two-sided ideal.

In other words, an ideal of an \(A\)-algebra is nothing special; it is simply the ideal of \(E\) when we forget the \(A\)-action defined on \(E\) and view \(E\) merely as 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 an arbitrary \(A\)-algebra \(E\) and a two-sided ideal \(\mathfrak{a}\) of \(E\), we call \(E/\mathfrak{a}\) the quotient algebra of \(E\) by \(\mathfrak{a}\).

Of course, we should prove that \(E/\mathfrak{a}\) is an \(A\)-algebra under this definition, but since it is the same as for rings, 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 exists an inclusion-preserving bijection between the set of two-sided ideals of \(E/\mathfrak{a}\) and the set of ideals of \(E\) containing \(\mathfrak{a}\).