Algebras

#algebra_theory

We've seen how the notion of a ring acting on an abelian group leads to the structure of a module. Can a ring act on another ring? Or, to phrase the question a bit differently, if a ring $R$ acts on a module $M$ , is it possible for $M$ to have a second operation that's both compatible with the given $R$ -action and makes $M$ into a ring? These two questions lead to the following two equivalent definitions of a structure known as an algebra.

Definition of an algebra (via modules)

Let $R$ be a commutative ring (with unity). An $R$ -algebra is an $R$ -module $A$ that is also equipped with a multiplication that makes $A$ into a ring (with unity), with the following compatibility property between the $R$ -action and the multiplication in $A$ :

r (a_{1} a_{2}) = (r a_{1}) a_{2} = a_{1} (r a_{2})

for all $r \in R$ and $a_{1}, a_{2} \in A$ .

Definition of an algebra (via rings)

Let $R$ be a commutative ring (with unity). An $R$ -algebra is a ring $A$ (with unity) together with a ring morphism^[1] $f : R \to A$ whose image is contained in the center of $A$ .

Let's quickly verify these two definitions are actually equivalent. First suppose $A$ is an $R$ -algebra in the first sense. For the sake of this analysis, let's use a $⋆$ to denote the action of an element $r \in R$ on an element $a \in A$ , and reserve a $\cdot$ (or no notation at all) for a product of elements in $A$ . Then $A$ is a ring (with unity) and we can consider the map $f : R \to A$ defined by $r \mapsto r ⋆ 1_{A}$ . We claim this is a ring morphism whose image is in the center of $A$ . First note we certainly have $f (1_{R}) = 1_{R} ⋆ 1_{A} = 1_{A}$ , since part of the assumption of the $R$ -action on the module $A$ is that the identity element $1_{R}$ acts as the identity on $A$ . Next note that properties of the $R$ -action on the module $A$ guarantee that

f (r + r^{'}) = (r + r^{'}) ⋆ 1_{A} = r ⋆ 1_{A} + r^{'} ⋆ 1_{A} = f (r) + f (r^{'}) .

Finally, observe that

\begin{aligned} f (r r^{'}) & = f (r^{'} r) (because R is commutative) \\ = (r^{'} r) ⋆ 1_{A} \\ = r^{'} ⋆ (r ⋆ 1_{A}) (by the properties of the R -action on the module A) \\ = r^{'} ⋆ f (r) \\ = r^{'} ⋆ (f (r) \cdot 1_{A}) \\ = f (r) \cdot (r^{'} ⋆ 1_{A}) (by compatibility of the R -action with the product in A) \\ = f (r) \cdot f (r^{'}) . \end{aligned}

So, our map $f : R \to A$ really is a ring morphism. Moreover, for every $r \in R$ the compatibility condition guarantees that for every $a \in A$ we have

f (r) a = (r ⋆ 1_{A}) a = r ⋆ (1_{A} \cdot a) = r ⋆ (a \cdot 1_{A}) = a \cdot (r ⋆ 1_{A}) = a f (r) .

Thus, $f (r)$ is in the center of $A$ .

Conversely, suppose $f : R \to A$ is a ring morphism whose image is contained in the center of $A$ . Then $A$ is an abelian group (under its additive operation) and we can define a set map $⋆ : R \times A \to A$ by $r ⋆ a = f (r) a$ . We claim this defines a left action of $R$ on $A$ . First note that

(r + r^{'}) ⋆ a = f (r + r^{'}) a = (f (r) + f (r^{'})) a = f (r) a + f (r^{'}) a = r ⋆ a + r^{'} ⋆ a,

and

(r r^{'}) ⋆ a = f (r r^{'}) a = f (r) f (r^{'}) a = r ⋆ f (r^{'}) a = r ⋆ (r^{'} ⋆ a) .

We also have

r ⋆ (a_{1} + a_{2}) = f (r) (a_{1} + a_{2}) = f (r) a_{1} + f (r) a_{2} = r ⋆ a_{1} + r ⋆ a_{2}

and

1_{R} ⋆ a = f (1_{R}) a = 1_{A} \cdot a = a .

So, we have indeed defined a left action of $R$ on $A$ , giving $A$ the structure of an $R$ -module. We also have

r ⋆ (a_{1} a_{2}) = f (r) (a_{1} a_{2}) = (f (r) a_{1}) a_{2} = (r ⋆ a_{1}) a_{2},

and also (since the image of $f$ is in the center of $A$ )

r ⋆ (a_{1} a_{2}) = f (r) (a_{1} a_{2}) = (f (r) a_{1}) a_{2} = (r_{1} f (r)) a_{2} = r_{1} (f (r) a_{2}) = r_{1} (r ⋆ a_{2}) .

Associative? Unital?

We assume rings have unity, which means we're assuming every algebra also has unity. There is an alternative definition without that assumption, which one would call a non-unital algebra.

There is also an alternative definition that results in a very similar structure to an algebra, with the notable exception that the multiplication in $A$ is not (assumed to be) associative. Such a structure (when the multiplication is not associative) is called a non-associative algebra.

We will not worry about these slightly more general structures.

Examples

Every ring (with unity) is a $Z$ -algebra. For each ring $A$ with unity,there is a unique ring morphism $Z \to A$ , and the image of that ring morphism is always contained in the center of $A$ .
If $A$ is a commutative ring (with unity), then $A$ is itself an $A$ -algebra. More generally, if $A$ is a ring (with unity) and $R \subseteq A$ is a subring of the center of $A$ , then $A$ is an $R$ -algebra.
The ring $M_{n} (R)$ of $n \times n$ matrices with entries in a commutative ring $R$ is an $R$ -algebra. The ring morphism $f : R \to M_{n} (R)$ sends each ring element $r$ to $r \cdot I_{n}$ , the diagonal matrix with $r$ on the diagonal.
More generally, if $M$ is an $R$ -module then the endomorphism ring ${End}_{R} (M)$ is an $R$ -algebra.
The field of complex numbers $C$ is a commutative $R$ -algebra, with $R \to C$ the usual inclusion map.
The quaternions $H$ is an $R$ -algebra but not a $C$ -algebra (as the complex numbers are not in the center of the quaternions).
The polynomial ring $R [x_{1}, \dots, x_{n}]$ is the free commutative $R$ -algebra on the set ${x_{1}, \dots, x_{n}}$ .
The tensor algebra $T (M)$ of an $R$ -module $M$ is an $R$ -algebra.
The symmetric algebra $S (M)$ and exterior algebra $⋀ (M)$ of an $R$ -module $M$ are both $R$ -algebras.

Category theory interpretation

Much like a ring is a monoid object in the category of abelian groups, an $R$ -algebra is a monoid object in the category of $R$ -modules.

Morphisms of algebras

As might be expected, morphisms of $R$ -algebras should be maps that respect "the algebraic structures." From the ring-centeric definition, that would mean:

Definition of an algebra morphism (via rings)

Suppose $A$ and $B$ are two $R$ -algebras. An $R$ -algebra morphism from $A$ to $B$ is a ring morphism $ϕ : A \to B$ such that $ϕ (r \cdot a) = r \cdot ϕ (a)$ for every $r \in R$ and $a \in A$ .

In other words, if $f : R \to A$ and $g : R \to B$ are the ring morphisms giving $A$ and $B$ their $R$ -actions, then we should have a commutative diagram

We always assume ring morphisms send $1_{R}$ to $1_{A}$ . ↩︎