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.

Rings have unity

Recall our standing assumption on rings, namely that all rings have a multiplicative identity ("unity") and that ring morphisms send unities to unities. In other words, our rings are always drawn from the category $Ring$ , of rings with unity together with unity-preserving ring morphisms.

Definition of an algebra (via modules)

Let $R$ be a commutative ring. An $R$ -algebra is an $R$ -module $A$ that is also equipped with a multiplication that makes $A$ into a ring, 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 $a_{1}, a_{2} \in A$ .

We can think of the above condition as the formalization of the idea that " $R$ can act before or after multiplication." In other words, given elements $r \in R$ and $a_{1}, a_{2} \in A$ , we can either multiply the elements in the algebra together first and then act by $r$ , or act by $r$ on one of the elements and then multiply the result by the other element.

In particular, notice that it looks symbolically like the element $r$ "commutes with" the elements of the algebra. This is not technically true, since the elements of the ring $R$ are not elements of the algebra $A$ , but it justifies the condition in the ring-focused definition, below.

Definition of an algebra (via rings)

Let $R$ be a commutative ring. An $R$ -algebra is a ring $A$ together with a ring morphism $f : R \to A$ 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^{'}) & = (r r^{'}) ⋆ 1_{A} \\ = r ⋆ (r^{'} ⋆ 1_{A}) (by the properties of the R -action on the module A) \\ = r ⋆ f (r^{'}) \\ = r ⋆ (1_{A} \cdot f (r^{'})) \\ = (r^{'} ⋆ 1_{A}) \cdot f (r^{'}) (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. When the multiplication is not associative such a structure is called a non-associative algebra.

We will not worry about these slightly more general structures.

Examples

Every ring is a $Z$ -algebra. For each ring $A$ , 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, then $A$ is itself an $A$ -algebra. More generally, if $A$ is a ring 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, via the inclusion morphism $R \to C$ .
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.

Note that in the above examples we often said things like "Every ring $A$ is a $Z$ -algebra ... ", which is somewhat incorrect. We should really be careful to specify the categories in which we are working, saying things like "There is a functor $F : Ring \to Z-Alg$ with object function sending each ring $A$ to the $Z$ -algebra defined by the unique ring morphism $Z \to A$ ." Can you figure out how to make each example above formally correct?

Morphisms of algebras

As might be expected, morphisms of $R$ -algebras should be maps that respect "the algebraic structures." From the ring-centric 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

Definition of algebra morphism (via modules)

Suppose $A$ and $B$ are two $R$ -algebras. An $R$ -algebra morphism from $A$ to $B$ is an $R$ -module morphism $ϕ : A \to B$ that is also a ring morphism of $A$ and $B$ as rings. In other words, it is a set map satisfying the following properties:

$ϕ (1_{A}) = 1_{B}$
$ϕ (a_{1} + a_{2}) = ϕ (a_{1}) + ϕ (a_{2})$ for all $a_{1}, a_{2} \in A$
$ϕ (a_{1} a_{2}) = ϕ (a_{1}) ϕ (a_{2})$ for all $a_{1}, a_{2} \in A$
$ϕ (r a) = r ϕ (a)$ for all $r \in R$ and $a \in A$ .

We can now form the category $R - Alg$ whose objects are $R$ -algebras and morphisms are morphisms of $R$ -algebras. Note that, by construction, we have forgetful functors both to the category $Ring$ of rings and to the category $R - Mod$ of $R$ -modules:

The forgetful functor to rings is not full (i.e., surjective on hom sets): for two $R$ -algebras $A$ and $B$ , there are ring morphisms $ψ : U_{2} (A) \to U_{2} (B)$ that are not compatible with the given $R$ -actions (and hence are not of the form $U_{2} (ϕ)$ for some $R$ -algebra morphism $ϕ : A \to B$ ).

The same is true for the forgetful functor to $R$ -modules, for analogous reasons.

Category theoretic 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. Since I haven't (yet) typed up any notes on monoidal categories and monoid objects, I won't say any more here.

Suggested next notes

Tensor algebras
Symmetric algebras
Exterior algebras