# The Unapologetic Mathematician

## Algebras

We have defined a ring as a $\mathbb{Z}$-module (abelian group) $R$ with a linear function $R\otimes R\rightarrow R$ satisfying certain properties. The concept of an algebra takes this definition and extends it to work over more general base rings than $\mathbb{Z}$.

Let $A$ be a module over a commutative ring $R$ with unit. Then $A$ has both a left and a right action by $R$, since $R$ is commutative. Thus, when we take the tensor product $A\otimes_RA$, the result is also an $R$ module. It makes sense, then, to talk about an $R$-module homomorphism $A\otimes_RA\rightarrow A$. Equivalently, this is a “multiplication” function $m:A\times A\rightarrow A$ such that

• $m(a,b_1+b_2)=m(a,b_1)+m(a,b_2)$
• $m(a_1+a_2,b)=m(a_1,b)+m(a_2,b)$
• $m(ra,b)=m(a,rb)=rm(a,b)$

An $R$-module equipped with such a multiplication is called an $R$-algebra. We will often write the multiplication as $m(a,b)=ab$. In many cases of interest, the base ring will be a field $\mathbb{F}$, but any ring is an algebra over $\mathbb{Z}$.

Usually the term “algebra” on its own will refer to an associative algebra. This imposes an additional condition like the one we had in the definition of a ring: $(ab)c=a(bc)$. An algebra may also have a unit $1$ so that $1a=a=a1$ for all $a\in A$. Algebras can also be commutative if $ab=ba$ for all elements $a,b\in A$. There are other kinds of algebras we’ll get to later that are not associative.

Pretty much everything I’ve said about rings works for associative algebras as well, substituting “$R$-module” for “abelian group”. An $R$-module $M$ is a left $A$-module if there is an $R$-linear function $A\otimes_RM\rightarrow M$, and a similar definition works for right $A$-modules. We can take direct sums and tensor products of $A$-modules, and we have an $R$-module of homomorphisms $\hom_A(M_1,M_2)$. All these constructions are clear from what we’ve said about modules over rings if we consider that $A$ is a ring, and that an $A$-module is an abelian group with actions of both $R$ and $A$ which commute with each other.

The standard constructions of rings also work for algebras. In particular, we can start with an $R$-module $M$ and build the free $R$-algebra on $M$ like we built the free ring on an abelian group. Just use $\bigoplus_{n\in\mathbb{N}}M^{\otimes_R n}$, where the tensor powers over $R$ make sense because $R$ is commutative.

We can also start with any semigroup $S$ and build the semigroup algebra $R[S]$ just like we did for the semigroup ring $\mathbb{Z}[S]$. As a special case, we can take $S$ to be the free commutative monoid on $n$ generators and get the algebra $R[x_1,...,x_n]$ of polynomials in $n$ variables over $R$. In fact, almost all of “high school algebra” is really about studying the algebra $\mathbb{Q}[x_1,...,x_n]$, where $\mathbb{Q}$ is the field of rational numbers I’m almost ready to define.

Another source of $R$-algebras extends the notion of the ring of endomorphisms. If $M$ is any $R$-module, then ${\rm End}_R(M)=\hom_R(M,M)$ is again an $R$-module, and composition is $R$-bilinear, making this into an $R$-algebra.

Algebras over more general commutative rings than $\mathbb{Z}$ — particularly over fields — are extremely useful objects of study mostly because the linear substrate can often be much simpler. Building everything on abelian groups can get complicated because abelian groups can be complicated, but building everything on vector spaces over a field is generally pretty straightforward since vector spaces and their linear transformations are so simple.

May 8, 2007 - Posted by | Ring theory

## 11 Comments »

1. […] one that we haven’t considered directly: let be a commutative ring with unit and let be an algebra over with unit. Then we have a homomorphism of rings sending to — the action of on the […]

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2. […] mention the Bracket polynomial and the Jones polynomial. Jones was studying a certain kind of algebra when he realized that the defining relations for these algebras were very much like those of the […]

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3. […] go any further into linear algebra. Specifically, we’ll need to know a few things about the algebra of polynomials. Specifically (and diverging from the polynomials discussed earlier) we’re […]

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4. […] And, of course, once we’ve got monoids and -linearity floating around, we’re inexorably drawn — Serge would way we have an irresistable compulsion — to consider monoid objects in the category of -modules. That is: -algebras. […]

Pingback by Algebra Representations « The Unapologetic Mathematician | October 24, 2008 | Reply

5. […] or not. We know that the linear maps from a vector space (of finite dimension ) to itself form an algebra over . We can pick a basis and associate a matrix to each of these linear transformations. It turns […]

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6. […] We’re about to talk about certain kinds of algebras that have the added structure of a “grading”. It’s not horribly important at the […]

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7. […] and Symmetric Algebras There are a few graded algebras we can construct with our symmetric and antisymmetric tensors, and at least one of them will be […]

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8. […] we proceed with the differential geometry: Lie algebras. These are like “regular” associative algebras in that we take a module (often a vector space) and define a bilinear operation on it. This much is […]

Pingback by Lie Algebras « The Unapologetic Mathematician | May 17, 2011 | Reply

9. […] Algebras from Associative Algebras There is a great source for generating many Lie algebras: associative algebras. Specifically, if we have an associative algebra we can build a lie algebra on the same […]

Pingback by Lie Algebras from Associative Algebras « The Unapologetic Mathematician | May 18, 2011 | Reply

10. […] called and give it a bilinear operation which we write as . We often require such operations to be associative, but this time we impose the following two […]

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11. […] off, we need an algebra over a field . This doesn’t have to be associative, as our algebras commonly are; all we […]

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