## Wordy Introduction, Motivation.

When you first start high school algebra, the big thing is FOIL-ing, right?  Factoring and factorizing quadratics.  When you get to calculus, the big things are derivatives and integrals.  Then when you get to college and start doing math, things get a little tougher.  We start learning about abstract structures, and these become increasingly specific and increasingly complex as we go along.

In an undergrad abstract algebra class, the big thing is groups.  Dihedral groups, general and special linear groups, permutation groups, fundamental groups, and so on.  From here, we go onto Rings, giving our groups a little bit of a multiplicative and additive structure.  Now we either give our ring some inverses to make it a field then make a vector space from this, or we take another route and stick with rings —

We study commutative rings, PIDs, ED’s, and other special types of rings; at some point we say, “Hey, let’s make something that’s like a vector space, but with a ring instead of a field.”  Thus, $R$-modules.  As you’ll recall, an $R$-module is almost exactly the same thing as a vector space, except we take out everything that has to do with the scalars having multiplicative inverses.

I wasn’t quite comfortable with modules until this year.  A little while after I sat down to really consider modules, I wrote this post which hopefully helped some other people think about modules as “more general vector spaces.”  I put this in quotes because we really don’t need to think about vector spaces at all to define modules, but, for me, it makes it a lot easier to think about them as “kind of like” vector spaces.  After starting Eisenbud’s commutative algebra book (which I highly recommend, by the by.) I realized that he wasn’t going to be doing a lot of module talk.  Almost right off the bat, we start to make use of algebras.

Well, what the hell is that?

Our good friends Dummit and Foote (p. 657) tell us:

Let $k$ be a field.  Recall that a ring $R$ is a k-algebra if k is contained in the center of $R$ and the identity of k is the identity of $R$.

This seems reasonable.  But it gives me absolutely no intuition regarding what this algebra actually is; I don’t instantly have a picture in my head, I need to work at it.  Also, what part of this is the algebra?  Is it $R$?  Is it $k$?  Or is it $k$ considered as part of $R$?

What does Wikipedia say?

Let $R$ be a commutative ring.  An algebra is an $R$-module $A$ together with a binary operation $[\cdot ,\cdot]\colon A\times A\rightarrow A$ called $A$-multiplication which [is bilinear].

This is getting a little clearer.  First, note that our algebra is now just an $R$-module, instead of picking out a field.  This is a little more general.  Also, we are given that it is an $R$-module plus some nice bilinear “multiplication” of elements in $A$.  So what does this say?  It says that we can multiply elements of our $R$-module, and not just the scalars.  Also, we can multiply them in nice ways that allow nice distributions.

Notice, also, that this is nearly the same definition as Dummit and Foote give, albeit slightly more general — under certain mappings (and thinking about what an $R$-module is) these definitions are equivalent.

One last source.  Today, I asked my algebraic geometry professor, and was told

It’s just an $R$-modules with a ring structure.

If $R$ is commutative (which I’m almost always considering), then this is a brilliant characterization.  Thinking about the elements of our $R$-module, $M$, we can write them as

$\displaystyle\sum_{i=1}^{n} r_{i}m_{i}$

where each $r_{i}\in R$ and $m_{i}\in M$.  We can multiply scalars from $R$ together, and we can even distribute in a nice way, but this structure is very much like an Abelian group: we can add and subtract the $r_{i}m_{i}$ nicely, but we cannot, under this structure, multiply them together.

But, hold on a second.  Why not?  $R$ is a ring, so it has a natural multiplicative structure, but what about $M$?  Recall that $M$ can be ANYTHING.  It could be the set of all different kinds of apples.  It could be the set of all integers.  It could be the permutation group on sixteen elements.  There is not necessarily a multiplicative structure on this.

Occasionally, though, there is a multiplicative structure that we can define which will be useful.  If $M$ is the set of all continuous functions on the interval $[0,1]$ then we may define the multiplication

$g\cdot f = g(x)\cdot f(x)$

for every $x\in [0,1]$.  If $M$ is the set of all polynomials in one variable over some ring, then we can multiply in a relatively obvious way:

$(ax + bx^{3})\cdot (c + dx^{4}) = acx + adx^{4} + bcx^{3} + bdx^{7}$

and so on.  Typical high school algebra.  And, in fact, this is more or less the idea behind an algebra: we create a multiplicative structure for our $R$-module.

Hence, an $R$algebra is an $R$module with a multiplicative structure.

(Little note: Then why was Dummit and Foote’s different?  We built our algebra our of an $R$-module, but sometimes we like to start with other things; for example, we can start from some ring $R$ and define an $R$-algebra via some sufficiently nice defining of the underlying set, or we could start with a ring $R$ and provide a ring homomorphism from our underlying set $A$ into our ring satisfying a few properties (this one is the one D+F use).  Ultimately, if I’m not going to be building them up and just want to think about them, then I prefer to use the vague bolded definition above.)

## The Prototypical Example.

So far, I haven’t seen too many examples of algebras; most of the time, the same ones crop up.  There are two related examples that I just want to detail before I end this post.

$R[x_{1}, \dots, x_{n}]$.

This is the polynomial ring over $n$ variables with coefficients in $R$.  Truth be told, I have very rarely seen anything but a field adjoined to this (making it $k[x_{1}, \dots, x_{n}]$) but for now it doesn’t really matter.  But if you’re like me, you’ll probably just want to think of polynomials with real coefficients.  Either way, a typical element of this looks like

$\displaystyle\sum ax_{1}^{i_{1}}x_{2}^{i_{2}}\cdots x_{n}^{i_{n}}$

where this sum is finite, and where $a\in R$.  This whole thing means “a bunch of powers of our variables multiplied together and then added to other stuff that looks like that.”  Well, what’s the multiplicative structure?  Yeah, just multiply together and add up powers to matching variables.  For example,

$x_{1}^{2}x_{2}^{3} \cdot x_{1}x_{2}^{4} = x_{1}^{3}x_{2}^{7}$

This multiplicative structure makes our $R$-module into an $R$-algebra.  Good job!

The second example is almost exactly the same.

$R[x_{1}, \dots, x_{n}]/ J$.

Note here that $J$ is some ideal of $R[x_{1}, \dots, x_{n}]$ and it essentially says, “make some stuff equal to zero.”  The multiplicative structure is the same, except after you multiply, you need to reduce depending on what $J$ is.  Let’s do a concrete example of this.

${\mathbb R}[x]/(x^{2} - 1)$.

This says to take elements of ${\mathbb R}[x]$ and then reduce them by taking

$x^{2} - 1 = 0$

or, equivalently,

$x^{2} = 1$

Using this equality is the key to reducing elements in this ring.  Thus the element

$x^{3} + 2x^{2} + x + 2$

in this ring is reduced to

$x\cdot 1 + 2\cdot 1 + x + 2 = 1 + 2 + x + 2 = x + 5$

where we simply replace $x^{2}$ with $1$, since this is exactly what modding out by the above ideal tells us to do.  Similarly, when multiplying, we may get

$(x + 2)\cdot(x - 4) = x^{2} + 2x - 4x - 8 = x^{2} - 2x - 8$

but we now need to reduce this to get

$1 - 2x - 8 = -2x - 7$

which is the most reduced form.

Equip with this multiplicative structure, our ${\mathbb R}$-module becomes an ${\mathbb R}$-algebra!  Nice!