## What the Hell is a Module?

### October 12, 2010

This post is going to be a gentle introduction to what a module is.  It isn’t hard, but, for me, modules were sort of just “thrown in” with a whole bunch of defining properties and no motivation for why I should care about them.  I’m hoping to motivate them at least a little bit so that you feel more comfortable thinking and working with them!

We’ve gone through vector spaces.  We’ve essentially done these to death!  We all know what the defining properties of the space are:

Vector Space Defining Properties.

• Associativity of addition of vectors.
• Commutativity of addition of vectors.
• Vector identity.
• Scalar Distributive Law into a sum of vectors.
• Vector Distributive Law into a sum of scalars.
• Scalar multiplication works nicely: $a(b{\bf v}) = (ab){\bf v}$ where $a,b$ are scalars, ${\bf v}$ is a vector.
• Scalars have a multiplicative identity.

Yes, it’s a lot to remember, but as soon as you’ve worked with vectors for a bit, you’ll realize that all of these are more-or-less intuitive.  Adding two vectors is no big deal, vectors commute, and scalars work nicely with vectors.

But what is a vector?  Given a vector space $V$ over a scalar field ${\mathbb R}$ (so, we’re working over the reals here), we have that $v\in V$ is a vector just means that it’s some element of $V$.  It doesn’t really need to be an arrow, but most of the time it’s nice to picture it as an arrow.  But, for example, we could have

$V = \{0, apple, pear\}$

and this can be our vector space over the reals.  We can, then, talk about things like

$2*(apple) + 3*(pear)$

and

$(apple) - (apple) = 0$

and these do not have intuitive “arrow” pictures.  Nonetheless, suitably defined, we can make this into a vector space.

It is, therefore, nice to think about vectors simply as elements of the set $V$ that work nicely with scalars.  Obviously, this is not an intuitive way to think about a vector space, but it is a nice way to lead up into modules.  In fact, modules are going to be quite similar to vector spaces; the only difference will be the scalars we use.

## So, what’s the deal with the Scalars?

Well, remember when I said that a vector space is a set of vectors that work nicely with scalars?  Well, what are the scalars?

In algebra, the scalars form what is called a field. If you don’t know what a field is, think about the real numbers: it’s closed under addition and multiplication (adding or multiplying two real numbers gives you a real number), it’s associative and commutative in multiplication and addition, there’s an additive and multiplicative identity, and there are additive and multiplicative inverses.  There’s also a nice distributive law that lets us distribute products over sums and sums over products.  Good.

So, this is pretty damn strict.  Not only does everything have an additive inverse, but everything has a multiplicative inverse, and this makes every element into a unit — which is actually a very strong condition on a set.  There are plenty of nice things which we use all the time that don’t have multiplicative inverses.

• ${\mathbb Z}$, the integers, do not have multiplicative inverses for any of their elements except 1 and -1.
• The set of all matrices doesn’t have multiplicative inverses for any of the matrices of determinant equal to 0.
• Any group under addition does not necessarily have a multiplicative inverse for any element besides the identity.

And these are pretty nice structures.  So, you know what?  We’re going to take out the requirement that every element has a multiplicative inverse. This structure is called a (commutative) Ring.

Most of the time, we’re going to want to use that multiplication is commutative — in particular, for commutative algbera — but sometimes we want to use things like matrices which don’t have such a nice commutative multiplication.  Therefore, we also take out the requirement that multiplication is commutative. This structure is now a general, not-necessarily-commutative, Ring.

In sum, a Ring is just a Field without commutative multiplication and multiplicative inverses.  Note that a ring can have these properties, but it is not necessary.

We know a lot of rings already.  The integers, the set of square matrices of a given size, and any field is automatically a ring: this gives us the real numbers, the rational numbers, the complex numbers, and so on.

## So?  Who cares about Rings?

Well, is it so hard to imagine a vector space that has scalars in a ring instead of a field?  Not really, right?  For example, we can just use the integers instead of real numbers to define vectors.  An obvious problem with this is that it would be difficult or impossible to normalize or scale these kinds of vectors.  But sometimes we don’t really care about normalizing, and we just want to have a nice space that we can talk about whose scalar field is a ring.

This is exactly what a module is.  We can think of a module as a “vector space” where our field of scalars is a ring instead of a field.  We say that for some ring $R$, that $M$ is an $R$-module when $M$ is our field of “vectors” (which we now just called elements, since it’s no longer a vector space) and where $R$ is our ring.

## Oh, wow, Neato.  Can I have an Example?

There was a joke back in my old school that it was annoying to teach undergraduates abstract algebra before linear algebra, because when you ask them what a vector space is they’d say, “a $k$-module for some field $k$.”

Of course, this is true.  If our ring is actually a field, our module becomes a vector space. Thus, an ${\mathbb R}$-module $V$ is actually a vector space.  So, for example, letting $V ={\mathbb R}^{2}$, we have that this is the “normal” 2-dimensional vector space over the reals.

A ${\mathbb Z}$-module is exactly an abelian group.  Think about this!  Really work it out.  It’ll blow your mind.

We can take our set $M = {\mathbb R}^{2}$ and take our scalars to be the set of all $2\times 2$ matrices.

There are a number of sweet examples that I’ll begin talking about as we go on to talk about commutative algebra.  For now, just live, breathe, and dream modules.  Do it!

### 7 Responses to “What the Hell is a Module?”

1. For example, we can just use the integers instead of real numbers to define vectors. An obvious problem with this is that it would be difficult or impossible to normalize or scale these kinds of vectors.

Not necessarily a problem. Sometimes you have to do integer optimisation so you want to restrict to just integers. Also a physicist I follow on tumblr, davidaedwards.tumblr.com, has some physics theories with adeles which are also not smooth.

Actually if you consider all of the philosophical problems with the reals it might just be preferable to stay away in general, if possible.

• James said

I guess a better way to have stated that would be, “An obvious difference with this…” In general, I’ll probably not discuss any philosophical problems with the reals since, as far as I can tell, it serves onto to ignite the fires of internet flame wars. :]

2. There was a joke back in my old school that it was annoying to teach undergraduates abstract algebra before linear algebra, because when you ask them what a vector space is they’d say, “a k-module for some field k.”

Speaking of doing that: http://www.math.miami.edu/~ec/book/ is very good so far as I’ve gotten.

• James said

It looks like a good text, though it seems a bit dense for an introductory text. I do see the category theory seeping in, though —

3. Do basis isomorphisms make sense in {0, apple, pear}≝V ? It seems like V the way you’re describing it looks really nice in one basis, but let’s say I chose the basis {0, apple+pear, apple−pear}, now it’s harder to think about.