# Mathematics Prelims

## November 28, 2009

### Modules

Filed under: Abstract Algebra,Algebra — cjohnson @ 9:59 pm

Anyone who has spent any amount of time in algebra or analysis has come across vector spaces: a set of elements, vectors, which we can add together or multiply by a scalar from some fixed field.  This idea is simple and natural enough that students in high-school physics classes become familiar with the basic idea of vectors, if only in an informal way.  However, in the axioms of a vector space there’s nothing that particularly requires that we pull our scalars from a field (though other results in vector space theory depend on our having an underlying field).  Indeed, we could just require that the scalars we multiply by simply be elements of a ring; this gives us a structure known as a module.  Though this seems like a simple enough generalization, module theory has some quirks that those of us more accustomed to vector spaces will find odd.  For instance, though every vector space has a basis, there are modules which do not; and even if a module has a basis, two different bases may have different cardinalities.

To be more precise, given a ring $R$ and an abelian group $M$ (written additively), we say that $M$ is a (left) $R$-module if there exists a ring homomorphism $\phi : R \to \text{End}(M)$.  (Recall that if $M$ is an abelian group, the collection of endomorphisms of $M$ forms a ring under piecewise addition and function composition.)  This says simply that given an $r \in R$ and an $x \in M$, we can define the multiplication of $x$ by $r$ as $rx = \phi(r)(x)$.  Given any $r, s \in R$ and any $x, y \in M$, the following are immediate consequences of our definition.

1. $(r + s)x = rx + sx$
2. $r(x + y) = rx + ry$
3. $(rs)x = r(sx)$

If in addition the ring $R$ has an identity, we may also require that $1x = x$ for all $x \in M$.  We call such a module unitary.  Though this isn’t strictly required, we will assume that all modules we deal with are unitary if the ring has identity.  Notice that we only perform multiplication on the left.  If we define multiplication on the right, we have a right $R$-module.  We will assume that all of our modules are left module unless otherwise specified.

The three properties listed above aren’t simply consequences of our definition of a module, but can actually be used as an alternative definition.  That is, if we have an abelian group $M$, a ring $R$, and some multiplication $R \times M \to M$ satisfying the above properties, we then have a module.  To see this, define for each $r \in R$ a $M$-endomorphism $\phi_r : M \to M$ by $x \mapsto rx$.  Notice that the second property above guarantees that $\phi_r$ is indeed an endomorphism: $\phi_r(x + y) = r(x + y) = rx + ry = \phi_r(x) + \phi_r(y)$.  Now defining a map $\phi : R \to \text{End}(M)$ as $r \mapsto \phi_r$, properties one and three guarantee that this is in fact a ring homomorphism.

As stated earlier, a module can be thought of as a generalization of a vector space.  In fact, if our ring $R$ is a field, then any $R$-module is simply a vector space over $R$.  A module can also be thought of as a generalization of an abelian group, in the sense that every abelian group is in fact a $\mathbb{Z}$-module.  Suppose that $A$ is an abelian group.  For each positive $n \in \mathbb{Z}$, define $na$ as $\sum_{i=1}^n a$.  If $n = 0$, define $na = 0$.  Finally, if $n < 0$, define $na$ as the inverse (in $A$) of $(-n)a$.

Notice that every ring can be viewed as a module over itself; $R$ is an $R$-module where scalar multiplication is simply the ring’s usual multiplication.  Additionally, if $S$ is a subring of $R$, then $R$ can be viewed as an $S$-module as the three properties of an $S$-module are satisfied by usual multiplication in the ring.  Similarly, if $I$ is left ideal of $R$, then $I$ is a left $R$-module; if $I$ is a right ideal, then it’s a right $R$-module.

As we may pull our scalars from a ring instead of a field, we can treat some more “interesting” objects are scalars.  For instance, suppose that $V$ is an $F$-vector space and $T : V \to V$ a linear transformation.  Then $F[x]$, the set of all polynomials with coefficients from $F$, forms a ring.  For any $f(x) \in F[x]$ with $f(x) = \sum_{i=0}^n f_i x^i$, we define $f(T) = \sum_{i=0}^n f_i T^i$ where $T^i$ means the $i$-fold composition of $T$ with itself.  For any $v \in V$ we define $f(x) \cdot v$ as $f(T)(v) = \sum_{i=0}^n f_i T^i(v)$.  Notice this satisfies the properties of a $F[x]$-module.

1. $(f(x) + g(x)) \cdot v = (f(T) + g(T)) \cdot v = f(T)(v) + g(T)(v) = f(x) \cdot v + g(x) \cdot v$
2. $f(x) \cdot (u + v) = f(T)(u + v) = f(T)(u) + f(T)(v) = f(x) \cdot u + f(x) \cdot v$
3. $f(x) \cdot (g(x) \cdot v) = f(T)(g(T)(v)) = (f(T) \circ g(T))(v) = (f(T) g(T))(v) = (f(x) g(x)) \cdot v$

Thus this module, which we’ll denote $V_T$, has polynomials as its scalars.