# Mathematics Prelims

## December 1, 2009

### Submodules and Homomorphisms

Filed under: Abstract Algebra,Algebra — cjohnson @ 8:57 pm

Just as a subgroup is a group within a group or a subfield is a field within a field, a submodule is a module within a module.  That is, if $M$ is an $R$-module, we say that $N$ is a submodule of $M$ if $N$ is a subgroup of $M$ and is closed under multiplication by elements of $R$.  This can be summarized by saying that for every $x, y \in N$ and every $r \in R$, the following two properties hold.

1. $x - y \in N$
2. $rx \in N$

In the case of groups, we could only form a quotient group if the subgroup we were modding out by as sufficiently “nice” (i.e., was a normal subgroup); likewise, in the case of rings we required that a subring be nice (an ideal) in order to form a quotient ring.  With modules however, we can always form the quotient module with a submodule.  We can do this since, as $M$ is an abelian group under addition, all subgroups are normal and we can form the quotient group $M/N$.  This is naturally an abelian group, so in order to turn this into an $R$-module we have to define multiplication of elements in $R$ and $M/N$, which we do in the most obvious way: $r(x + N) = rx + N$.  To check that this is in fact an $R$-module, we simply verify that the three axioms of a module hold.

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

Finally, if $M$ is a unitary module, then so too is $M/N$: $1(x + N) = 1x + N = x + N$.

We say a map $\phi : M \to N$ between two $R$-modules is a homomorphism if $\phi$ is a group homomorphism with the additional property that for each $r \in R$, $\phi(rx) = r \phi(x)$.  As you would expect, the kernel of this homomorphism is a submodule of $M$, and the image is a submodule of $N$.

Supposing that $S \subseteq M$, we define $R\langle S \rangle$ to be the smallest submodule of $M$ containing $S$, which is naturally the intersection of all submodules containing $S$

$\displaystyle R\langle S \rangle = \bigcap \left\{ N : N \text{ a submodule of } M \text{ and } S \subseteq N \right\}$

If $S$ is a finite set, we may write $R\langle s_1, s_2, \ldots, s_n \rangle$ in place of $R \langle S \rangle$, and in the event that $S$ is a singleton, we say that $R\langle s_1 \rangle$ is a cyclic submodule of $M$.  We call the elements of $S$ the generating set of the $R \langle S \rangle$ submodule, and call the elements of $S$ the generators of this submodule.