# The Unapologetic Mathematician

## Submodules, quotient modules, and the First Isomorphism Theorem for modules

Okay, getting a little back down to Earth now. Just like we had for groups and rings, we have an isomorphism theorem for modules.

First off, a submodule of a left $R$-module $M$ is just an abelian subgroup $N\subseteq M$ that’s closed under the action of $R$. That is, for any $r\in R$ and $m\in N$, we have $r\cdot m\in N$. A submodule $N$ comes with an inclusion homomorphism $\iota_{M,N}:N\rightarrow M$.

Now if we take an $R$-module $M$ and a submodule $N\subseteq N$ we can just consider $M$ and $N$ as abelian groups and form the quotient group. Remember that every subgroup of an abelian group is normal, so the quotient is again an abelian group $M/N$ made up of the “slices” $m+N$. It turns out that this is again an $R$-module. Just use the action $r\cdot(m+N)=(r\cdot m)+N$. If we chose a different representative $m+n+N$ of the slice, then we’d get $(r\cdot m+r\cdot n)+N$, which represents the same image, so this $R$ action is well-defined. Quotient modules come with projection homomorphisms $\pi_{M,N}:M\rightarrow M/N$.

Actually there’s one sort of submodule we’ve already seen. Remember that every ring $R$ is a left module over itself. Then the left submodules of $R$ are exactly the left ideals of $R$! If we have a two-sided ideal $I\subseteq R$ then we get a left action of $R$ on the quotient module $R/I$ and a right action as well, since $I$ is also a right submodule. Then we can take the tensor product $R/I\otimes_RR/I$ and get a linear function $R/I\otimes_RR/I\rightarrow R/I$ from multiplying representatives. Presto! Quotient ring!

We can use ideals to give submodules and quotient modules of other $R$-modules too. Take a ring $R$ with left module $M$ and a left ideal $I\subseteq R$. Then we can restrict the action of $R$ on $M$ to the ideal to get $IM=\{im|i\in I,m\in M\}$. This is clearly an abelian subgroup of $M$, and it turns out to be a submodule too. Indeed, we see that $r\cdot(im)=(ri)\cdot m\in IM$. Then we can make the quotient $R$-module $M/IM$ as above. Even better, if $I$ is two-sided this is actually a module over $R/I$: use the right $R$-module structure on $R/I$ and the left $R$-module structure on $M/IM$ and tensor to get $R/I\otimes M/IM$. Then we can get a linear function $R/I\otimes M/IM\rightarrow M/IM$ by choosing representatives and showing that the choice is immaterial.

I promised an isomorphism theorem. Well, I’ll state it, but the proof is pretty much exactly the same as the two we’ve seen before so I’ll leave you to review those. Any homomorphism $M\rightarrow N$ of left (right) $R$-modules factors as the composition $M\rightarrow M/M'\rightarrow N'\rightarrow N$, where the first arrow is the projection homomorphism, the third is the inclusion homomorphism, and the middle arrow is an isomorphism. We call the submodule $M'$ the kernel of the homomorphism and the submodule $N'$ the image of the homomorphism. Notice that there’s no restriction on the sorts of submodules that can be kernels here. For groups a kernel is a normal subgroup, and for rings a kernel is a two-sided ideal, but any submodule can be the kernel of a module homomorphism. This leads to a few more definitions that come in handy. The quotient module $M/M'$ is called the coimage, and the quotient module $N/N'$ is called the cokernel. Thus we see that the coimage and the image of any homomorphism of modules are isomorphic.

May 9, 2007 Posted by | Ring theory | 10 Comments