# 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

1. Can this get proven in categorical generality? What do we need from the category for the isomorphism theorem to hold?

Comment by Mikael Johansson | May 10, 2007 | Reply

I’ll leave you to think about that for now, and to ponder the second and third isomorphism theorems. The first requires less than those do from the category, but there is a good answer to your question.

Comment by John Armstrong | May 10, 2007 | Reply

3. There is quite an interesting report here: http://cse.lmu.edu/mathematics/frugonithesis.pdf that I found whilst searching the web for information about the First Isomorphism Theorem. Towards the end, it partially answers the category theoretical question (I think – I don’t really know any category theory and originally read it for the ring-theoretic setting of the theorem.)

Comment by Jake | May 10, 2007 | Reply

4. That looks like a good paper, Jake. And it gets to some of the setup in abelian categories, which I figure are most of what Mikael is interested in. The result is even more general than that, though. Groups, for instance, do not form an abelian category, yet they have an isomorphism theorem.

Comment by John Armstrong | May 10, 2007 | Reply

5. Actually, y’know, I do care about non-abelian categories too from time to time. Even though I do concentrate, in my research, on R-Mod and weird things we can figure out from R-Mod.

The fact that we have isomorphism theorems for groups, rings, vector spaces, algebras, Lie algebras, et.c, et.c was one of the things that first made me start valuing category theory as a concept. I never have gotten around to investigate the isomorphism theorems as such, but it remains one of the most powerful thought germs I’ve ever seen.

Comment by Mikael Johansson | May 11, 2007 | Reply

6. I just skimmed the paper, and the argument seems to be that since abelian categories have short exact sequences, it’s natural to talk about the isomorphism theorem there.

Which immediately leads me to ask: “What about triangulated categories?” – where we don’t have short exact sequences, but we do have the next best thing: triangles (corresponding vaguely to the short exact sequences in homology you get in beginning homological algebra)

Comment by Mikael Johansson | May 11, 2007 | Reply

7. Yeah, that’s still not quite it. To give a hint as to why this happens in abelian categories, I’ll tell you that the technical statement of the isomorphism theorem for a category C is that C is (regular mono, epi)-factorizable.

Oh, and I did say “most” of what you’re interested in, not “all” :P.

Comment by John Armstrong | May 11, 2007 | Reply

8. This isn’t my area, but do you mean (mono, regular epi)-factorizable?

Comment by Walt | May 14, 2007 | Reply

9. You know, I might have swapped it around.. I wasn’t planning on doing this generality for a while yet, so I haven’t reviewed it recently.

Yeah, you’re right, Walt.

Comment by John Armstrong | May 14, 2007 | Reply

10. doing m phil pls sent me detail of module theory

Comment by anil | August 20, 2007 | Reply