The Unapologetic Mathematician

Mathematics for the interested outsider

More modules, more ideals

The first construction I want to run through today is related to the amalgamated free product from group theory. Here’s the diagram in modules:
Pushout diagram for modules
Remember we read it as follows: If we have modules N, M_1, and M_2, and homomorphisms from N into each of M_1 and M_2, then the “amalgamated direct sum” M_1\underset{N}\oplus M_2 is another module with homomorphisms into it from each of M_1 and M_2 making the square commute. Further, for any other module X and pair of homomorphisms into it, there is a unique homomorphism from M_1\underset{N}\oplus M_2 to X.

How do we know that such a thing exists? Well, we can take the direct sum of M_1 and M_2, which comes with homomorphisms into it from M_1 and M_2. Then we can follow both paths from N to M_1\oplus M_2. In general they’re different homomorphisms, since one has an image completely in M_1 and the other completely in M_2. But since we’re looking at module homomorphisms we can subtract one from the other. To make the square commute we want the image of this difference to be zero, and we can make it be zero by forming the quotient module!

This turns out to be useful right away. Let’s say that M_1 and M_2 are both submodules of a module M. They definitely share the zero element of M, but they might share a larger submodule than that. It’s easily verified that their intersection N is a submodule, and it comes with inclusion homomorphisms into each of M_1 and M_2. Now if we want to “add” the submodules M_1 and M_2, we had better not treat elements in their intersection differently depending on which submodule module we pick them from, since they’re all just submodules of M, so the direct sum isn’t what we want.

Instead, we find that the submodule M_1+M_2 of all elements of M of the form m_1+m_2 with m_1\in M_1 and m_2\in M_2 is isomorphic to the direct sum amalgamated over their intersection. In particular, we can apply this to submodules of our base ring R itself — ideals! We define the sum I+J of two ideals as this sum of submodules of R.

There’s one more thing we can do for ideals. If we have a left ideal I\subseteq R and an abelian subgroup A\subseteq R then we can form their tensor product over \mathbb{Z}: I\otimes A. Since I is a left R-submodule, this turns out to be a left R-submodule of R\otimes R. Then we can use the multiplication on R to get a homomorphism I\otimes A\rightarrow R. We denote its image as IA, and it is a left ideal of R. In terms of elements, it’s the set of all sums of products in R: \sum\limits_{k=1}^ni_ka_k with i_k\in I and a_k\in A. In particular, we could choose A to be another ideal J and get the product of ideals IJ.

Now here’s where it gets really fun. Start with a ring R and consider the collection of all its left ideals I. There are a bunch of things we can show about these operations on ideals, which I’ll leave as exercises. If it’s easier, use the descriptions in terms of elements, but I think it’s more satisfying to work with the diagrams and universal properties. Here I, J, and K are ideals, and \mathbf{0} is the ideal consisting of only the zero element.

  • (I+J)+K=I+(J+K)
  • I+J=J+I
  • I+\mathbf{0}=I
  • (IJ)K=I(JK)
  • I\mathbf{0}=\mathbf{0}=\mathbf{0}I
  • I(J+K)=IJ+IK
  • (I+J)K=IK+JK

What does all this mean? The collection of left ideals of R form a rig, like the natural numbers! Further if R has a unit, then we find RI=I=IR, so this rig has a unit. If R is commutative, then IJ=JI so the rig is too.

May 10, 2007 Posted by | Ring theory | 2 Comments