# The Unapologetic Mathematician

## 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: 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.