If $a$ and $b$ are two real numbers and $ax = bx$, then we can't conclude that $a = b$ because $x$ may be zero. The same is true for tensor products of modules: if $A$ and $B$ are two left $R$-modules and $X$ is a right $R$-module, then an isomorphism $X\otimes_R A\cong A\otimes_R B$ does not necessarily mean that $A\cong B$. Of course, $X$ not even need be zero for this to happen.

Addition for real numbers is a little different. If $a$ and $b$ are two real numbers then $x + a = x + b$ is equivalent to $a = b$. What about for direct sums? If $A$ and $B$ are two $R$-modules, and $X$ is a third $R$ module, what if $X\oplus A\cong A\oplus B$? Is it true that $A\cong B$?

The answer is no. Perhaps this is surprising from the way direct sums work. After all, in a direct sum $X\oplus A$, it "feels like" what happens in $X$ is independent from what happens in $A$. And for vector spaces, this is true: for $k$-modules where $k$ is a field, if $X\oplus A\cong X\oplus B$, then certainly $A\cong B$, because $A$ and $B$ have the same dimension.

Of course, this blog is all about examples, so we'll see an example.

To keep things simple, we'll stick with the case where $X$ is a finitely-generated free module. Even here there are interesting things happening. So, consider a ring $R$ and two modules $A$ and $B$. We say that $A$ and $B$ are **stably isomorphic** if there exists a natural number $n$ such that $R^n\oplus A\cong R^n\oplus B$.

It's a strange definition, but it's actually very natural from a certain context, which we'll think about a little later. Of course, isomorphic modules are stably isomorphic, but the converse is not true. There are stably isomorphic modules that are not isomorphic.

## The Example

So what's the example? Consider the division ring $\mathbb{H}$ of quaternions. It is the $\R$-algebra, which as a vector space has a basis $\{1,i,j,ij\}$. As an algebra it satisfies the relations $i^2 = j^2 = (ij)^2$. So it's indeed four-dimensional as an $\R$-vector space.

Now consider the ring $R = \mathbb{H}[x,y]$ of two-variable polynomials over $\mathbb{H}$. Here, the variables commute with the quaternions and commute with each other.

Now we'll produce a left $R$-module that is stably isomorphic to $R$ but not isomorphic to it. How do we do this?

We'll start with a simple observation. Suppose $P$ is stably isomorphic to a finitely generated free module. Then $P\oplus R^k \cong R^n$ for some $n$ and $k$. This means that $P$ is the kernel of a surjective $R$-module homomorphism $R^n\to R^k$. Conversely, such a surjective homomorphism splits and so gives rise to a module stably isomorphic to a finitely generated free module.

To put it informally: modules that are stably isomorphic to free modules are pretty much the same thing as surjective homomorphisms of finitely-generated free modules.

Back to the two-variable polynomial ring over the quaternions, $R = \mathbb{H}[x,y]$. Let's produce a map $R^2\to R$. Luckily, I have one handy: define $f:R^2\to R$ by

$$(a,b)\longmapsto a(x + j) – b(y + i).$$

It's easy to prove the following facts, that I will leave to the reader:

- The map $f$ is surjective
- Via the first projection, the kernel is isomorphic to the left ideal $J = \{ a\in R : a(x + j)\in R(y+i)\}$.

We have an exact sequence $0\to J\to R^2\to R\to 0$. Since $R$ is free (hence projective), this sequence splits and we have an isomorphism $R^2\cong R\oplus J$. Therefore, $J$ is stably isomorphic to $R$.

## Proof that the example given is not free

So, we have our ideal $J = \{a\in R : a(x + j)\in R(y+i)\}$, and we know that $R^2\cong R\oplus J$. Therefore, $J$ is stably isomorphic to $R$. We claim that $J$ is **not** free. Now, $R$ is a finite-dimensional algebra over a division ring and hence free modules have a unique rank (the invariant basis property). Therefore, if $J$ is a free module, it actually has to be isomorphic to $R$. That is, $J = Rz$ for some $z\in J$.

By direct calculation, $(y + i)(x – j)\in J$. Therefore, the $y$-degree of $z$ is less than or equal to one. And all elements of $J$ have $y$-degree at least one, so therefore we can write $z = cy + d$ where $c,d\in \mathbb{H}[x]$.

Next, you should verify that $y^2 + 1\in J$. Therefore, $y^2 + 1 = r(cy + d)$ for some $r\in R$. This means that $r$ also has to have $y$-degree exactly one, so write $r = c'y + d'$. Comparing coefficients in the equation $y^2 + 1 – (c'y + d')(cy + d)$ gives $cc' = 1$ and $dd' = 1$. Therefore, we can also write $J = R(y + c'd)$.

Using $(y + i)(x – j)\in J$ again:

$$(y + i)(x-j) = s(y + c'd),$$

and comparing the $y$ coefficient of both sides gives $s = x – j$. This then gives us:

$$(x-j)c'd = i(x-j).$$

The $x$-coefficients give us $c'd = i$. But then the constant coefficients give us $-ji = -ij$ or $ij = ji$, which is false in $\mathbb{H}$. This is a contradiction that we obtained by assuming that $J$ was free.

Hence, $J$ is not free, but it is stably isomorphic to a free module.

Direct sums are strange indeed somtimes…

## Connection to the Grothendieck group

The concept of stable isomorphism is actually not just a random curiosity, though even if it were it'd still be cool. Actually, this concept is closely related to the Grothendieck group.

Recall that if you have a ring $R$, the Grothendieck group of $R$, denoted by $K_0(R)$, is an abelian group defined by generators and relations. Its set of generators is a set of isomorphisms classes of finitely-generated projective $R$-modules. For each such finitely-generated projective $P$, we write $[P]$ for the corresponding generator.

The relations are something you'd expect: for each pair $P,Q$ of finitely-generated projective $R$-modules, there is one relation $[P] + [Q] = [P\oplus Q]$.

The Grothendieck group is a fundamental invariant of a ring and is the first in a series of functors called the $K$-theory functors. For example, any commutative ring whose projectives are free (like fields, local rings, and the integers) the Grothendieck group is the abelian group $\Z$ of integers.

If $K_0(R)$ is the Grothendieck group of a ring, then two generators $[P]$ and $[Q]$ are equal in $K_0(R)$ if and only if….$P$ and $Q$ are stably isomorphic! That's the connection. So for our ring $\mathbb{H}[x,y]$ we showed that $[R] = [J]$ where $J$ is the ideal $\{ a\in R : a(x+j)\in R(y + i)\}$.

## One last remark

I can't help but saying that the ideal $J$ we constructed is also a neat example in another way: our proof showed that $J$ is a projective module that isn't free. Therefore, $J$ is an example of a projective module over a polynomial ring over a division ring that's not free.

The cool thing about this is that modules over finitely-many-variable polynomial rings over *fields* are free – that's the Quillen-Suslin theorem. Not easy at all, though the two-variable case was proved long before Quillen and Suslin solved it (independently, I might add). Thus, the Quillen-Suslin theorem is no longer true if we add just a slight dash of noncommutativity in the form of quaternions.