Stable Isomorphisms, Grothendieck Groups: Example

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:

  1. The map $f$ is surjective
  2. 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.

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