Grothendieck Group of the integers under multiplication

Question

I want to construct the group $\mathbb{Q}\setminus 0$, the nonzero rational numbers under multiplication, from the monoid $\mathbb{Z}\setminus 0$ using the Grothendieck group construction.

First I form the free abelian group $F\langle\mathbb{Z}\setminus 0\rangle$. I'll use [x] to denote the symbols in this group. The Grothendieck group construction is $$F\langle\mathbb{Z}\setminus 0\rangle/B$$ $$\text{B generated by }[xy]-[x]-[y]$$

I did a lot of playing with the elements of this factor group, and I got that my eventual isomorphism with $\mathbb{Q}\setminus 0$ does these things:

(a) For $m\in \mathbb Z\setminus 0, [m]B\rightarrow m$ and $-1[m]B\rightarrow \frac{1}{m}$

(b) For $m\in \mathbb Z\setminus 0,k\in\mathbb{N}, k[m]B=[m]B$ and $ -k[m]B=-1[m]B$

(c) For $m,n\in \mathbb Z\setminus 0, ([m]+ -1[n])B \rightarrow \frac{m}{n}$

Now, I am fairly certain this is enough for me to show that the mapping with this property is an isomorphism, IF all the cosets in the factor group look like this. That's where I am having trouble.

First, I want to show that for any $m\neq n$, $[m]B\neq[n]B$, So I need to show there is no $b\in B$ such that $b=[m]-[n]$. I can definitely show that there are no $x,y$ such that $[m]-[n]=[xy]-[x]-[y]$, but $B$ contains all sorts of combinations of these generating elements, and I am unsure how to prove that there is no magic set of choices of these generating elements such that their sum comes out to $[m] - [n]$

Answer 1

I would suggest a different approach here. If one already guessed the answer for a universal construction like a Grothendieck group, it's often easier to check the universal property and not to construct an isomorphism to the explicit construction.

So, what one has to check here is that there is a map $i\colon\mathbb Z\setminus\{0\}\to \mathbb Q^\times$ such that for every multiplicative homomorphism $\varphi\colon\mathbb Z\setminus\{0\}\to A$ there is a unique homomorphims $\overline\varphi\colon \mathbb Q^\times\to A$ with $\overline\varphi\circ i = \varphi$.

Well, $i$ is pretty obviously just the inclusion map, and constructing $\overline\varphi$ given $\varphi$ is not difficult either: $\overline\varphi(m/n) := \varphi(m)\cdot\varphi(n)^{-1}$ has the desired properties:

• it is obviously a homomorphism;
• it clearly extends $\varphi$.

Moreover, it is indeed unique with these properties (if $\varphi$ is given), because in this case one has to have $\overline\varphi(m) = \varphi(m)$ for $m\in \mathbb Z$, and $\overline\varphi(1/m) = \varphi(m)^{-1}$ because a homomorphism has to preserve inverses.

Answer 2

The key piece of missing insight was that for any free abelian group generated by a set $S$, any mapping $\phi$ from $S$ into an abelian group, say $G$ has a corresponding map that goes from the free abelian group into $G$ (map $k[x]$ to $\phi(x)^k$). This is a homomorphism of groups$^1$. If our $\phi$ goes into an abelian monoid, we still define a map from the free abelian group if we restrict the domain of this new map to $k[x]$ with $k$ nonnegative. Then the map is a homomorphism of monoids. We'll call it $\bar{\phi}$

All this to say, that in this case, the map $k[x] \rightarrow k^x, k\geq0$ is a homomorphism of monoids. To see this, use the above with $\phi:\mathbb{Z}\setminus 0 \rightarrow \mathbb{Z}\setminus 0$ the trivial map.

Now for the question at hand. Let $m,n\in\mathbb{Z}\setminus 0$. Suppose $[m]B=[n]B$, then there is $b\in B$ so that $[m] = [n] + b$. All $b$ have the form $\sum([x_i y_i] - [x_i] - [y_i]) - \sum([z_j w_j] - [z_j] - [w_j]) $. Because we are operating in a group ($F\langle\mathbb{Z}\setminus 0\rangle$) we can rearrange the above to get this$^2$: $$ [m] + \sum([z_j w_j]) + \sum([x_i]+[y_i]) = [n] + \sum([x_i y_i]) + \sum([z_j] +[w_j]) $$

Now we apply our homomorphism of monoids. Note that these ugly sums map to the same number, since our map has the property $\bar{\phi}([xy]) =\bar{\phi}([x]+[y])$. So we have $mx = nx$ for some $x \in\mathbb{Z}\setminus 0$. So we conclude that $m=n$ and each $[m]B$ is a distinct element. Similarly we conclude the $-[m]B$ are not only distinct, but they are also distinct from the $[m]B$ cosets as well, as long as $m\neq1$.

An easy observation shows that $[mn]B = [m]B + [n]B$. So any element of our quotient group can be expressed as $\sum[p_i]B - \sum[\pi_j]B$, with $p_i,\pi_j$ prime. Recombining and cancelling any terms which show up in both sums leaves us with all elements of the quotient group expressable as $[m]B - [n]B$, $m$ and $n$ coprime.

Finally, let $I: F\langle\mathbb{Z}\setminus 0\rangle / B \rightarrow \mathbb{Q}\setminus 0$ be the map such that $I([m]B - [n]B) = \frac{m}{n}$. Because we already went to the trouble to ensure that the quotient group $m$ and $n$ are coprime, it's easy to see that this map is bijective and distributes over the group operation (cancelling common factors in $\mathbb{Q}\setminus 0$ is the same as cancelling cosets that appear in both the positive and negative sums in the quotient group). So $I$ is the desired isomorphism.

1: Serge Lange, Algebra 3rd edition, pg. 38

2: I saw this particular way of rearranging the equation in Charles Weibel's The K-book 2nd chapter, pg. 2