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I'm trying to verify the universal property of the Grothendieck group. Let $\overline{C}$ be the set of isomorphism classes of finitely generated $R$-modules over say a Noetherian ring $R$ and let $C$ denote the set of all f.g. $R$-modules. Let $A$ be the free abelian group over $\overline{C}$. Let $S$ be the set $S = \{ [M] - [M^\prime] + [M^{\prime \prime}] \mid 0 \to M \to M^\prime \to M^{\prime \prime} \to 0 \text{ exact}, [M], [M^\prime], [M^{\prime \prime}] \in \overline{C} \}$. Let $K = A / \langle S \rangle$.

Added: The universal property $(K, f)$ has to satisfy is the following (according to my current understanding): For any abelian group $G$ and additive function $\lambda : C \to G$ there exists a unique group homomorphism $h: K \to G$ such that $h \circ f = \lambda$.

In a diagram: $$\begin{matrix} C &\xrightarrow{f} & A / \langle S \rangle = K \\ \left\updownarrow{=}\vphantom{\int}\right. & & \left\uparrow{i^\prime_m}\vphantom{\int}\right.\\ C &\xrightarrow{\lambda} & G \end{matrix}$$

Sorry, I don't know how to draw diagonal arrows in latex, if anyone knows how to fix this please go ahead.

Now I want to show that for any abelian group $G$ and any additive function $\lambda : C \to G$ there exists a unique group homomorphism $h$ such that $h \circ f = \lambda$ where $f : C \to A / \langle S \rangle$.

So I define a function $h: K \to G$ as $[M] + \langle S \rangle \mapsto \lambda (M)$. What I'm stuck with is: how do I show that $h$ is a group homomorphism that is, $h(a + b) = h(a) + h(b)$? It does not follow from the definition since $\lambda$ is any function. But I need it to show that $h$ is well-defined.


What I have so far:

If $[M] + \langle S \rangle, [N] + \langle S \rangle \in K / \langle S \rangle$ then an element in $A$ mapping to $[M] + \langle S \rangle$ looks like $[M] + [P] - [P^\prime] + [P^{\prime \prime}]$ where the $P$s form an exact sequence. Similarly for $N$. Then $$ h([M] + \langle S \rangle + [N] + \langle S \rangle) = h \circ f (M + P - P^\prime + P^{\prime \prime } + N + S - S^\prime + S^{\prime \prime}) = \lambda (M + P - P^\prime + P^{\prime \prime } + N + S - S^\prime + S^{\prime \prime}) = \lambda (M) + \lambda (N)$$ How do we achieve that? Now we want to do $$ \dots = \lambda (M + P - P^\prime + P^{\prime \prime }) + \lambda( N + S - S^\prime + S^{\prime \prime}) = \lambda (M) + \lambda(P) - \dots + \lambda(S^{\prime \prime}) $$

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@Andrew $\lambda$ is not a homomorphism it is any (additive) function. See here. – Rudy the Reindeer Aug 6 '12 at 6:37
This does not seem to be the correct universal property. Surely, if $G = C$ and $\lambda$ is the identity then it does not factor through $K$? – Tobias Kildetoft Aug 6 '12 at 6:44
@Tobias You are right. Maybe I have to replace $C$ with $E$ in the diagram. But that still wouldn't make $h$ into a homomorphism. – Rudy the Reindeer Aug 6 '12 at 7:02
In $K$ two isomorphism classes $[A], [B]$ are equal if there is an isomorphism class $[C]$ such that $0 \to A \to C \to B \to 0 $ is exact. – Rudy the Reindeer Aug 6 '12 at 7:12
@Tobias: If $G=C$ and $\lambda :E\to C$ is the "identity", then $\lambda$ is not an additive function, since $C$ is free on generators $[M]$ (we have no relations yet). – Andrew Aug 6 '12 at 7:12
up vote 1 down vote accepted

We should define $h(\sum_ja_j[M_j]+\langle S\rangle):= \sum_ja_j\lambda([M_j]),$ for any element $\sum_ja_j[M_j]\in A,a_j\in\mathbb Z.$ To see that this is well-defined, suppose $\sum_ja_j[M_j]+\langle S\rangle=\sum_kb_k[N_k]+\langle S\rangle,$ so that $\sum_ja_j[M_j]- \sum_kb_k[N_k]\in\langle S\rangle.$ This implies


is an identity in $A$ for some finite sum over generators in $S.$ Hence,

$$h(\sum_ja_j[M_j]-\sum_kb_k[N_k]+\langle S\rangle)=h(\sum_{i}c_i([M_i']-[M_i]+[M_i''])+\langle S\rangle)$$

which implies by definition


Since the RHS is generated by short exact sequences and $\lambda$ is additive, we find that


or in other words

$$h(\sum_ja_j[M_j]+\langle S\rangle)=h(\sum_kb_k[N_k]+\langle S\rangle).$$

Thus $h:A/\langle S\rangle\to G$ is indeed well-defined. To see that $h$ is a homomorphism is now trivial from the definition of $h.$

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Sorry, how did you get that $i = 1$? – Rudy the Reindeer Aug 6 '12 at 9:33
Dear Andrew, your displayed equality is not true because some elements in the sum over i will be preceded by a factor of $(-1)$. So the sum will have more than one term in general, which addresses Clark's comment. – Georges Elencwajg Aug 6 '12 at 9:50
Dear @GeorgesElencwajg, thank you for the correction. I have tried to correct & improve my answer. – Andrew Aug 6 '12 at 15:33
Dear Andrew, thank you very much for this excellent answer! Of course, my definition of $h$ was flawed. Now that I know the mistake it seems obvious. – Rudy the Reindeer Aug 8 '12 at 4:51
Dear @ClarkKent, you're welcome. Indeed, I also initially made the same mistake! Glad to help, and to work through it. – Andrew Aug 8 '12 at 5:29

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