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For a commutative ring $R$, let $\mathrm{mod-}R$ denote the category of finitely generated $R$-modules. Let $\mathcal C$ be an abelian, full, isomorphism closed (i.e. $M\cong N$ in $\mathrm{mod-}R$ and $M\in Ob(\mathcal C)\implies N\in Ob(\mathcal C)$) subcategory of $\mathrm{mod-}R$.

The Grothendieck group of $\mathcal C$, usually denoted by $G(\mathcal C)$, is defined as $$\dfrac{\bigoplus_{X\in \operatorname{Ob}(\mathcal C)}\mathbb Z[X]}{\langle [A]-[B]+[C] \mid 0\to A \to B \to C\to 0 \text{ is exact} \rangle }$$ where $[X]$ denotes the isomorphism class of $X$.

Now if for every $M,N\in Ob(\mathcal C)$, we also have $M\otimes_R N\in Ob(\mathcal C)$, then the $\mathbb Z$-module $$\bigoplus_{X\in \operatorname{Ob}(\mathcal C)}\mathbb Z[X]$$ can be given a $\mathbb Z$-algebra structure where the ring multiplication is induced by the tensor product. If every module in $\mathcal C$ is flat, then the $\mathbb Z$-submodule $${\langle [A]-[B]+[C] \mid 0\to A \to B \to C\to 0 \text{ is exact} \rangle }$$ of $\bigoplus_{X\in \operatorname{Ob}(\mathcal C)}\mathbb Z[X]$ is also an ideal, so the Grothendieck group actually becomes a commutative ring where the product structure is induced from tensor product.

My question is: What happens in case not every module of $\mathcal C$ necessarily flat? Does the Grothendieck group necessarily become a commutative ring where the product structure is induced from tensor product? I'm particularly interested in the case of $G(\mathrm{mod-}R)$: is it necessarily a commutative ring?

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3 Answers 3

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Suppose $R$ is a ring such that for every $R$-module $M$ there is a projective resolution $P_\bullet \to M$ of finite length (actually, this condition can be loosened quite a bit as I will note at the end). Then we can define a ring structure on $G(R)$ the Grothendieck group of finitely generated $R$-modules by defining the following product $$ [A] \cdot [B] = \sum_i (-1)^i \text{Tor}_i(A,B). $$ Note that if $P$ is a projective $R$-module, then $[A]\cdot[P] = [A \otimes P]$ since the $\text{Tor}_i(A,P)$ vanish for $i >0$.

The way to show this is a little painful. We need to show that

  1. This product is independent of choice of resolution,
  2. this product is well-defined,
  3. this product is symmetric,
  4. this product satisfies the ring axioms.

It is in my opinion easier to define this product as a map on $G(R)$ first, though tastes may vary. Fix a finitely generated $R$-module $M$ and let $$ - \cdot M : G(R) \to G(R),\; [A] \mapsto [A] \cdot M := \sum_i (-1)^i [\text{Tor}_i(A,M)]. $$

Then one can show that this map is independent of the choice of resolution by showing $\text{Tor}_i(-,M)$ is. The well-definedness of the map follows since for any short exact sequende $0 \to A \to B \to C \to 0$ we have a long exact sequence of Tor, which shows $[A] \cdot M - [B] \cdot M + [C] \cdot M = 0$. The symmetry follows from the symmetry of Tor, so $[A] \cdot B = [B] \cdot A$.

From this point it is therefore acceptable to write $[A] \cdot [B]$, and we can prove the ring axioms one by one. For this it is good to note that for an $R$-module $M$ with projective resolution $P_\bullet \to M$ we have $$ [M] \cdot [N] = \sum_i (-1)^i [\text{Tor}_i(M,N)] = \sum_i(-1)^i H_i(P_\bullet \otimes N) = \sum_i (-1)^i [P_i \otimes N]. $$

This computation involves the nontrivial lemma that in the Grothendieck group we have $\sum_i(-1)^i [C_i] = \sum_i (-1)^i [H_i(C_\bullet)]$ for any finite complex $C_\bullet$ in the category.

Then the ring axioms can easily be checked by using properties of the tensor product.

The requirement that every object in the category have a finite projective resolution is actually too strong. If every object has a finite resolution that is $F$-acyclic for a functor $F$ on the category, then it induces a derived morphism on Grothendieck groups. Actually, the resolution need not even be in the category itself. For example for categories of coherent sheaves we almost never have projective resolutions, but in some situations (details omitted on purpose) we may have locally free resolutions, which are $\text{Tor}_i(-, *)$ acyclic, so that we can define the product.

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    $\begingroup$ well thanks , but I do not understand what is the relation of your derived tensor product with my question ... does this derived tensor product naturally turns the Grothendieck group into a commutative ring ? $\endgroup$
    – user102248
    Jun 30, 2019 at 16:08
  • $\begingroup$ @user102248 Yes. $\endgroup$
    – Ruben
    Jun 30, 2019 at 16:09
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    $\begingroup$ thanks, I will think why ... btw, more generally, for regular rings of finite Krull dimension, every module has finite projective dimension $\endgroup$
    – user102248
    Jun 30, 2019 at 16:22
  • $\begingroup$ I will take time reading your answer carefully ... but for now, could you kindly give a reference for the fact you mention that in the Grothendieck group, $\sum_i(-1)^i [C_i] = \sum_i (-1)^i [H_i(C_\bullet)]$, for any complex $C_\bullet$ ? also , if possible, could you give some reference as to where I can find the multiplication structure you mention (may be explained in slightly more detail ) ? Thanks $\endgroup$
    – user102248
    Jul 1, 2019 at 15:17
  • $\begingroup$ @user102248 That one is not so hard to see. Suppose $C_\bullet$ is a complex with maps $d_i : C_i \to C_{i - 1}$. Then for every $i$ we have an exact sequence $0 \to \text{ker} d_i \to C_i \to \text{im} d_i \to 0$ which shows that $[C_i] = [\text{ker} d_i] + [\text{im} d_i]$ in the Grothendieck group. But also $[H^i(C_\bullet)] = [\text{ker} d_i] - [\text{im} d_{i + 1}]$ by definition. Play around with the indices of the two sums a bit and you will see that it works. $\endgroup$
    – Ruben
    Jul 1, 2019 at 18:42
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No. For instance, let $k$ be a field and let $R=k[x]/(x^2)$. Then every finitely generated $R$-module is a direct sum of copies of $R$ and $R/(x)$, and it is easy to see that $G(\mathrm{mod}R)$ is freely generated as an abelian group by $[R/(x)]$, with $[R]=2[R/(x)]$ (you can see this since there is a homomorphism $G(\mathrm{mod}R)\to\mathbb{Z}$ sending a module to its dimension as a $k$-vector space). But this group does not form a ring under the tensor product, since $R\otimes R\cong R$ and $R/(x)\otimes R/(x)\cong R/(x)$ so we would need to have both $[R]^2=[R]=2[R/(x)]$ and $[R]^2=(2[R/(x)])^2=4[R/(x)]$.

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  • $\begingroup$ thanks ... this might be a stupid question, but is there a chance that $G($mod$R)$ would become a ring with tensor multiplication structure if we took some other coefficient ring than $\mathbb Z$ ? $\endgroup$
    – user102248
    Jun 30, 2019 at 16:01
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    $\begingroup$ In a sense you're not done, as far as I can see you still need to see why $[R] \neq 0$ $\endgroup$ Jun 30, 2019 at 16:01
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    $\begingroup$ @Max: That's implied by the fact I mentioned that the group is freely generated by $[R/(x)]$. $\endgroup$ Jun 30, 2019 at 16:05
  • $\begingroup$ Oh my bad I hadn't noticed the word freely ! $\endgroup$ Jun 30, 2019 at 16:12
  • $\begingroup$ @user102248: Not unless the coefficient ring is trivial. You will always get a module freely generated by $[R/(x)]$, and then the relations $[R/(x)]^2=[R/(x)]$, $[R][R/(x)]=[R/(x)]$, and $[R]=2[R/(x)]$ are easily seen to imply that everything is $0$. $\endgroup$ Jun 30, 2019 at 16:34
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Take $R=\mathbb {Z/4}$, and assume $\otimes$ induces a distributive multiplication.

Then we have $\mathbb{Z/2} + \mathbb{Z/2}= \mathbb Z/4$ from the usual extension.

However, if you multiply by $\mathbb Z/2$, you get on the LHS $\mathbb Z/2 + \mathbb Z/2$ and on the RHS $\mathbb Z/2$, which would imply that $\mathbb Z/2 = 0$.

However, if you look at the map defined on finitely generated $\mathbb Z/4$-modules (which are therefore finite) by $f:X\mapsto \ln |X|$, then it extends linearly to $\bigoplus_{X} \mathbb Z \cdot[X]\to (\mathbb R, +)$ and whenever $0\to A\to B\to C\to 0$ is exact, we have $|B| = |C||A|$ so $f(A)-f(B)+f(C)=0$, therefore $f$ actually induces a map $G(\mathrm{mod}-R)\to (\mathbb{R},+)$ that sends $\mathbb Z/2$ to $\log 2 \neq 0$

Therefore $\otimes$ does not induce a distributive multiplication on $G(\mathrm{mod}-R)$ (actually if you look at it closely you don't even need to assume distributivity, because it would automatically follow)

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