## Projective Modules and the Grothendieck Group

This is a continuation of the previous article. Throughout this article, R is an artinian ring (and hence noetherian) and all modules are finitely-generated. Let K(R) be the Grothendieck group of all finitely-generated R-modules; K(R) is the free abelian group generated by [M] for simple modules M.

Now let P(R) be the Grothendieck group of all finitely-generated projective R-modules. Thus, P(R) is the free abelian group generated by [P] for finitely-generated P, modulo [P] = [Q] + [Q’] for each short exact 0 → Q’ → P → Q → 0. By the lemma here, this means Q is a direct summand of P so P ≅ QQ’.

Theorem. The group P(R) is the free abelian group generated by [P] for indecomposable finitely-generated projective modules P. Furthermore, if $[P] = [Q_1] + \ldots + [Q_r]$ in P(R) for projective modules $P, Q_1, \ldots, Q_r,$ then $P \cong Q_1 \oplus \ldots \oplus Q_r.$

Proof

By Krull-Schmidt’s theorem, every finitely-generated module P is uniquely written as a direct sum of indecomposable modules; if P is projective, so is each direct summand. Hence [P] is a sum of [Q] for indecomposable projective Q. Furthermore, each short exact 0 → Q’ → P → Q → 0 gives a decomposition P ≅ QQ’ so [P] = [Q] + [Q’] if and only if the modules on the LHS and RHS match after decomposition. The general case of r>2 follows by induction on r. ♦

Now we define a map:

$c_R : P(R) \to K(R)$

which takes a projective module P to its class [P] in K(R). Note that this is a well-defined group homomorphism. The next map we define is:

$f_R : P(R) \to K(R)$

which is given by $f_R([P]) := [P/JP]$, where J := J(R) is the Jacobian radical of R. Note that this is well-defined since a short exact sequence 0 → Q’ → P → Q → 0 of projective modules splits to give P ≅ QQ’ and so P/JP ≅ (Q/JQ) ⊕ (Q’/JQ’), and we get [P/JP] = [Q/JQ] + [Q’/JQ’] in K(R). Furthermore, we saw earlier that $P \mapsto P/JP$ gives a bijection between projective indecomposable modules and simple modules. Since P(R) is freely generated by the projective indecomposable modules while K(R) is freely generated by the simple modules, we have:

Theorem. The map $f_R : P(R) \to K(R), [P] \mapsto [P/JP]$ is a group isomorphism.

The map $c_R\circ f_R^{-1} : K(R) \to K(R)$ can be represented by a matrix with integer entries; let’s compute this for a simple example.

## Example

Let R be the ring of upper 3 × 3 matrices with real entries. We thus see that:

$R = \left\{ \begin{pmatrix} * & * & * \\ 0 & * & * \\ 0 & 0 & *\end{pmatrix} \right\} =\overbrace{\left \{ \begin{pmatrix} * & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{pmatrix}\right \}}^I \oplus \overbrace{\left \{ \begin{pmatrix} 0 & * & 0 \\ 0 & * & 0 \\ 0 & 0 & 0 \end{pmatrix} \right\}}^J \oplus \overbrace{ \left\{ \begin{pmatrix} 0 & 0 & * \\ 0& 0 & * \\ 0 & 0 & *\end{pmatrix} \right\}}^K$

is a direct sum of indecomposable projective modules. On the other hand, K is isomorphic to the column of 3-vectors, which has a composition series: $0 \subset \mathbf{R} e_1 \subset \mathbf{R} e_1 \oplus \mathbf{R} e_2 \subset \mathbf{R}^3.$ Denoting the consecutive factors by ABC, we see that in K(R), we have [I] = [A],  [J] = [A]+[B],  [K] = [A]+[B]+[C], so the matrix is:

$\begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 1\\ 0 & 0& 1\end{pmatrix}.$

Exercise

Calculate the corresponding matrix for the ring:

$R=\begin{pmatrix} \mathbf{Q} & \mathbf{Q}(\sqrt 2) & \mathbf{Q}(\sqrt[4]2)\\ 0 & \mathbf{Q}(\sqrt 2) & \mathbf{Q}(\sqrt[4]2) \\ 0 & 0 & \mathbf{Q}(\sqrt[4]2)\end{pmatrix}$

## Pairing

Next, consider the pairing given by:

$\left<-, -\right> : P(R) \times K(R) \to K(\mathbf{Z}), \qquad \left<[P], [M]\right> :=[\text{Hom}_R(P, M)].$

We claim that this is well-defined; indeed for the first term, an exact sequence of projective modules splits as P ≅ QQ’ so Hom(PM) is the direct sum of Hom(QM) and Hom(Q’M), and the RHS gives:

$[\text{Hom}_R(P,M)] =[\text{Hom}_R(Q, M)]+ [\text{Hom}_R(Q', M)].$

On the other hand, if 0 → M’ → M → M” → 0 is an exact sequence of modules, then since P is projective, so is the resulting:

$0 \to \text{Hom}_R(P, M') \to\text{Hom}_R(P, M)\to \text{Hom}_R(P, M'')\to 0$

and we get $[\text{Hom}_R(P, M)] = [\text{Hom}_R(P, M')] + [\text{Hom}_R(P, M'')]$ as well.

This entry was posted in Notes and tagged , , , . Bookmark the permalink.