Let *K* be a field and *G* a finite group. We know that when char(*K*) does not divide |*G*|, the group algebra *K*[*G*] is semisimple. Conversely we have:

Proposition. If char(K) divides |G|, then K[G] is not semisimple.

**Proof**

Let , a two-sided ideal of *K*[*G*]. If *K*[*G*] = *I*⊕*J* for some left ideal *J* of *K*[*G*], then dim *J* = 1 so it is spanned by, say, . Since α*g* is a scalar multiple of α for each *g*∈*G*, all the *b _{g}*‘s must be equal to some

*b*∈

*K*. But this means lies in

*I*since its sum of coefficients is ♦

For the case where char(*K*) divides |*G*|, we have *modular representationt theory*. Here, we’ll require some knowledge of complete discrete valuation rings; throughout this article, we adopt the following notations and assumptions.

*R*is a complete discrete valuation ring with maximal ideal generated by π∈*R*.*K*is the field of fractions of*R*.*k*:=*R*/(π) is the residue field.*p*:= char(*k*) divides |*G*| and char(*K*) = 0.

Representations of *G* over *k* are studied by first lifting them to *R*, then projecting to *K*. Let be the Grothendieck group of finitely-generated modules over *k*[*G*], *K*[*G*] respectively. Since char(*K*) = 0, *K*[*G*] is semisimple. Since *k*[*G*] is of finite dimension over *k*, it is artinian; we let *J* be its Jacobian radical. Finally, let be the Grothendieck group of finitely-generated projective modules over *k*[*G*]. From earlier results:

- are free abelian groups, with the following bases,
*which we will fix from now on*.- Basis of or : [
*M*] for simple modules*M*. - Basis of : [
*P*] for finitely-generated indecomposable projective modules*P*.

- Basis of or : [
- We have a map taking [
*P*] to [*P*]. - We have an isomorphism , taking [
*P*] to [*P*/*JP*]; this corresponds to identity matrix (under the above bases).

*Note*: if , then *M* and *N* have identical composition factors but *M* and *N* may not be isomorphic. On the other hand, if , then the decomposition factors of *M* and *N* are the same and thus *M* ≅ *N*.

Next, we introduce the following pairings:

- The pairing taking is a bi-additive map. The above basis for is orthogonal by Schur’s lemma. Extending
*K*if necessary, we may assume it is*orthonormal*. - The pairing taking is a bi-additive map. The given bases for and are dual with respect to this pairing. [ Indeed, for with
*M*simple, we have Hom(*P*,*M*) = Hom(*P*/*JP*,*M*) since*JM*= 0; if*P*is projective indecomposable, then*P*/*JP*is simple and the result follows from Schur’s lemma. Assuming*k*is large enough, we may assume that if*P*/*JP*≅*M*. ]

The next step is to define the maps *d* and *e* in the following diagram, such that *c* = *de*:

## Definition of *d*

To define , let *M* be a finitely-generated *K*[*G*]-module.

Proposition. There is an R[G]-module N such that . The class of this module depends only on M.

**Proof**

**Step 1.** First prove the existence of *N*.

Let *N’* ⊂ *M* be a free *R*-module with a basis spanning *M*, so as *K*-vector spaces. Now let which is also a free *R*-module since it is torsion-free, and has a basis spanning *M*. Thus Since *N* is *G*-invariant it is an *R*[*G*]-module.

**Step 2**. Proof of uniqueness: initial step.

Suppose are *R*[*G*]-modules such that Replacing by a scalar multiple (this does not affect ), we may assume Now for each we have for some *r*; since has a finite basis, we have for some *r*>0.

**Step 3**. Now we show: if , then in

This is by induction on *r*. When *r*=1, let ; we have an exact sequence of *k*[*G*]-modules:

Since we have in the group For *r*>1, let we have:

By induction hypothesis, and we’re done. ♦

We thus define via: given pick an *R*[*G*]-module *N* such that We then define

**Note**

An exact sequence of *K*[*G*]-modules splits as *M* ≅ *M’* ⊕ *M”*; picking modules *N’* and *N”* for *M’* and *M”* via the above proposition, *N* := *N’* ⊕ *N”* also satisfies the proposition for *M*, and we obtain *d*([*M*]) = *d*([*M’*]) + *d*([*M”*]) as desired, so *d* is well-defined.

**Example**

Suppose *G* = {*e*, *g*} is of order 2. Let *R* = **Z**_{2}, *K* = **Q**_{2} and *k* = **F**_{2}. Take *M* = *K*[*G*] itself. One can obtain the *R*[*G*]-module *N* (of rank 2 over *R*) in different ways, e.g.:

Indeed, they’re isomorphic over *K*[*G*] since the above two matrices are diagonalizable with eigenvalues -1, +1. However, reducing modulo 2 gives non-isomorphic *k*[*G*]-modules:

Note that they have the same composition factors, although one is semisimple and the other is not.

## Definition of *e*

Next, we will define . This is given in two steps: first we consider the map , where is the Grothendieck group of finitely-generated projective *R*[*G*]-modules. [ Warning: since *R*[*G*] is not artinian, we have to avoid using results from the previous two articles. ] Note that Ψ is well-defined:

- If
*P*is projective over*R*[*G*], then for some*n*so and so*P*/π*P*is projective over*k*[*G*]. - An exact sequence of projective modules must split, so Ψ is a well-defined homomorphism of abelian groups.

The next step is to show that Ψ is an isomorphism. First, injectivity:

Lemma. If P, Q are finitely-generated and projective such that as k[G]-modules, then as R[G]-modules.

**Proof**

Let be an isomorphism of *k*[*G*]-modules. Since *P* is projective, the map lifts to a map of *R*[*G*]-modules such that the reduction of *f*_{1} mod π is *f*_{0}. To show that *f*_{1} is bijective, it suffices to show it is an isomorphism of *R*-modules. But *P*, *Q* are finite *R*-free since they are projective over *R*[*G*] and thus torsion-free. So we only need to show that det(*f*_{1}) is invertible in *R*, i.e. ≠0 modulo π. But det(*f*_{1}) mod π = det(*f*_{0}) ≠ 0. ♦

Hence, is injective. Finally, we have the following:

Lemma. For any finitely-generated projective k[G]-module N, there is a finitely-generated projective R[G]-module M such that

**Proof**

We may assume *N* is indecomposable, so *N* is a direct summand of *k*[*G*], i.e. The image of 1∈*k*[*G*] in *N* then gives an idempotent *x*∈*N* (i.e. *x*^{2}=*x*). We wish to lift *x* to an idempotent *y*∈*R*[*G*] such that *x* = *y* mod π.

**Claim**: if*S*is any ring and*I*⊂*S*is an ideal satisfying*I*^{2}=0, then any idempotent*y*∈*S*/*I*lifts to an idempotent*x*∈*S*.**Proof**: let*z*∈*S*be any lift of*y*. Then*x*:= 3*z*^{2}-2*z*^{3}satisfies, modulo*I*,*x*≡ 3*z*^{2}-2*z*^{3}≡ 3*z*– 2*z*=*z*so it is another lift of*y*. Now*x*^{2}–*x*= 0 since it is divisible by (*z*^{2}–*z*)^{2}∈*I*^{2 }= 0. ♦

Now consider the ring with ideal This satisfies *I*^{2 }= 0 and *S*/*I* ≅ *k*[*G*] so by the claim, *x* lifts to an idempotent *y’*∈*S*.

Repeat the process with the ring with ideal Again and *S’*/*I’* ≅ *S* so *y’* lifts to an idempotent *y”*∈*S’*. Repeating this process, since *R* is a complete discrete valuation ring, we obtain an idempotent *y*∈*R*[*G*] such that *x* = *y* mod π.

Then *R*[*G*] is the direct sum of left modules *M* := *R*[*G*]*y* and *M’* := *R*[*G*](1-*y*), and ♦

Thus, Ψ is an isomorphism and we can define:

where the second map takes to

**Relations Between d and e**

Finally we wish to prove that *c* = *de*. Indeed, let and so that Then the element *e*([*P*]) is given by To obtain we recover *Q* as the *R*[*G*]-module in the definition of *d* and take: as desired.

## Concrete Example

Let’s compute an explicit example for *G* = *S*_{3}, *R* = **Z**_{3}, *K* = **Q**_{3}, *k* = **F**_{3}. It’s easy to calculate since *K*[*G*] is semisimple. From elementary character theory, there are three simple modules:

*trivial rep.*:*g*acts as trivially on*K*;*alternating rep.*:*g*acts via sgn(*g*) on*K*;*other*: take the subspace of*K*^{3}given by (*x*,*y*,*z*) satisfying*x*+*y*+*z*= 0;*G*acts by permuting the coordinates.

This gives a basis of . On the other hand, let’s consider all simple *k*[*G*]-modules. We still have the trivial and alternating representations, clearly simple. But now the third representation has a submodule:

since char(*k*) = 3. This gives a subspace which is the trivial representation, and the quotient is alternating. Hence, has two simple modules, given by the trivial and alternating modules, and the matrix for is:

Next we compute *e*; to do that we need to find the indecomposable projective *k*[*G*]-modules. These can be found by decomposing *k*[*G*] itself. From above, we see that the key is to find idempotents of *k*[*G*] (specifically indecomposable projectives correspond to primitive idempotents, but we won’t use that here). Clearly is idempotent, and we obtain:

We leave it to the reader to prove that these are indecomposable. To compute *e*([*W*]) and *e*([*W’*]), we lift them to projective *R*[*G*]-modules then tensor with *K*: this still gives

which has character values (3, 1, 0) and (3, -1, 0), i.e. and So the matrix for *e* is: