Injective Modules over Group Rings Given the group ring $\mathbb{Z}[G]$ of a finite group $G$ over $\mathbb{Z}$, is there a way to generalize the notion of the "frobenius algebra" in some cases? One can show that every group ring $\mathbb{Q}[G]$ is a frobenius algebra and thus the projektive and injective modules correspond. This seems to be wrong in general over $\mathbb{Z}$ since even for the trivial group $G$ the module $\mathbb{Z}[G]$ is projective, but not injective as $\mathbb{Z}[G]$-module.
Is it still possible to obtain a relation between projective and injective $\mathbb{Z}[G]$-modules?
like stated in the comments below, i would like to understand in particular if and why $\mathbb{Z}[G]$ for a cyclic group $G$ of prime order is an injective object in the category of finitely generated $\mathbb{Z}[G]$-modules.
(edit: it turned out that this statement is wrong - for further details, see the discussion below.)
 A: I claim that if $R$ is a left Noetherian ring and $M$ a finitely generated left $R$-module, then $M$ is injective as an $R$-module iff it is injective in the category of finitely generated $R$-modules.
Suppose $M$ is injective in the category of f.g. $R$-modules.  Then for any left ideal $I$ of $R$ and any left $R$-module homomorphism $g: I \rightarrow M$, then since both $I$ and $R$ are finite $R$-modules and $M$ is injective over that category, $g$ extends to a map from $R$ to $M$.  But then by Baer's criterion, $M$ is injective in the category of $R$-modules.  The converse is trivial.
So when $R=\mathbb{Z}[G]$, since $R$ is not injective as an $R$-module, it is also not injective in the category of f.g. $R$-modules.
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I suspect the correct generalization should be "$R=\mathbb{Z}[G]$ has finite injective dimension as an $R$-module''.  In other words, $R$ is a Gorenstein ring (assuming $R$ is Noetherian, which in this case it will be).
I don't know about the nonabelian case.  However, if $G$ is a finite abelian group, then in Bass's seminal article (H. Bass, "On the ubiquity of Gorenstein rings" Math. Z. , 82 (1963) pp. 8–28), he shows that $R=\mathbb{Z}[G]$ is a Gorenstein ring of Krull dimension 1.  In particular, he shows that $R$ has injective dimension $1$ as an $R$-module.
You can't do much better than this (again, in the commutative case).  There's a classical result (cf. the book Cohen-Macaulay rings, by Bruns and Herzog, chapter 3) that says that if $R$ is a commutative Noetherian local ring and $M$ is a finitely generated $R$-module of finite injective dimension, then the injective dimension of $M$ equals the depth of $R$. (!)  However, another classical result (same source) says that if $R$ is Gorenstein, then it must be Cohen-Macaulay -- that is, the depth of $R$ equals the Krull dimension of $R$.  Put together, then, we have that if $R$ is Gorenstein, the injective dimension of $R$ over itself equals the Krull dimension of $R$.
Now, the rings you are looking at are not local; however, everything reduces to the local case.  So again, as Jason was saying, the fact that $\mathbb{Z}$ has positive Krull dimension is what gets in the way of $\mathbb{Z}[G]$ being injective. It does, however, have injective dimension 1.
A: The notion of Frobenius algebra is still useful in the general case, but then group rings that are Frobenius algebras aren't necessarily quasi-Frobenius rings, as your example notes. However if $M$ is a noetherian Frobenius $A$-algebra where $A$ is a commutative and self-injective ring then $M$ itself is a quasifrobenius ring so the notions of projective and injective coincide again. This is the content of Corollary 19 in [1].
If $A$ is not self-injective then your method of constructing a counterexample for group rings shows that the conclusion cannot hold in general for $A[G]$, since $A$ itself is not quasifrobenius.
[1].Eilenberg and Nakayama. "On the dimension of modules and algebras. II". Nagoya Mathematical Journal, 9. pp 1-16, 1955
