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.)


2 Answers 2


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.


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.

  • $\begingroup$ I just realized that the original version of my answer doesn't fully answer your question, as you also wanted to know whether $R=\mathbb{Z}[G]$ can be an injective object in the category of finite R-modules. I have edited the answer accordingly to eliminate this possibility. $\endgroup$ Commented Feb 10, 2013 at 2:56
  • $\begingroup$ This confuses me. My goal is to show that if there is a projective submodule $P$ of rank $n \geq 1$ of a finitely generated (torsionfree) $\mathbb{Z}[G]$-module $M$, i can find a direct summand which is a projective module of rank n (or a direct summand which is an invertible module - the previous statement would follow by induction since the projective modules in this case are the sums of invertible modules). However i believed that $P$ is already a direct summand of $M$ which would make projective modules in turn injective since all extensions of projective modules are split. $\endgroup$
    – GBW
    Commented Feb 10, 2013 at 11:22
  • $\begingroup$ Which part of what I said was confusing? As for your goal, it is impossible in general, even in the cyclic case you mention. Consider the case of the group $G=C_2$, the cyclic group of order 2; then $R=\mathbb{Z}[x] / (x^2-1)$. Let $M = (2, x-1)$ (a [maximal] ideal of $R$, considered as a submodule). Then $M$ contains the rank 1 projective module $(2)$, but $M$ is direct-sum indecomposable and is not projective. In particular, the only projective direct summand $M$ contains is the zero submodule, which has rank 0, not 1. $\endgroup$ Commented Feb 10, 2013 at 12:54
  • $\begingroup$ It was not so much your formulation that confused me, but rather the result that surprised me since this doesn't seem fit into my intuition regarding this topic. How can you tell that $M$ is direct-sum indecomposable? I would expect this to be a direct sum of a $\mathbb{Z}[G]$-module isomorph to $\mathbb{Z}$ and a projective module, but i didn't find concrete generators yet. $\endgroup$
    – GBW
    Commented Feb 10, 2013 at 14:14
  • $\begingroup$ Hm.. You may be right about this particular $M$. It seems to me that $M = (x-1) \oplus (x+1)$\ldots Sorry to be so hasty in my earlier comment. $\endgroup$ Commented Feb 10, 2013 at 16:29

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

  • $\begingroup$ This was very helpful already! I would like to rephrase my question though since i realized that the previous question was not specific enough. I would like to understand particularly if the following statement is true after all. "Every finitely generated projective $\mathbb{Z}[G]$-module is injective in the category of finitely generated $\mathbb{Z}[G]$-modules" (we can assume G cyclic of prime order for now). Since $\mathbb{Z}$ is not self-injective, $\mathbb{Z}[G]$ is not self-injective either due to the article, but working in the category of finitely-generated modules might fix this. $\endgroup$
    – GBW
    Commented Feb 9, 2013 at 11:23
  • $\begingroup$ Basically if i can show that $\mathbb{Z}[G]$ is injective in the category of finitely-generated $\mathbb{Z}[G]$-modules, i could conclude that all the finitely generated projective modules are as well since they are direct summands of an injective module (namely a finitely generated, free $\mathbb{Z}[G]$-module). $\endgroup$
    – GBW
    Commented Feb 9, 2013 at 11:30

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