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4 added 59 characters in body

For an inverse system {$G_i$} of finite groups, and a fixed field $\mathbb{k}$, one can consider the corresponding group algebras $\mathbb{k}[G_i]$. The latter form an inverse system of $\mathbb{k}$-algebras {$\mathbb{k}[G_i]$} (unless I miss something obvious). Is it true that the inverse limit of {$\mathbb{k}[G_i]$} is the group algebra $\mathbb{k}[G]$, for $G=\lim\limits_\leftarrow${$G_i$} ?

In my case $\mathbb{k}=\mathbb{Z}/p\mathbb{Z}$, and $G_i$ are abelian $p$-groups, if this helps.

Added: I see that the answer is much less trivial than I expected. What about the simplest case, perhaps, when $G_i=\mathbb{Z}/p^i\mathbb{Z}$, for $i\geq 1$, and thus $G$ is the additive group of $\mathbb{Z}_p$? Are there fields for which the complete group algebra $[[\mathbb{k}G]]$ is easy to describe (particularly interesting for me would be the case $\mathbb{k}=\mathbb{Z}/p\mathbb{Z}$.) ? [This is answered by Simon Wadsley in the comment below.]

P.S. This is a spill-over of an innocently looking final year project on invertible circulant matrices of an undergraduate student of mine---I am unfamiliar with profinite things...

3 added 190 characters in body

For an inverse system {$G_i$} of finite groups, and a fixed field $\mathbb{k}$, one can consider the corresponding group algebras $\mathbb{k}[G_i]$. The latter form an inverse system of $\mathbb{k}$-algebras {$\mathbb{k}[G_i]$} (unless I miss something obvious). Is it true that the inverse limit of {$\mathbb{k}[G_i]$} is the group algebra $\mathbb{k}[G]$, for $G=\lim\limits_\leftarrow${$G_i$} ?

In my case $\mathbb{k}=\mathbb{Z}/p\mathbb{Z}$, and $G_i$ are abelian $p$-groups, if this helps.

Added: I see that the answer is much less trivial than I expected. What about the simplest case, perhaps, when $G_i=\mathbb{Z}/p^i\mathbb{Z}$, for $i\geq 1$, and thus $G$ is the additive group of $\mathbb{Z}_p$? Are there fields for which the complete group algebra $[[\mathbb{k}G]]$ is easy to describe (particularly interesting for me would be the case $\mathbb{k}=\mathbb{Z}/p\mathbb{Z}$.) ?

P.S. This is a spill-over of an innocently looking final year project on invertible circulant matrices of an undergraduate student of mine---I am unfamiliar with profinite things...

2 added a question regarding a particular case

For an inverse system {$G_i$} of finite groups, and a fixed field $\mathbb{k}$, one can consider the corresponding group algebras $\mathbb{k}[G_i]$. The latter form an inverse system of $\mathbb{k}$-algebras {$\mathbb{k}[G_i]$} (unless I miss something obvious). Is it true that the inverse limit of {$\mathbb{k}[G_i]$} is the group algebra $\mathbb{k}[G]$, for $G=\lim\limits_\leftarrow${$G_i$} ?

In my case $\mathbb{k}=\mathbb{Z}/p\mathbb{Z}$, and $G_i$ are abelian $p$-groups, if this helps.

Added: I see that the answer is much less trivial than I expected. What about the simplest case, perhaps, when $G_i=\mathbb{Z}/p^i\mathbb{Z}$, for $i\geq 1$, and thus $G$ is the additive group of $\mathbb{Z}_p$? Are there fields for which the complete group algebra $[[\mathbb{k}G]]$ is easy to describe (particularly interesting for me would be the case $\mathbb{k}=\mathbb{Z}/p\mathbb{Z}$.) ?

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