Let $p$ be a prime number and $n,l$ be natural numbers. I'm interested in the conditions under which the general linear groups of degree $n$ over the following two length $l$ finite discrete valuation rings with residue field $\mathbb{F}_p$ are isomorphic:

$$GL(n,\mathbb{Z}/p^l\mathbb{Z}) \cong GL(n,\mathbb{F}_p[t]/(t^l))$$

I've worked out the following cases:

  • The case $n = 1$ and $l = 2$: In this case, both groups are isomorphic to $\mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/(p-1)\mathbb{Z}$.
  • The case $n = 1$ and $l > 2$: In this case, the only situation that the groups are isomorphic seems to be where $p = 2$ and ($l = 4$ or $l = 5$).
  • The case $n > 1$ and $l = 2$: The isomorphism seems to be equivalent to the requirement that the quotient map $GL(n,\mathbb{Z}/p^2\mathbb{Z}) \to GL(n,\mathbb{Z}/p\mathbb{Z})$ split. It seems that, for $n = 2$, this happens when $p = 2$ and when $p = 3$, but not when $p = 5$. This in turn seems to have something to do with modular/local representation theory. In particular, it seems that $GL(2,\mathbb{Z}/4\mathbb{Z}) \cong GL(2,\mathbb{F}_2[t]/(t^2))$ and $GL(2,\mathbb{Z}/9\mathbb{Z}) \cong GL(2,\mathbb{F}_3[t]/(t^2))$, but $GL(2,\mathbb{Z}/25\mathbb{Z}) \not \cong GL(2,\mathbb{F}_5[t]/(t^2))$. Is there a general way of figuring out what $p$ work and what don't?
  • The case $n > 1$ and $l > 2$: Don't know what happens here.

The question can be generalized somewhat to comparing the general linear group over a Galois ring with characteristic $p^l$ with residue field of size $q = p^r$ versus the ring $\mathbb{F}_q[t]/(t^l)$.


For $l=2$, the only cases where the extension splits are $n=2$ and $p=2,3$ and $n=3$ and $p=2$. In her PhD thesis, on page 3, Pooja Singla attributes this to [Chih Han Sah. Cohomology of split group extensions. II. J. Algebra, 45(1):17–68, 1977] and [Yuval Ginosar. On splitting of the canonical map: mod(p): GL(n,(p)) → GL(n,p). Comm. Algebra, 29(12):5879– 5881, 2001].

  • $\begingroup$ Thanks! I think that mostly settles the question (the l > 2 case is probably a finite amount of additional checking). $\endgroup$ – Vipul Naik Sep 12 '12 at 17:24

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