Let $A$ be a commutative ring, let $A[[t]]$ be the ring of formal power series and consider the ring of formal Laurent series $A((t)) = A[[t]][t^{-1}]$. I would like to know:

What is $Gl_n(A((t)))$, the group of invertible matrices with entries in the formal Laurent series over A?

Let me give some background to clarify what kind of answer I am looking for: One can verify that for any $n\geq 1$, the ring of matrices $M_n$ with entries in $A[[t]]$ is isomorphic to the formal power series with coefficients in $M_n(A)$, via \begin{align} M_n(A[[t]]) \cong & \; M_n(A)[[t]] \\ \left (\sum_k a^{(k)}_{i,j}t^k \right)_{i,j} \mapsto & \sum_k (a_{i,j}^{(k)})_{i,j} \,t^k.\end{align} Using the well known result that a power series over any ring is invertible if and only if the first coefficient is invertible, we see from this that the invertible matrices $Gl_n(A[[t]])$ are precisely those for which the first coefficient of the corresponding series is invertible: $$Gl_n(A[[t]]) = Gl_n(A) + t\cdot M_n(A)[[t]]$$ One could see this as a generalization of the mentioned one-dimensional case $A[[t]]^\times=A^\times + tA[[t]]$.

I am interested in finding a similar description for the situation where $A[[t]]$ is replaced by the ring of formal Laurent series $A((t)) = A[[t]][t^{-1}]$. In particular, I would like to get a feeling for:

How "large" is $Gl_n(A((t)))$ compared to $Gl_n(A[[t]])$?

Here are some thoughts: While I think we still get an isomorphism $$M_n(A((t))) \cong M_n(A)((t)),$$ it seems much harder to determine what the units of the right hand side are, because $M_n(A)$ has many zero divisors. As an example, consider the matrix $$\pmatrix{1&0\\ 0&t} \mapsto \pmatrix{1&0\\ 0&0} + \pmatrix{0&0\\ 0&1} t.$$ This element of $M_n(A)((t))$ clearly has an inverse, although all coefficients are zero-divisors.

As for the second question, in case that $A$ is an integral domain, for $n=1$ we know that $A((t))^\times/A[[t]]^\times=\mathbb{Z}$. In general, $Gl_n(A[[t]])$ is not a normal subgroup anymore, but I was hoping that one could still say something about coset representatives of $Gl_n(A((t)))/Gl_n(A[[t]])$.

I came to this question when I was trying to find out which finite locally free $A((t))$-modules come from locally free $A[[t]]$-modules.

  • $\begingroup$ Note that this is not a duplicate of math.stackexchange.com/questions/533071/…, because there the ring is commutative whereas $M_n(A)$ for $n> 1$ is not. $\endgroup$ – benh Apr 8 '16 at 17:10
  • $\begingroup$ See also: affine Grassmannian $\endgroup$ – Minseon Shin Sep 20 '18 at 21:20

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