I noticed that starting with a $2 {\times} 2$ matrix $M$ with a handful of polynomial entries in two variables such that $\det M$ is invertible in $\mathbb C [x] [[y]]$ I can add infinitely many terms of higher orders in $y$ to the entries of $M$ to make $\det M$ a constant.

Further, I noticed that with a bit of tinkering it's actually possible to add finitely many terms to $M$ and still achieve that $\det M$ is constant.

I am now wondering if this is true in general:

Let $M$ be a $2 {\times} 2$ matrix with entries in the ring $\mathbb C [x][[y]]$ and constant determinant. Given $k > 0$, does there exist a matrix $N$ with entries in $\mathbb C [x, y]$ such that $M \,\mathrm{mod} \, (y^{k+1}) = N \,\mathrm{mod} \, (y^{k+1})$ and $\det M = \det N$?

(If it helps, I would be content with assuming that one of the entries is a multiple of $y$ — e.g. if the $(1,2)$ entry is a multiple of $y$, then the determinant being constant implies that the $(1,1)$ and $(2,2)$ entries have a constant term and are both of the form $1 + O (y)$.)

I give a simple example to illustrate:

Example. Let $$ M = \begin{pmatrix} 1 - y & 0 \\ 0 & 1 + y + y^2 + \dotsb \end{pmatrix} $$ Then $\det M = 1$ (as the $(1,1)$ entry is invertible in $\mathbb C [[y]]$ with inverse $(2,2)$).

Now let $$ N = \begin{pmatrix} 1-y & -y^{k+1} \\ y^{k+1} & 1 + y + \dotsb + y^{2k+1} \end{pmatrix} $$ Then $\det M = \det N = 1$ and $$ M \, \mathrm{mod} (y^{k+1}) = N \, \mathrm{mod} (y^{k+1}) = \begin{pmatrix} 1 - y & 0 \\ 0 & 1 + y + \dotsb + y^k \end{pmatrix}. \qquad \diamond $$

This example works equally well if you replace $y$ by a polynomial $p = p (x,y)$ which is a multiple of $y$.

I would also be interested to hear about (textbook/paper) examples computing with matrices of formal power series to see what kind of techniques can be used.

Edit. I have discovered a truly tedious proof of this, which this edit section is too narrow to contain.

It is a "proof by brute force" (solving linear equations for each monomial that may appear in the expression of $\det M \mod (y^{n+1})$), and I think writing up the details should best be left as an Exercise To The Reader.

There ought to be a more elegant proof and although I haven't disclosed the full proof, at least I now believe the result to be true, which might be a reason to look for such a proof.

  • $\begingroup$ Is $u=y$? If so, what have you tried? $\endgroup$ – Mohan Jan 11 '18 at 15:19
  • $\begingroup$ @Mohan Yes, $u$ was a typo for $y$. For a range of "simple" matrices (2-3 summands per entry up to $y^3$ or $y^4$) I was able to do it by hand. I tried to translate the general case into a different way of thinking about it — e.g. do there exist polynomials $a,b,c,d$ in $x,y$ such that for given fixed polynomials $A,B,C,D,P$ we have that $P = ad-bc + Ad - Bc + aD - bC$ — but the problem doesn't appear any simpler in this formulation... $\endgroup$ – Earthliŋ Jan 11 '18 at 17:43
  • $\begingroup$ "it's actually possible to add finitely many terms to $M$ and still achieve that $\det M$ is constant". Let $M=\begin{pmatrix} 1-y & 0 \\ 0 & 1 \end{pmatrix} $. Then $\det M=1-y=\left(\sum_{n=0}^\infty y^n\right)^{-1}$ is invertible in $\mathbb C [x] [\mkern-3mu[y]\mkern-3mu]$. Let $P=\|p_{ij}\|$ be an arbitrary $2\times 2$ matrix over $\mathbb C[x,y]$. Then $\det (M+y^2P)=1-y+y^2p(x,y)$, where $p\in\Bbb C[x,y]$, so it is not invertible in $C[x,y]$. $\endgroup$ – Alex Ravsky Jan 20 '18 at 1:58
  • $\begingroup$ Of course, $M+\begin{pmatrix} y & 0 \\ 0 & 0 \end{pmatrix}=I$, but this carry no deep idea, because if $M=M_0+yM_1$, where $M_0\in\Bbb C[x]$ and $M_1\in \mathbb C [x] [\mkern-3mu[y]\mkern-3mu]$ are arbitrary matrices such that $\det M$ is invertible in $\mathbb C [x] [\mkern-3mu[y]\mkern-3mu]$ then $\det M_0$ is a non-zero constant. $\endgroup$ – Alex Ravsky Jan 20 '18 at 1:58

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