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.
Notation: For a ring $A$ and ideal $\mathfrak{a}$, let $\operatorname{SL}_{r}(A,\mathfrak{a})$ denote the kernel of the group homomorphism $\operatorname{SL}_{r}(A) \to \operatorname{SL}_{r}(A/\mathfrak{a})$.
Lemma: Let $r \ge 2$, let $A$ be a ring, let $\mathfrak{a}$ be an ideal of $A$. For any $\ell \ge 1$, the map \begin{align} \varphi_{\ell} : \operatorname{SL}_{r}(A,\mathfrak{a}) \to \operatorname{SL}_{r}(A/\mathfrak{a}^{\ell},\mathfrak{a}/\mathfrak{a}^{\ell}) \end{align} is surjective.
Proof: We proceed by induction on $\ell$; for $\ell = 1$, we have $\operatorname{SL}_{r}(A/\mathfrak{a},\mathfrak{a}) = \{\mathsf{id}_{r}\}$. Suppose $\ell \ge 2$ and let $\mathsf{W}_{\ell} \in \operatorname{SL}_{r}(A/\mathfrak{a}^{\ell},\mathfrak{a}/\mathfrak{a}^{\ell})$; let $\mathsf{W}_{\ell-1} \in \operatorname{SL}_{r}(A/\mathfrak{a}^{\ell-1},\mathfrak{a}/\mathfrak{a}^{\ell-1})$ be the image of $\mathsf{W}_{\ell}$; by induction on $\ell$, there exists $\mathbf{M}_{\ell-1} \in \operatorname{SL}_{r}(A,\mathfrak{a})$ such that $\varphi_{\ell-1}(\mathbf{M}_{\ell-1}) = \mathsf{W}_{\ell-1}$; after replacing $\mathsf{W}_{\ell}$ by $(\varphi_{\ell}(\mathbf{M}_{\ell-1}))^{-1} \cdot \mathsf{W}_{\ell}$, we may assume that $\mathsf{W}_{\ell} \in \operatorname{SL}_{r}(A/\mathfrak{a}^{\ell} , \mathfrak{a}^{\ell-1}/\mathfrak{a}^{\ell})$. Write $\mathsf{W}_{\ell} = \mathsf{id}_{r} + \mathsf{F}$ with $\mathsf{F} = (\mathsf{F}_{i,j}) \in \operatorname{Mat}_{r \times r}(\mathfrak{a}^{\ell-1}/\mathfrak{a}^{\ell})$. Suppose $\mathsf{F}_{1,1} \ne 0$. Choose a lift $\mathbf{f}_{1,1} \in \mathfrak{a}^{\ell-1}$ of $\mathsf{F}_{1,1}$; we may replace $\mathsf{W}_{\ell}$ by $\varphi_{\ell}(\mathbf{E}_{1}) \cdot \mathsf{W}_{\ell}$ where $\mathbf{E}_{1} \in \operatorname{SL}_{r}(A)$ is the matrix which differs from the identity $\mathrm{id}_{r}$ by \begin{align} \begin{bmatrix} 1 - \mathbf{f}_{11} & \mathbf{f}_{11} \\ -\mathbf{f}_{11} & 1+\mathbf{f}_{11} \end{bmatrix} \end{align} in the upperleft $2 \times 2$ submatrix. This ensures that $\mathsf{F}_{1,1} = 0$. If $\mathsf{F}_{1,1} = 0$, then we may add multiples of the first row (i.e. multiply on the left by elementary matrices) to ensure that the first column is equal to the 1st standard basis vector. In this way we may assume that (for $i = 1,\dotsc,r-1$) the $i$th column of $\mathsf{W}_{\ell}$ is equal to the $i$th standard basis vector; for $\mathsf{F}_{r,r}$, the condition $\det \mathsf{W}_{\ell} = 1$ implies that necessarily $\mathsf{F}_{r,r} = 0$ as well; then as before we can make the $r$th column the $r$th standard basis vector using elementary row operations.
Remark: IIRC I got the idea for the above from Cohn's example $\begin{bmatrix} 1+xy & x^{2} \\ -y^{2} & 1-xy \end{bmatrix}$ which is a matrix that cannot be expressed as a product of elementary matrices with coefficients in $\mathbb{Q}[x,y]$.
Corollary: Let $r \ge 2$, let $A$ be a ring, let $\mathfrak{a}$ be an ideal of $A$, let $A^{\wedge}$ be the $\mathfrak{a}$-adic completion of $A$. Then the group homomorphism $\operatorname{SL}_{r}(A,\mathfrak{a}) \to \operatorname{SL}_{r}(A^{\wedge},\mathfrak{a}A^{\wedge})$ is dense for the $\mathfrak{a}$-adic topology.