Approximating finite type algebras over a formal power series ring

Question

Let $k$ be a ring, let $A := k[x_{1},\dotsc,x_{d}]$ be the polynomial ring and let $A^{\wedge} := k[[x_{1},\dotsc,x_{d}]]$ be the formal power series ring. For a $d$-tuple $\mathbf{e} = (e_{1},\dotsc,e_{d}) \in \mathbb{Z}_{\ge 0}^{\oplus d}$ of nonnegative integers, let me denote $|\mathbf{e}| := e_{1} + \dotsb + e_{d}$ and $x^{\mathbf{e}} := x_{1}^{e_{1}} \dotsb x_{d}^{e_{d}}$. We may write any $a \in A^{\wedge}$ as \begin{align} \textstyle a = \sum_{\mathbf{e} \in \mathbb{Z}_{\ge 0}^{\oplus d}} c_{\mathbf{e}}x^{\mathbf{e}} \end{align} with $c_{\mathbf{e}} \in k$. For any nonnegative integer $\ell \ge 0$, let me denote \begin{align} \textstyle a^{\le \ell} := \sum_{\mathbf{e} \in \mathbb{Z}_{\ge 0}^{\oplus d} ; |\mathbf{e}| \le \ell} c_{\mathbf{e}}x^{\mathbf{e}} \end{align} the "truncation of $a$ at degree $\ell$".

To any polynomial $f \in A^{\wedge}[t_{1},\dotsc,t_{n}]$, we denote $f^{\le \ell} \in A[t_{1},\dotsc,t_{n}]$ the polynomial obtained by applying the truncation operation $(-)^{\le \ell}$ to the coefficients of $f$.

Let $f_{1},\dotsc,f_{m} \in A^{\wedge}[t_{1},\dotsc,t_{n}]$ be a collection of polynomials and set \begin{align} B &:= A^{\wedge}[t_{1},\dotsc,t_{n}]/(f_{1},\dotsc,f_{m}) \\ B^{\le \ell} &:= A^{\wedge}[t_{1},\dotsc,t_{n}]/(f_{1}^{\le \ell},\dotsc,f_{m}^{\le \ell}) \end{align} for all $\ell \ge 0$.

Is there a way to "approximate" $B$ (resp. $\operatorname{Spec}(B)$) by its "truncations" $B^{\le \ell}$ (resp. $\operatorname{Spec}(B^{\le \ell})$)? Is there a way to "approximate" the category of (e.g. finitely generated projective) modules $\mathrm{Mod}(B)$ by the $\mathrm{Mod}(B^{\le \ell})$?

Answer 1

I'm not sure if this constitutes an answer, but it is an example to show that the $B^{\leq \ell}$ can be quite different from $B$.

Example. Let $k = \mathbb C$ (for simplicity), and $A^\wedge = k[[x]]$. In $A^\wedge[s,t]$, consider the polynomials \begin{align*} f &= s-e^x,\\ g &= t-e^{2x},\\ h &= s^2-t. \end{align*} Then the ring $B = A^\wedge[s,t]/(f,g,h)$ is isomorphic to $A^\wedge$ by sending $s$ to $e^x$ and $t$ to $e^{2x}$.

To compute $B^{\leq \ell}$, note that $A^\wedge[s,t]/(f^{\leq \ell},g^{\leq \ell}) \cong A^\wedge$, by sending $s$ and $t$ to $(e^x)^{\leq \ell}$ and $(e^{2x})^{\leq \ell}$ respectively. But then we further divide by the degree $2\ell$ (nonzero) polynomial (in $x$) $$\left(\left(e^x\right)^{\leq \ell}\right)^2 - \left(e^{2x}\right)^{\leq \ell}.$$ Thus, $B^{\leq \ell}$ is a finite $k$-algebra of length $2\ell$, whereas $B$ is $k[[x]]$. Moreover, it is not the case that there are natural transition maps between the $B^{\leq \ell}$ making $B$ into the limit in any way.

Remark. In nicer situations, what one hopes is that $B$ is a deformation of the $B^{\leq \ell}$, at least for $\ell \gg 0$. But the example above shows that is not always the case. If there is any relationship, it's going to be a bit tricky.

Remark. On the other hand, it is true that $B^{\leq \ell}/\mathfrak m^{\ell + 1}B^{\leq \ell} \cong B/\mathfrak m^{\ell + 1}B$, where $\mathfrak m \subseteq A^\wedge$ is the maximal ideal. Then one can recover the $\mathfrak m$-adic completion $B^\wedge$ of $B$ from the $B^{\leq \ell}$ as $$B^\wedge = \lim_{\substack{\longleftarrow\\ \ell}} B/\mathfrak m^\ell B \cong \lim_{\substack{\longleftarrow\\ \ell}} B^{\leq \ell-1}/\mathfrak m^\ell B^{\leq \ell-1}.$$