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While working with flat degenerations of flag varieties and Schubert varieties I've noticed that among the numerous known constructions there doesn't seem to be a single one that doesn't turn out to be a Gröbner degeneration. I've started asking myself if there's a concrete mathematical reason for this. In general terms my question is: is there some reasonably large class of situations in which all flat degenerations are Gröbner degenerations? To pose a precise question I will use a very simple and perhaps naive setting (mostly because I'm not really an algebraic geometer).

Consider an ideal $I$ in the polynomial ring $R=\mathbb C[x_1,\dots,x_n,t]$. I will assume that $I$ is homogeneous in the $x_i$ (but not $t$). To keep things nice and geometric I will also require $I$ to be a radical ideal. Denote $R_0=\mathbb C[x_1,\dots,x_n]$, then for every complex $c$ we have an ideal $I(c)\subset R_0$ which is obtained from $I$ by setting $t=c$. Suppose that the algebras $R_0/I(c)$ are pairwise isomorphic for all $c\neq 0$. The composition map in $\mathbb C[t]\hookrightarrow R\twoheadrightarrow R/I$ turns $R/I$ into a $\mathbb C[t]$-module, suppose this module is flat. (What I'm going for is a flat family of projective subvarieties in $\mathbb P^{n-1}$ over $\mathbb A^1$ with a constant fiber outside of $0$.)

Let me call the above construction a "flat family". In particular, let $I_1\subset R_0$ be a homogeneous ideal and $I_0\subset R_0$ is a (radical) initial ideal of $I_1$ with respect to some monomial order or grading on $R_0$. There's a standard way of constructing a flat family with $I(1)=I_1$, $R/I(c)\simeq R/I_1$ for $c\neq 0$ and $I(0)=I_0$. I will call this a "Gröbner family".

  1. Is there an example of a flat family that is not (at least fiberwise) isomorphic to a Gröbner family?
  2. Is there some natural set of conditions under which a flat family is necessarily (fiberwise?) isomorphic to a Gröbner family?

Update. I have realized that a simple (and not very interesting) way of obtaining flat families that are not Gröbner per se is to take a Gröbner degeneration and apply, say, a linear change of variables. This led me to give the questions an "up to isomorphism" flavor. It's been a year but I'm still very curious about this and still have no idea.

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  • $\begingroup$ You would probably want it up to a "local" isomorphism, since any flat family with more than one special fiber would be a trivial counter example. $\endgroup$
    – pinaki
    Commented Feb 12, 2021 at 13:57
  • $\begingroup$ @auniket Yes, that is why I give a very restrictive definition of a flat family which, in particular, requires exactly one special fiber. Although I must say I don't know any examples with multiple special fibers either (where for at least one special fiber we don't have a Gröbner family such that...). $\endgroup$ Commented Feb 12, 2021 at 14:58

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