Let $X$ be a projective scheme flat over a local Artinian ring $A$, the residue field of $A$ is algebraically closed, and the special fiber of $X$ (under the natural morphism from $X$ to $A$) is smooth, projective. Let $E$ be a locally free globally generated sheaf over $X$ of rank $r$. As far as I understand for a general choice of $t$ global sections $\sigma\_1, ..., \sigma\_t$ for $t \le r$ the degeneracy locus, say $V(\sigma\_1,...,\sigma\_t)$ corresponding to these $t$ sections is of codimension $r-t+1$. The question is whether $V(\sigma\_1,...,\sigma\_t)$ is flat over $\mathrm{Spec}(A)$? If not true in general is there any known condition under which it holds true?



Degeneracy locus and flatness over local Artinian ring
There are some necessary basic facts about flatness. First of all, let $R$ be a local Noetherian, flat $A$-algebra, and let $x\_1,\dots,x\_c \in \mathfrak{m}\_R$ be a collection of elements. Then, by the [Local Flatness Criterion](http://stacks.math.columbia.edu/tag/00MD), if the images $\overline{x}\_1,\dots,\overline{x}\_c \in \mathfrak{m}\_R/\mathfrak{m}\_AR$ form a regular sequence in $R/\mathfrak{m}\_AR$, then $x\_1,\dots,x\_c$ is a regular sequence in $R$. So, if $R$ is the local ring of $X$ at a closed point, then the hypothesis of Eagon-Hochster about regular sequences can be checked on $R/\mathfrak{m}\_AR$. Thus, if $V(\sigma\_1,\dots,\sigma\_t)$ has the appropriate codimension $c$, then the local ring $R/I$ has an $R$-projective resolution of length $c$ by Eagon-Hochster. 


Okay, now I have to run, so I will return to this later. It *should* follow from the local flatness criterion that an $R$-algebra $R/I$ is $A$-flat if $R/I$ has codimension $c$ in $R$, has a length $c$ $R$-projective resolution, $R/(I+\mathfrak{m}\_AR)$ has codimension $c$ in $R/\mathfrak{m}\_AR$, and $R/(I+\mathfrak{m}\_AR)$ has a length $c$ $R/\mathfrak{m}\_AR$-projective resolution.


**Edit.** Okay, I looked through EGA IV, Section 6, and I could not find quite what I claimed in the previous paragraph (although I suspect that it is true). At any rate, here is a kludge that reduces to the case of a regular ring. 


First, there is a standard reduction to the case that $A$ is a $\mathbb{Z}$-algebra or $\mathbb{Q}$-algebra that is essentially of finite type, using the fact that smooth schemes are locally finite type, etc. Thus, we may assume that $A$ is the quotient of $B/J$, where $B$ is a regular, complete, local Noetherian ring. 


Since the problem is local, and since $X$ is smooth, we may as well pass to the completion of $R$, which is now isomorphic to the complete local ring $A[[t\_1,\dots,t\_n]]$. This is the quotient $R\_B/JR\_B = R\_B\otimes\_B A$, where $R\_B$ is the completed power series ring $B[[t\_1,\dots,t\_n]]$.
Since $R$ is local, the stalk of $E$ is free, say $\widetilde{R}^{\oplus r}$, so that the defining elements $\sigma\_1,\dots,\sigma\_t$ are equivalent to an $r\times t$ matrix of elements in $R$. The entries of this matrix lift to elements $\sigma\_{B,1}, \dots, \sigma\_{B,t}$ of $R\_B^{\oplus n}$, i.e., the matrix lifts to an $r\times t$ matrix of elements in $R\_B$. Denote by $I\_B\subset R\_B$ the ideal generated by the $t\times t$-minors of this matrix. The image ideal $I\_B (R\_B/JR\_B)$ equals the ideal $I$ generated by the $t\times t$-minors of the original matrix. Thus, to prove that $R/I$ is $A$-flat, it suffices to prove that $R\_B/I\_B$ is $B$-flat. Now, since $B$ is a regular local ring, you can directly apply the Eagon-Hochster theorem.


**Second edit.** I checked with an expert, and the general flatness result I claimed above is true. The kludge in the first edit works, but the general flatness result is probably the "right" way to approach this type of result.


