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 $rt+1$. 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?

$\begingroup$ The codimension should increase with $t$. $\endgroup$ – Laurent MoretBailly Dec 11 '14 at 16:21

$\begingroup$ @MoretBailly: Are you saying I have got the codimension formula wrong? $\endgroup$ – user46578 Dec 11 '14 at 16:26

$\begingroup$ If $X$ is "relatively CohenMacaulay" over $A$ in the sense of EGA IV Section 6, and if the codimension of the closed fiber equals the "expected" codimension, then the vanishing scheme "should" be flat. I expect this follows from EagonHochster. $\endgroup$ – Jason Starr Dec 11 '14 at 16:50

$\begingroup$ @user46578 : yes. It is correct if $t=1$, but clearly we have $\dim V(\sigma_1,\dots,\sigma_t)\leq\dim V(\sigma_1,\dots,\sigma_{t1})$. My guess is $rt$ for the expected codimension. Also "globally generated" is probably not enough. $\endgroup$ – Laurent MoretBailly Dec 11 '14 at 17:01

$\begingroup$ @MoretBailly: May be I am understanding something wrong, but I got the statement from the book on "Intersection theory" by Eisenbud and Harris isites.harvard.edu/fs/docs/icb.topic720403.files/book.pdf Lemma $7.1$ on page $254$. $\endgroup$ – user46578 Dec 11 '14 at 17:12
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, 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 EagonHochster 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 EagonHochster.
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 EagonHochster 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.