Let $R$ be a commutative ring (zero characteristic). Take a skew-symmetric matrix  $A\in Mat^{skew-sym}(n,R)$.

1. If $n$ is even, then $\det(A)=Pf^2(A)$ and there exists the "Pfaffian adjugate/adjoint" matrix satisfying: $A\times adj^{Pf}(A)=Pf(A)Id$. What is the standard notation for this $adj^{Pf}(A)$?  (And the standard reference? preferably some linear-algebra book)

2. If $n$ is odd then $\det(A)=0=Pf(A)$. Still, one would like to distinguish between the generic matrices (of corank=1) and more degenerate (of corank$\ge2$). There are various ways to distinguish these two sets, i.e. various scheme structures on the stratum of corank$\ge2$ matrices, i.e. various ideals that measure the degeneracy.
One such ideal is: $\sum\limits_i \det(A_{ii})$, where $A_{ii}$ is the block obtained by erasing the i'th row and column. 
 Note that each $A_{ii}$ is  skew-symmetric and of even size. Thus one might  use the refined ideal: 
$\sum\limits_i Pf(A_{ii})$.  (The two ideals have the same radical.)
Any other related ideals/scheme structures?

There is no direct "Pfaffian-adjugate" in the odd case.  In my case  I consider the submodule $Span(UA+AU^T)\subseteq Mat(n,R)$, where $U$ runs over all the possible matrices (not necessarily skew-symmetric). I need the largest ideal $J$ satisfying: $Span(UA+AU^T)\supseteq Mat^{skew-sym}(n,J)$. For $n=3$ one easily sees: $J=\sum\limits_i Pf(A_{ii})$. But for $n\ge5$ I can only prove: 
$J\supseteq\sum\limits_i Pf^2(A_{ii})$.

Any better bounds? (And the references? Probably this is well known.)

ps. To the moderators: maybe it's good to add the tag "Pfaffians"?