Equation of motion of an auxiliary field I'm a newbie in the field of QFT and SUSY, so I'm warning you: this might be a stupid question.
I'm working with auxiliary fields to describe supersymmetric models and I understand that upon eliminating the auxiliary field $A$ in the (first order) Lagrangian $\mathcal{L}_{1}$ you need to solve his equation of motion. Substituting this in $\mathcal{L}_{1}$ gives you the second-order Lagrangian $\mathcal{L}_{2}$.
Now, assume
$$\mathcal{L}_{1} = \int^{A}\!\mathrm{d}q ~F(q) - Ay(\phi)$$
The equation of motion for $A$ then gives:
$$\partial_{A}\mathcal{L}_{1} = F(A) - y(\phi) = 0 \quad \Rightarrow \quad A = F^{-1}(y)$$
Now, where I get confused is: can you also say that $\partial^{2}_{A}\mathcal{L}_{1} = 0$ because
$$\partial^{2}_{A}\mathcal{L}_{1} = \partial_{A} F(A) = \partial_{A}y(\phi) = 0~  ?$$
I'm not comfortable with substituting the EOM of the auxiliary field in the second step because this is maybe something from the second order Lagrangian which you cannot use here...?
 A: Perhaps the most enlightening is just to show how it goes in OP's example.


*

*If the Lagrangian reads
$${\cal L}_1(A,\phi)~:= ~{\cal F}(A)- Ay(\phi),\qquad  F~=~{\cal F}^{\prime}(A),\tag{1}$$
then the eom for the "auxiliary" variable $A$ reads
$$ F(A)~\approx~ y(\phi) \qquad \Leftrightarrow\qquad A~\approx~ F^{-1}(y(\phi)),\tag{2}$$
where we have assume that $F$ is an invertible function. [The equation of motion (eom) means the EL equation. The $\approx$ sign means here equal modulo the eom.]
Note that eom is not preserved under differentiation wrt. to field variables.  [E.g. assume the opposite is the case. Then differentiation of eq. (2)
wrt. $A$ leads to the contradiction $1=0$.]

*Eliminating the "auxiliary" $A$-variable in the Lagrangian (1) leads to a new Lagrangian for the remaining variable $\phi$:
$${\cal L}_2(\phi)~:=~ {\cal L}_1(F^{-1}(y(\phi)),\phi)
~=~{\cal F}(F^{-1}(y(\phi)))- y(\phi)F^{-1}(y(\phi)).\tag{3}$$
The main point is that the new eom for $\phi$ using the new Lagrangian (3) remains the same as the old eom for $\phi$ using the old Lagrangian (1) if we eliminate the "auxiliary" $A$-variable from the old eom using eq. (2).
