In Quantum Field Theory and the Standard Model by Schwartz the author starts with the QED bare Lagrangian, defines $A_\mu^0 = \sqrt{Z_3}A_\mu$, $\psi^0 = \sqrt{Z_2}\psi$, $m_0=Z_m m_R$ and $e_0 = Z_e e_R$. Further defining $Z_1 = Z_eZ_2\sqrt{Z_3}$ the bare Lagrangian gets written as \begin{eqnarray} {\cal L} &=& -\frac{1}{4}Z_3 F_{\mu\nu}F^{\mu\nu}+iZ_2\bar \psi\gamma^\mu{\partial_\mu}\psi- Z_2Z_m m_R\bar \psi\psi - e_R Z_1 \bar \psi \gamma^\mu A_\mu \psi.\tag{1} \end{eqnarray}\begin{eqnarray} {\cal L} &=& -\frac{1}{4}Z_3 F_{\mu\nu}F^{\mu\nu}+iZ_2\bar \psi\gamma^\mu{\partial_\mu}\psi- Z_2Z_m m_R\bar \psi\psi - e_R Z_1 \bar \psi \gamma^\mu A_\mu \psi.\tag{19.8} \end{eqnarray} Now he writes $Z_i = 1+\delta_i$.$$Z_i = 1+\delta_i.\tag{19.9+10}$$ The problem is that in the term $Z_2Z_m$ we would also have one term $\delta_2\delta_m$ in eq. (19.12) and this term is discarded by Schwartz.
I don't see how to justify it. I mean, even if you write a series expansion for each $\delta_i$ in powers of $e_R^2$ why would we just keep the leading result?
Being more explicit. Each $\delta_i$ starts at order $e_R^2$. Now say we want a result of order $e_R^4$. Clearly $\delta_2\delta_m$ should contribute because combining the two leading $O(e_R^2)$ contributions will give one $O(e_R^4)$ contribution.
So why one may just drop the $\delta_2\delta_m$ contribution as Schwartz seems to be doing?
