# The BRST construction for YM with or without auxiliary field

I'm learning BRST symmetry for Yang-Mills theory and I see that there are two ways of writing BRST differential. In some books (for example Ryder's and Ramond's textbooks) BRST differential acts as \begin{gather} \delta A_\mu^a =-D_\mu c^a, \\ \delta c^a= -\frac{1}{2} f^a_{bc} c^b c^c ,\\ \delta \bar{c}^a= f^a, \end{gather} where I skipped coupling constant, and $f^a$ is a gauge-fixing function, for example $f^a=\partial^\mu A_\mu^a$.

But in Srednicki's or Peskin and Schoeder's textbooks differential $\delta$ acts on $\bar{c}$ as $$\delta \bar{c}^a= B^a,$$ where $B^a$ is the auxiliary field.

For me it seems that first approach is a simple elimination of the auxiliary field $B^a$ from the differential and from the action using condition $f^a=B^a$. Is that so? I just want to be sure I'm not missing something.

What form of BRST is "preferable" i.e. what are reasons to choose BRST transformation with or without auxiliary field?

On one hand, by including the Lautrup-Nakanishi field $$B^a$$, we have an off-shell BRST formulation, i.e. we can prove the nilpotency of the BRST transformation without using the (Euler-Lagrange) equations of motion.
On the other hand, for some applications, a simpler on-shell BRST formulation (where the Lautrup-Nakanishi field $$B^a$$ has been integrated out/eliminated) would suffice.