= Solution
Choose Hermitian generators $T^a$ of the gauge <Lie algebra>, with $[T^a,T^b]=if^{abc}T^c$, and write $A_\mu=A_\mu^aT^a$. For the usual compact gauge algebra choose $\operatorname{tr}(T^aT^b)=\delta^{ab}/2$. More generally use a nondegenerate invariant <symmetric bilinear form> on the algebra in place of this trace contraction; a completely arbitrary component metric would not give gauge invariance.
A matter vector is useful to fix conventions: with $\psi'=U(x)\psi$ and $D_\mu=\partial_\mu-igA_\mu$, require $D_\mu'\psi'=U D_\mu\psi$. Acting on an arbitrary vector gives the <Yang-Mills gauge transformation>
$$
\boxed{A_\mu'=UA_\mu U^{-1}+\frac{i}{g}U\partial_\mu U^{-1},\qquad
D_\mu'=U D_\mu U^{-1}.}
$$
For $U=1+ig\theta+O(\theta^2)$ this becomes
$$
\delta A_\mu=D_\mu\theta=\partial_\mu\theta-ig[A_\mu,\theta],\qquad
(D_\mu\theta)^a=\partial_\mu\theta^a+gf^{abc}A_\mu^b\theta^c.
$$
Here the second $D_\mu$ denotes the induced <derivative> in the <Adjoint representation>.
Define the curvature by $[D_\mu,D_\nu]=-igF_{\mu\nu}$. Expanding the <commutator> derives
$$
F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu-ig[A_\mu,A_\nu],\qquad
F_{\mu\nu}^a=\partial_\mu A_\nu^a-\partial_\nu A_\mu^a+gf^{abc}A_\mu^bA_\nu^c.
$$
Since <commutators> of the transformed <derivatives> conjugate, $F_{\mu\nu}'=UF_{\mu\nu}U^{-1}$. Cyclicity of the trace proves the <invariant bilinear-form construction of a Yang-Mills action>:
$$
\boxed{\mathcal L_{\rm YM}=-\frac12\operatorname{tr}(F_{\mu\nu}F^{\mu\nu})
=-\frac14F_{\mu\nu}^aF^{a\mu\nu}.}
$$
For a general invariant form $h$, the corresponding proof is $h([\theta,X],Y)+h(X,[\theta,Y])=0$, applied to the two curvature factors. No matter field is required in this pure-gauge Lagrangian.
Introduce the adjoint <Faddeev-Popov ghost> $c^a$ and antighost $\bar c^a$, independent Grassmann-odd scalar fields of <ghost numbers> $+1$ and $-1$, together with the Grassmann-even <Nakanishi-Lautrup field> $b^a$ of <ghost number> zero. The bar labels the independent antighost, not an additional propagating complex-conjugate matter field. Choose a left <BRST symmetry> differential:
$$
\boxed{sA_\mu^a=(D_\mu c)^a,\qquad
sc^a=-\frac g2f^{abc}c^bc^c,\qquad
s\bar c^a=b^a,\qquad sb^a=0.}
$$
It is an odd derivation: $s(XY)=(sX)Y+(-1)^{|X|}X(sY)$ for homogeneous <Grassmann parity> $|X|$. The two relevant properties are $s\mathcal L_{\rm YM}=0$ and off-shell nilpotence $s^2=0$. The first is gauge invariance with parameter replaced by the ghost, and the second follows from the <Lie algebra>'s graded Jacobi identities. Thus
$$
s(\mathcal L_{\rm YM}+s\Psi)=s\mathcal L_{\rm YM}+s^2\Psi=0.
$$
Algebraically this holds for any $\Psi$; a physical gauge-fixing addition uses an odd <gauge-fixing fermion> of <ghost number> $-1$ so that $s\Psi$ is even and has <ghost number> zero.
For <BRST-exact covariant gauge fixing>, choose
$$
\Psi=\bar c^a\left(\partial^\mu A_\mu^a+\frac12b^a\right).
$$
The <graded Leibniz rule> supplies the essential minus sign:
$$
s\Psi=b^a\partial^\mu A_\mu^a+\frac12b^ab^a-\bar c^a\partial^\mu(D_\mu c)^a.
$$
Use <Gaussian gauge fixing with an auxiliary field> to integrate $b$. Pointwise,
$$
b^aF^a+\frac12(b^a)^2=\frac12(b^a+F^a)^2-\frac12(F^a)^2,
\qquad F^a=\partial^\mu A_\mu^a.
$$
Translation of the regulated oscillatory Gaussian leaves only a field-independent normalization. The remaining Lagrangian is therefore
$$
\boxed{\mathcal L_{\rm YM}-\frac12(\partial^\mu A_\mu^a)^2
-\bar c^a\partial_\mu(D^\mu c)^a.}
$$
This is covariant <Feynman gauge>, with gauge parameter one. Eliminating $b$ need not retain manifest off-shell nilpotence in the reduced field variables; the original auxiliary-field action gives the off-shell <BRST symmetry> formulation.
Finally, the ungauge-fixed quadratic <momentum> kernel is
$$
Q_{\mu\nu}^{ab}(p)=\delta^{ab}(-p^2\eta_{\mu\nu}+p_\mu p_\nu),\qquad
Q_{\mu\nu}^{ab}p^\nu=0.
$$
Gauge directions are null vectors, so there is no inverse defining a vector <propagator>, and the functional integral also counts an infinite <gauge orbit> volume. The added gauge-fixing term cancels the longitudinal part, leaving $Q_{\mu\nu}^{ab}=-\delta^{ab}p^2\eta_{\mu\nu}$. With the vacuum prescription, the <propagators> are
$$
\boxed{\langle A_\mu^a A_\nu^b\rangle(p)=-\frac{i\delta^{ab}\eta_{\mu\nu}}{p^2+i0},\qquad
\langle c^a\bar c^b\rangle(p)=\frac{i\delta^{ab}}{p^2+i0}.}
$$
The ghost action represents the <Faddeev-Popov determinant> and supplies ghost-vector vertices; closed ghost loops have the fermionic minus sign. Expanding the cubic and quartic Yang-Mills terms and this ghost interaction now defines consistent perturbative <Feynman rules>. Physical mass-shell poles remain, whereas the gauge degeneracy has been removed. Possible global gauge-fixing obstructions and boundary zero modes are separate from the perturbative inverse around the trivial vacuum.
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