= Solution
The <auxiliary field> $\sigma$ has quadratic kernel $-1$ and no <kinetic term>. Inverting that kernel, rather than introducing an artificial <scalar mass>, gives the <momentum space> <Feynman rules>
$$
\boxed{\sigma\text{ line: }-i,\qquad
\psi\text{ line: }\frac{i(\not p+m)}{p^2-m^2+i0},\qquad
\sigma\bar\psi\psi\text{ vertex: }ig.}
$$
The <auxiliary-field propagator> is independent of <momentum>; in position space it is $-i\delta^{(4)}(x-y)$. At each <interaction vertex> attach $(2\pi)^4\delta^{(4)}(\sum p)$ for <momentum conservation>. Integrate independent <loop momenta> with $d^4p/(2\pi)^4$, respect <fermion flow>, include the <closed fermion loop sign> $-1$, and use the appropriate <symmetry factor>. Incoming and outgoing <fermions> carry $u_s(p)$ and $\bar u_s(p)$; incoming and outgoing <antiparticles> carry $\bar v_s(p)$ and $v_s(p)$. Exchanging external <fermions> introduces the <fermionic sign>. Here $\not p=\gamma^\mu p_\mu$ is <Feynman slash notation>.
For a dimensional check, $[\psi]=3/2$, $[\sigma]=2$, $[g]=-1$ in <mass dimensions>. Eliminating $\sigma$ consequently produces a <four-fermion interaction> of negative coupling <mass dimension>, not a propagating scalar degree of freedom.
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