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Let $f\in R[X]$ have degree $n$. A useful reduction prime must preserve its degree and make its reduction separable. Thus exclude the leading coefficient and discriminant from $P$; after localizing at those elements one may make $f$ monic. The residue base field is $k=\mathrm{Frac}(R/P)$, not necessarily the ring $R/P$ itself. Then the Galois group of the reduced polynomial embeds in the generic group, with the same action on the corresponding roots.

Here is the unramified reduction of a polynomial Galois group argument. A UFD is integrally closed. Let $E$ be the splitting field over $F$, and use the root algebra $B$ obtained by adjoining all roots after inverting the discriminant. It is finite étale over the resulting normal base: successive simple-root adjunctions are unramified, and the distinct generic components separate over a normal ring. Choose a prime $Q$ above $P$. The residue field $k_Q$ is generated over $k$ by the reduced roots, so it is their splitting field. The decomposition groupacts on $k_Q$. Its residue action is onto $\mathrm{Gal}(k_Q/k)$; this is the local decomposition-group description of the unramified fiber. Equivalently, the geometric points of that fiber form a free transitive Galois-group set, and the commuting residue-field action identifies the residue group with a stabilizer subgroup of the generic group.

The kernel is the inertia group. An element in that kernel fixes every reduced root. Since the discriminant is a unit, distinct roots have distinct reductions; consequently it fixes every original root and is the identity. ThereforeChoosing another prime changes this subgroup by conjugacy, which does not change the cycle types available for computing the generic group.

When $k={\mathbb{F}}_q$, its splitting-field group is generated by Frobenius. A reduced irreducible factor of degree $d_i$ contributes one $d_i$-cycle. Hence a squarefree factorization with degrees $d_1,\dots ,d_r$ proves that the generic group contains that cycle type. Irreducible degree-preserving reduction also proves generic irreducibility by Gauss's lemma: a nontrivial monic factorization would reduce to one. Several good primes, a discriminant-square test and permutation-group arguments can then identify the group. A ramified or degree-dropping prime does not justify the same cycle test.