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Write $G=\mathrm{Gal}(F/K)$ andEvery member of this decomposition group preserves ${\mathcal{O}}_F$ and $\mathfrak{q}$, so reduction defines a homomorphismIts kernel is the inertia group. We prove surjectivity of decomposition group reduction without presupposing a lift of Frobenius.

If the two residue fields coincide, surjectivity is immediate. Otherwise choose a nonzero $\overline{a}\in k_{\mathfrak{q}}$ generating it over $k_{\mathfrak{p}}$; a generator exists by the primitive element theorem for finite fields. Use the Chinese remainder theorem in ${\mathcal{O}}_F$ to choose $a$ whose residue is $\overline{a}$ at $\mathfrak{q}$ and zero at every other prime above $\mathfrak{p}$. ConsiderIts coefficients are integral and fixed by $G$, hence lie in ${\mathcal{O}}_K$: an element of $K$ integral over ${\mathcal{O}}_K$ belongs to this integrally closed ring of integers.

Reduce at $\mathfrak{q}$. When $\sigma \notin D_{\mathfrak{q}}$, the chosen residue of $a$ at ${\sigma }^{-1}\mathfrak{q}$ is zero, so $\sigma a$ reduces to zero at $\mathfrak{q}$. When $\sigma \in D_{\mathfrak{q}}$, its residue is $\overline{\sigma }(\overline{a})$, which is nonzero. ThereforeAlthough this product is written in $k_{\mathfrak{q}}[T]$, its coefficients belong to $k_{\mathfrak{p}}$. Put $q=|k_{\mathfrak{p}}|$. The finite-field Frobenius automorphism $b\mapsto b^q$ therefore permutes its nonzero roots. In particular, ${\overline{a}}^q$ is $\overline{\sigma }(\overline{a})$ for some $\sigma \in D_{\mathfrak{q}}$. Since $\overline{a}$ generates $k_{\mathfrak{q}}$, that $\overline{\sigma }$ agrees with Frobenius on the whole field. The Galois group of a finite field extension is cyclic and generated by Frobenius, so its generator lies in the image. Thus

The splitting field of $f\in K[t]$ is the smallest extension $L/K$ in which $f$ is a product of linear factors; equivalently, $L$ is generated over $K$ by all its roots of a polynomial. To prove existence and uniqueness of splitting fields, adjoin a root of an irreducible polynomial dividing $f$ and repeat until $f$ splits. If $L$ and $L^{\prime }$ are two resulting fields, an embedding between the fields generated so far extends by sending each newly adjoined root to a root of its transformed minimal polynomial. Repetition gives a $K$-embedding $L\to L^{\prime }$, whose image contains every root of $f$ and is therefore all of $L^{\prime }$. Thus the splitting field exists and is unique up to $K$-isomorphism.

Now write $K={\mathbb{F}}_q$, and let the distinct irreducible factors of $f$ have degrees $d_1,\dots ,d_r$. The irreducible factors of a finite-field Frobenius polynomial show that a root of the degree-$d_i$ factor lies in ${\mathbb{F}}_{q^n}$ exactly when $d_i\mid n$. Hence the splitting field over a finite field isEvery finite field is a perfect field, so $L/K$ is separable; it is normal because it is a splitting field. It is consequently a Galois extension. More explicitly, the Galois group of a finite field extension is the cyclic group of order $m$ generated by the finite-field Frobenius automorphism $x\mapsto x^q$.

Finally, let $N_2(n)$ denote the number of monic irreducible polynomials over a finite field of degree $n$ over ${\mathbb{F}}_2$. Sinceand $\mu (1)=1$, $\mu (3)=-1$, $\mu (9)=0$, we obtain