Decomposition Inertia and Frobenius — Examples

Given: The data and hypotheses of the statement.

Galois prime decomposition efg: For a finite Galois extension L/K and nonzero prime p, every P above p has the same ramification index e and residue degree f. If there are g such primes, then $efg=[L:K]$.

Orders of decomposition and inertia groups: For finite Galois L/K and nonzero $P\mid p$, writing e and f for its ramification index and residue degree, $\mathrm{\mid }D(P\mathrm{/}p)\mathrm{\mid }=ef,\mathrm{\mid }I(P\mathrm{/}p)\mathrm{\mid }=e,\mathrm{\mid }D(P\mathrm{/}p)\mathrm{/}I(P\mathrm{/}p)\mathrm{\mid }=f\mathrm{.}$ The prime P is unramified over p if and only if its inertia group is trivial.

Frobenius order is residue degree: For finite Galois L/K and nonzero $P\mid p$, the arithmetic Frobenius coset has order $f(P\mathrm{/}p)$ in D/I. If P is unramified, ${\mathrm{Frob}}_P$ has the same order in D.

Given: The data and hypotheses of the statement.

Frobenius cycle type and prime splitting: Let $F\in \mathbb{Z}[T]$ be monic separable with splitting field L, and let p be a rational prime not dividing $\mathrm{Disc}(F)$. Then p is unramified in L and $\bar{F}\in {\mathbb{F}}_p[T]$ is squarefree. The degrees of its monic irreducible factors, with each distinct factor counted once, are exactly the cycle lengths of arithmetic Frobenius on the roots of F.

Decomposition inertia in a quadratic field: In a quadratic Galois extension of number fields, let $G=C_2$. For any nonzero base prime the three possibilities are: split: $(e,f,g)=(1,1,2)$, $D=I=1$, Frobenius identity; inert: $(1,2,1)$, $D=C_2$, $I=1$, Frobenius the nonidentity element; ramified: $(2,1,1)$, $D=I=C_2$, arithmetic Frobenius coset identity in D/I.

The multiplicative group ${\mathbb{F}}_q^{\times }$ of a finite field is cyclic: The multiplicative group $F^{\times }=F\setminus \{0\}$ of every finite field $F$ is cyclic.

Integers in a quadratic field: For squarefree $d\mathrm{\ne }1$, ${\mathcal{O}}_{\mathbb{Q}(\sqrt{d})}=\mathbb{Z}[(1+\sqrt{d})\mathrm{/}2]$ if $d\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}4)$, and $\mathbb{Z}[\sqrt{d}]$ otherwise.

Given: The data and hypotheses of the statement.

Frobenius cycle type and prime splitting: Let $F\in \mathbb{Z}[T]$ be monic separable with splitting field L, and let p be a rational prime not dividing $\mathrm{Disc}(F)$. Then p is unramified in L and $\bar{F}\in {\mathbb{F}}_p[T]$ is squarefree. The degrees of its monic irreducible factors, with each distinct factor counted once, are exactly the cycle lengths of arithmetic Frobenius on the roots of F.

Frobenius order is residue degree: For finite Galois L/K and nonzero $P\mid p$, the arithmetic Frobenius coset has order $f(P\mathrm{/}p)$ in D/I. If P is unramified, ${\mathrm{Frob}}_P$ has the same order in D.

Galois prime decomposition efg: For a finite Galois extension L/K and nonzero prime p, every P above p has the same ramification index e and residue degree f. If there are g such primes, then $efg=[L:K]$.

Eisenstein criterion over the integers: Let $f=a_n{x}^n+\cdots +a_0\in \mathbb{Z}[x]$ be primitive with $n\ge 1$. If there is a prime $p$ such that $p\nmid a_n,p\mid a_i\text{ for every }i<n,p^2\nmid a_0,$ then $f$ is irreducible in $\mathbb{Q}[x]$.

The discriminant of a monic polynomial as the coefficient expression of ${\mathrm{\Delta }}_n^2$: By prop-vandermonde-square-is-symmetric and thm-fundamental-theorem-of-symmetric-polynomials, there is a unique polynomial $D_n\in \mathbb{Z}[T_1,\dots ,T_n]$ such that ${\mathrm{\Delta }}_n(x_1,\dots ,x_n{)}^2=D_n(e_1,\dots ,e_n)\mathrm{.}$ For a monic polynomial $f(t)=t^n+a_1{t}^{n-1}+\cdots +a_n$ over a commutative ring, its discriminant is $\mathrm{Disc}(f):=D_n(-a_1,a_2,\dots ,(-1{)}^n{a}_n)\mathrm{.}$ Equivalently, in any algebra in which $f$ splits with roots ${\alpha }_1,\dots ,{\alpha }_n$, this coefficient expression evaluates to ${\mathrm{\Delta }}_n({\alpha }_1,\dots ,{\alpha }_n{)}^2$. The definition therefore depends only on the coefficients and not on a choice or ordering of roots. For a monic constant polynomial, $\mathrm{Disc}(1)=1$.

Given: The data and hypotheses of the statement.

Frobenius cycle type and prime splitting: Let $F\in \mathbb{Z}[T]$ be monic separable with splitting field L, and let p be a rational prime not dividing $\mathrm{Disc}(F)$. Then p is unramified in L and $\bar{F}\in {\mathbb{F}}_p[T]$ is squarefree. The degrees of its monic irreducible factors, with each distinct factor counted once, are exactly the cycle lengths of arithmetic Frobenius on the roots of F.

Frobenius elements above a prime are conjugate: In a finite Galois extension L/K let the nonzero prime p be unramified. If $\sigma P=P^{\mathrm{\prime }}$ above p, then ${\mathrm{Frob}}_{P^{\mathrm{\prime }}}=\sigma {\mathrm{Frob}}_P{\sigma }^{-1}\mathrm{.}$ Thus p determines one conjugacy class. If the Galois group is abelian, the element is independent of P.

Eisenstein criterion over the integers: Let $f=a_n{x}^n+\cdots +a_0\in \mathbb{Z}[x]$ be primitive with $n\ge 1$. If there is a prime $p$ such that $p\nmid a_n,p\mid a_i\text{ for every }i<n,p^2\nmid a_0,$ then $f$ is irreducible in $\mathbb{Q}[x]$.

Given: The data and hypotheses of the statement.

Frobenius in a small cyclotomic field: Let $\zeta ={\zeta }_5$ be a primitive fifth root of unity and $L=\mathbb{Q}(\zeta )$. Then $\mathrm{Gal}(L\mathrm{/}\mathbb{Q})\cong (\mathbb{Z}\mathrm{/}5\mathbb{Z}{)}^{\times }=C_4$ by ${\sigma }_a(\zeta )={\zeta }^a$. For every prime $p\mathrm{\ne }5$, p is unramified and its arithmetic Frobenius is ${\sigma }_{p<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>5}$. In particular 2 is inert, with $(e,f,g)=(1,4,1)$.

Decomposition and inertia in towers: Let M/L/K be a tower of number fields with M/K and L/K finite Galois, and fix nonzero primes $Q\mid P\mid p$. With $H=\mathrm{Gal}(M\mathrm{/}L)$, $D(Q\mathrm{/}P)=D(Q\mathrm{/}p)\cap H,I(Q\mathrm{/}P)=I(Q\mathrm{/}p)\cap H\mathrm{.}$ Restriction gives exact sequences $1\to D(Q\mathrm{/}P)\to D(Q\mathrm{/}p)\to D(P\mathrm{/}p)\to 1,$ $1\to I(Q\mathrm{/}P)\to I(Q\mathrm{/}p)\to I(P\mathrm{/}p)\to 1.$ The intersection identities also hold without L/K Galois; the displayed quotient assertions use that hypothesis.

Frobenius compatibility in finite towers: Let M/L/K have M/K and L/K finite Galois, and let $Q\mid P\mid p$ be nonzero primes with Q unramified over p. Then $\mathrm{Frob}(Q\mathrm{/}p){\mathrm{\mid }}_L=\mathrm{Frob}(P\mathrm{/}p),\mathrm{Frob}(Q\mathrm{/}P)=\mathrm{Frob}(Q\mathrm{/}p{)}^{f(P\mathrm{/}p)}\mathrm{.}$ If $M=L_1{L}_2$ with both $L_i\mathrm{/}K$ finite Galois and Q unramified over p, the Frobenius elements at its two contractions determine $\mathrm{Frob}(Q\mathrm{/}p)$ uniquely by restriction.

Given: The data and hypotheses of the statement.

Gaussian and eisenstein frobenius: In $\mathbb{Q}(i)$, an odd prime p splits if $p\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}4)$ and is inert if $p\equiv 3(\mathrm{m}\mathrm{o}\mathrm{d}4)$; arithmetic Frobenius sends $i\mapsto i^p$. The prime 2 ramifies. In $\mathbb{Q}({\zeta }_3)$, a prime $p\mathrm{\ne }3$ splits if $p\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}3)$ and is inert if $p\equiv 2(\mathrm{m}\mathrm{o}\mathrm{d}3)$; arithmetic Frobenius sends ${\zeta }_3\mapsto {\zeta }_3^p$. The prime 3 ramifies.

Arithmetic frobenius coset: For finite Galois L/K and nonzero $P\mid p$, the arithmetic Frobenius coset is the unique element of $D(P\mathrm{/}p)\mathrm{/}I(P\mathrm{/}p)$ corresponding under the residue isomorphism to $x\mapsto x^{Np}$ on $\kappa (P)$, where $Np=\mathrm{\mid }\kappa (p)\mathrm{\mid }$. It is defined also when P is ramified. Its inverse is called geometric Frobenius. The quotient element is distinguished; a representative in D need not be unique.

Given: The data and hypotheses of the statement.

Frobenius cycle type and prime splitting: Let $F\in \mathbb{Z}[T]$ be monic separable with splitting field L, and let p be a rational prime not dividing $\mathrm{Disc}(F)$. Then p is unramified in L and $\bar{F}\in {\mathbb{F}}_p[T]$ is squarefree. The degrees of its monic irreducible factors, with each distinct factor counted once, are exactly the cycle lengths of arithmetic Frobenius on the roots of F.

Dedekind--Kummer prime factorisation: Let $L\mathrm{/}K$ be a finite extension of number fields, let $\alpha \in {\mathcal{O}}_L$ with ${\mathcal{O}}_L={\mathcal{O}}_K[\alpha ]$, and let $F\in {\mathcal{O}}_K[X]$ be its monic minimal polynomial over $K$. Let $\mathfrak{p}$ be a nonzero prime ideal of ${\mathcal{O}}_K$, not dividing the index of this power order (the index is $1$ under the stated monogeneity hypothesis). If $\bar{F}={\prod }_i{\bar{g}}_i^{e_i}$ with distinct monic irreducibles over ${\mathcal{O}}_K\mathrm{/}\mathfrak{p}$, then $\mathfrak{p}{\mathcal{O}}_L={\prod }_i{\mathfrak{P}}_i^{e_i},{\mathfrak{P}}_i=(\mathfrak{p},g_i(\alpha )),f({\mathfrak{P}}_i\mathrm{/}\mathfrak{p})=\mathrm{deg}{\bar{g}}_i,$ Here $g_i\in {\mathcal{O}}_K[X]$ are any monic lifts of ${\bar{g}}_i$, and the last $f$ denotes the residue degree from def-prime-above-and-residue-degree.

Ramification is detected by the number-field discriminant: A rational prime $p$ ramifies in $K\mathrm{/}\mathbb{Q}$ if and only if $p\mid d_K$.

Integers in a quadratic field: For squarefree $d\mathrm{\ne }1$, ${\mathcal{O}}_{\mathbb{Q}(\sqrt{d})}=\mathbb{Z}[(1+\sqrt{d})\mathrm{/}2]$ if $d\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}4)$, and $\mathbb{Z}[\sqrt{d}]$ otherwise.

Power-basis and polynomial discriminants: Let $K=\mathbb{Q}(\alpha )$. If $f$ is the degree-$n$ monic minimal polynomial of $\alpha$, then $\mathrm{disc}(1,\alpha ,\dots ,{\alpha }^{n-1})=(-1{)}^{n(n-1)\mathrm{/}2}{N}_{K\mathrm{/}\mathbb{Q}}(f^{\mathrm{\prime }}(\alpha ))=\mathrm{disc}(f)\mathrm{.}$