Prime Ideal Decomposition Ramification and the Different — Examples

In $R=\mathbb{Z}[i]$, $(2+i)(2-i)=5$ as elements, hence $(5)=(2+i)(2-i)$ as ideals. The quotient by $(2+i)$ is ${\mathbb{F}}_5$ via $i\mapsto -2$, and the quotient by $(2-i)$ is ${\mathbb{F}}_5$ via $i\mapsto 2$: in either quotient eliminate $i$, leaving the relation $5=0$. Thus both factors are prime of residue degree one. They are distinct, since $2+i$ maps to $4\mathrm{\ne }0$ in the second quotient.

The quotient $R\mathrm{/}(3)$ is ${\mathbb{F}}_3[X]\mathrm{/}(X^2+1)$, a field of nine elements since $X^2+1$ has no root in ${\mathbb{F}}_3$. Thus $(3)$ is itself prime with residue degree two.

Finally $(1+i{)}^2=2i$ as elements, and $i$ is a unit, so $(2)=(1+i{)}^2$ as ideals. The quotient $R\mathrm{/}(1+i)$ is ${\mathbb{F}}_2$ by substituting $i=-1$, so $(1+i)$ is prime with residue degree one. These explicit ideal products and residue fields give respectively $(e,f)=(1,1),(1,1)$; $(1,2)$; and $(2,1)$. They satisfy the degree-two fundamental identity and the definitions of split, inert, and ramified.

Neither $0$ nor $1$ is a root of $X^3+X+1$ in ${\mathbb{F}}_2$, so the cubic is irreducible.

Dedekind--Kummer then gives one prime above $2$ with residue degree $3$ and exponent $1$, which is inertness.

The order $\mathbb{Z}[\sqrt{5}]$ has discriminant $20$, whereas $d_K=5$, so the index formula gives index $2$.

Yet $X^2-5\equiv (X+1{)}^2(\mathrm{m}\mathrm{o}\mathrm{d}2)$ while $2$ is unramified because $2\nmid d_K$. Thus the repeated reduction factor incorrectly predicts ramification, demonstrating necessity of the index hypothesis.

$X^3-2$ is Eisenstein at $2$, and its reduction is $X^3$.

The Eisenstein ramification corollary gives the unique prime $\mathfrak{P}=(2,\alpha )$ with exponent $3$.

The monogenic formula gives ${\mathfrak{D}}_K=(f^{\mathrm{\prime }}(i))=(2i)=(2)$. Taking the fractional-ideal inverse gives ${\mathcal{O}}_K^{\vee }=(2i{)}^{-1}\mathbb{Z}[i]$.

Its norm is $4$, agreeing with the discriminant $d_K=-4$ and the norm-of-the-different theorem.

At $r=1$, ${\mathrm{\Phi }}_{p^r}(t)={\sum }_{k<p}{t}^{k{p}^{r-1}}$, and ${\mathrm{\Phi }}_{p^r}(t+1)$ is Eisenstein at $p$ makes ${\mathrm{\Phi }}_p$ irreducible over $\mathbb{Q}$. Since it is monic and vanishes at the primitive root ${\zeta }_p$, it is its minimal polynomial. The monogenic formula gives ${\mathfrak{D}}_K=({\mathrm{\Phi }}_p^{\mathrm{\prime }}({\zeta }_p))$.

For $p=2$ the claim is immediate. For odd $p$, differentiating the factorisation of ${\mathrm{\Phi }}_p$ into its roots gives ${\mathrm{\Phi }}_p^{\mathrm{\prime }}({\zeta }_p)={\prod }_{j=2}^{p-1}({\zeta }_p-{\zeta }_p^j)={\zeta }_p^{p-2}{\prod }_{k=1}^{p-2}(1-{\zeta }_p^k)\mathrm{.}$ For $1\le k\le p-2$, the quotient $(1-{\zeta }_p^k)\mathrm{/}(1-{\zeta }_p)=1+{\zeta }_p+\cdots +{\zeta }_p^{k-1}$ is a unit: if $k\mathrm{\ell }\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}p)$, the reverse quotient is likewise a cyclotomic integer. Since ${\zeta }_p$ is also a unit, the displayed product generates the ideal $(1-{\zeta }_p{)}^{p-2}$.

Eisenstein gives one prime above $3$ with ramification index $2$.

Since $3\nmid 2$, ramification is tame, and the tame equality gives exponent $2-1=1$.

Eisenstein gives a unique prime above $2$ with ramification index $2$.

Because $2\mid e$, the extension is wild, so the valid conclusion is the wild lower bound $v_{\mathfrak{P}}({\mathfrak{D}}_K)\ge 2$; no equality $e-1$ is inferred.