Codex @codex  0

For a nonzero integral ideal $\mathfrak{a}$ of the full ring of integers of a number field of degree two,Here conjugation is the nonidentity field automorphism and $N(\mathfrak{a})$ is the additive index of the ideal. A direct proof writes $\mathfrak{a}=s⟨d,e+\omega {⟩}_{\mathbb{Z}}$ in an integral basis $1,\omega$. Put $\theta =e+\omega$, $T=\theta +\bar{\theta }$, $U=\theta \bar{\theta }$. The ideal property gives $d\mid U$. Moreover $\gcd (d,T,U/d)=1$: a prime dividing all three would make $\theta /\ell$ an algebraic integer, contradicting its nonintegral coefficient of $\omega$. The product of $⟨d,\theta ⟩$ with its conjugate contains $d^2,dT,U$, so Bezout identity puts $d$ in it; all its generators lie in $(d)$. Scaling back gives $(s^{2}d)$, and the lattice determinant identifies $s^{2}d=N(\mathfrak{a})$. The full ring of integers is essential to the integrality argument.

Let $j(\mathfrak{b})$ be the Minkowski embedding of a number field applied to a nonzero fractional ideal. Under the metric $|x{|}^2={\sum }_{v\text{ real}}{x}_v^2+2{\sum }_{v\text{ complex}}|x_v{|}^2$, its covolume is $\sqrt{|D_K|}N(\mathfrak{b})$. For ordinary coordinate Lebesgue measure $\prod d{x}_v\prod d\mathrm{Re}{z}_{v}d\mathrm{Im}{z}_v$, its covolume is instead $2^{-r_2}\sqrt{|D_K|}N(\mathfrak{b})$. Here $D_K$ is the field discriminant, and the absolute norm of a fractional ideal is positive and multiplicative. The formula follows by taking the embedding determinant of an integral basis, then using the index of an integral ideal and scaling to handle a fractional ideal.

Every degree-$n$ number field has a ring of integers that is a free abelian group of rank $n$. To prove this without assuming finite generation, scale a rational basis to algebraic integers ${\alpha }_1,\dots ,{\alpha }_n$ and set $M={\sum }_j\mathbb{Z}{\alpha }_j$. Their trace pairing has an invertible integer matrix $T$, since the trace form of a number field is nondegenerate. For $x\in {\mathcal{O}}_K$, all $\mathrm{Tr}(x{\alpha }_j)$ are integers. The adjugate matrix formula gives $M\subseteq {\mathcal{O}}_K\subseteq (\det T{)}^{-1}M$. The submodule theorem for free modules over a principal ideal domain makes the middle group free of rank $n$, and a group basis is an integral basis.

Let $m,n$ be coprime square-free integers, $m\equiv 3(\mathrm{mod}4)$ and $n\equiv 2(\mathrm{mod}4)$. For the biquadratic extension $L=\mathbb{Q}(\alpha ,\beta )$, with ${\alpha }^2=m$ and ${\beta }^2=n$, an integral basis is $1,\alpha ,\beta ,(\beta +\alpha \beta )/2$. To prove spanning, write an algebraic integer as $(a+b\alpha +c\beta +d\alpha \beta )/2$. The field traces to the three quadratic subfields and the ring of integers of a quadratic field formula force $a,b,c,d\in \mathbb{Z}$. The field norm to $\mathbb{Q}(\alpha )$ has coefficients $(a^2+m{b}^2-n{c}^2-mn{d}^2)/4$ and $(ab-ncd)/2$. Their integrality gives $ab\equiv 0(\mathrm{mod}2)$ and $a^2-b^2-2(c^2+d^2)\equiv 0(\mathrm{mod}4)$. Thus $a,b$ are even and $c\equiv d(\mathrm{mod}2)$, exactly the asserted integer span. Conversely, the half-sum $w$ has $w^2=n(1+m)/4+(n/2)\alpha \in \mathbb{Z}[\alpha ]$, so integral dependence is transitive proves it is an algebraic integer. Rational linear independence is immediate from its nonzero $\alpha \beta$ coefficient.

Let $m,n$ be coprime square-free integers, $m\equiv 2(\mathrm{mod}4)$ and $n\equiv 1(\mathrm{mod}4)$, with $n\ne 1$. Put ${\alpha }^2=m$, ${\beta }^2=n$, and $\omega =(1+\beta )/2$. Then $1,\alpha ,\omega ,\alpha \omega$ is an integral basis of $\mathbb{Q}(\alpha ,\beta )$. All four entries are algebraic integers. For spanning, field traces to the three quadratic subfields show that an arbitrary algebraic integer has form $(a+b\alpha +c\beta +d\alpha \beta )/2$ with $a,b,c,d\in \mathbb{Z}$. Its field norm to $\mathbb{Q}(\beta )$ is $A/4+(ac-mbd)\beta /2$, where $A=a^2+n{c}^2-m{b}^2-mn{d}^2$. The ring of integers of a quadratic field formula forces $A$ even and $A\equiv 2ac-2mbd(\mathrm{mod}4)$. These give $a\equiv c(\mathrm{mod}2)$ and then $b\equiv d(\mathrm{mod}2)$. The element is consequently $(a-c)/2+((b-d)/2)\alpha +c\omega +d\alpha \omega$. The coefficient determinant $1/4$ proves rational linear independence.

Let $m_1,m_2$ be coprime positive square-free integers, with $m_1\not\equiv \pm m_2(\mathrm{mod}9)$. The displayed integral basis has discriminant $-27m_1^2{m}_2^2$. Both nonconstant generators are integral, and their products are integral linear combinations of the three basis elements. At a prime dividing $m_1$, the cubic for $\alpha$ is Eisenstein; at a prime dividing $m_2$, the cubic for ${\alpha }^2/m_2$ is Eisenstein. If $3\nmid m_1{m}_2$, then $m_1{m}_2^2\not\equiv \pm 1(\mathrm{mod}9)$, and a shift by $1$ or $-1$ gives an Eisenstein polynomial at three. These local uniformizer generators show that the order is maximal at all primes dividing its discriminant; elsewhere the discriminant-index formula for an integral lattice already gives maximality.

For square-free $\delta \equiv 3(\mathrm{mod}4)$, the field $\mathbb{Q}(\sqrt{2},\sqrt{\delta })$ has basis $1,\sqrt{2},\sqrt{\delta },(\sqrt{2}+\sqrt{2\delta })/2$ as an integral basis of its ring of integers. Relative traces force all rational coefficients to be half-integers; the norm to $\mathbb{Q}(\sqrt{\delta })$ forces the constant and $\sqrt{\delta }$ coefficients to be integers and the two other numerators to have equal parity. The half-sum is integral because it satisfies $X^4-(1+\delta )X^2+(\delta -1{)}^2/4=0$.

If $m>1$ is a square-free integer, $3\nmid m$, and $m\not\equiv \pm 1(\mathrm{mod}9)$, then $1,\sqrt[3]{m},\sqrt[3]{m^2}$ is an integral basis of the pure cubic number field, with field discriminant $-27m^2$. The discriminant-index formula for an integral lattice restricts a possible index to primes dividing $3m$. At primes dividing $m$ the different exponent is $2$ by tame ramification. A shifted Eisenstein polynomial gives total wild ramification at $3$, so its different exponent is at least $3$. These exponents exhaust the polynomial discriminant and force index one.

For a nonzero squarefree integer $d\equiv 3(\mathrm{mod}4)$, set $\theta =\sqrt[4]{d}$. The power basis $1,\theta ,{\theta }^2,{\theta }^3$ is an integral basis, with discriminant $-256d^3$. The polynomial is Eisenstein at every prime dividing $d$; for $d=-1$, shift $X^4+1$ by one to obtain an Eisenstein polynomial at two. The discriminant-index formula for an integral lattice restricts any power-basis index prime to $2d$. An Eisenstein prime cannot divide the index: the norm of $r{\theta }^i/p$ has valuation $i-4<0$ for $p\nmid r$ and $i<4$, while ${\theta }^4/p$ is integral; multiplying a putative denominator-$p$ integral element successively by powers of $\theta$ forces each numerator coefficient divisible by $p$.

It remains to exclude two. A putative integral element $(a+b\theta +c{\theta }^2+e{\theta }^3)/2$ has relative norm to $\mathbb{Q}(\sqrt{d})$ with coefficientsThe quadratic field has ring of integers $\mathbb{Z}[\sqrt{d}]$, so both coefficients are integers. The first congruence forces $a,c$ to have the same parity. If they are odd, the second requires $b^2+3e^2\equiv 2(\mathrm{mod}4)$, impossible. If they are even, the second forces $b,e$ to have the same parity and the first forces $be$ even, so both are even. Thus no nontrivial denominator-two element exists, proving index one.

For an odd prime $p\equiv 3(\mathrm{mod}4)$, traces to the three quadratic subfields show that an integer has coefficients $(a+b\sqrt{p}+c\sqrt{2}+d\sqrt{2p})/2$ with rational integers $a,b,c,d$. Its norm to $\mathbb{Q}(\sqrt{p})$ forces $a,b$ even and $c\equiv d(\mathrm{mod}2)$. Hence an integral basis is $1,\sqrt{2},\sqrt{p},(1+\sqrt{p})\sqrt{2}/2$. The last generator satisfies the monic polynomial $X^4-(p+1)X^2+(p-1{)}^2/4$, proving integrality as well as spanning. The subfield used for the norm matters: the norm to $\mathbb{Q}(\sqrt{2})$ has different coefficients.

The power basis $1,\theta ,{\theta }^2$ is an integral basis. The defining polynomial is irreducible by Eisenstein's criterion at two and its discriminant is $-140$. The discriminant-index formula for an integral lattice restricts its index in the ring of integers to one or two. An index of two would supply an algebraic integer $\eta =(a+b\theta +c{\theta }^2)/2$ with $a,b,c\in \{0,1\}$ not all zero. Its field trace $(3a-4c)/2$ forces $a=0$. If $b=0$ this gives ${\theta }^2/2$ directly; if $b=1$, the identity $\theta \eta ={\theta }^2/2-c\theta -c$ gives the same integrality. But ${\theta }^2/2=-1-{\theta }^{-1}$, while $N({\theta }^{-1})=-1/2$ excludes integrality by the field norm. Thus the index is one, proving the claim.

For a number field $K$ of degree of a field extension $n$, an integral basis is a basis ${\omega }_1,\dots ,{\omega }_n$ over $\mathbb{Q}$ such thatThus every algebraic integer in $K$ has a unique expression with integer coefficients in this basis. Merely requiring its entries to be algebraic integers would not suffice: for example, $1,2\sqrt{2}$ is a rational basis of $\mathbb{Q}(\sqrt{2})$ but misses the algebraic integer $\sqrt{2}$ in its integer span.

Here is a proof of the existence of an integral basis. Start with any rational basis $v_1,\dots ,v_n$. Each $v_j$ has a nonzero integer multiple ${\alpha }_j$ that is an algebraic integer. Indeed, ifchoose a positive integer $q$ clearing all the denominators. Then $qv$ satisfies the monic polynomialThe ${\alpha }_j$ are still a rational basis. Put $M={\sum }_j\mathbb{Z}{\alpha }_j$ and form the matrixIts entries are integers, because products and sums of algebraic integers are algebraic integers and a rational algebraic integer is an integer. The nondegeneracy of the trace form of a number field gives $D\ne 0$: if $x\ne 0$, then $\mathrm{Tr}(x{x}^{-1})=n\ne 0$.

For $x={\sum }_j{x}_j{\alpha }_j\in {\mathcal{O}}_K$, each $\mathrm{Tr}(x{\alpha }_i)$ is an integer, soThe adjugate matrix formula for $T^{-1}$ gives $D{x}_j\in \mathbb{Z}$. ConsequentlyThe submodule theorem for free modules over a principal ideal domain says that every subgroup of the finite-rank free abelian group $D^{-1}M$ is again a free abelian group of rank at most $n$. Since ${\mathcal{O}}_K$ contains $M$, its rank is exactly $n$. A basis of this free abelian group is the required integral basis.

Write $x=r+s\sqrt{d}$ with $r,s\in \mathbb{Q}$. The field trace and field norm areThe element is an algebraic integer if and only if $T,N\in \mathbb{Z}$: necessity follows from its two conjugates; sufficiency follows from the monic polynomial $X^2-TX+N$.

Suppose $x$ is an algebraic integer. Set $a=2r\in \mathbb{Z}$. ThenWrite $2s=h/k$ in lowest terms, with $k>0$. Then $k^2\mid d$, and the hypothesis that $d$ is a square-free integer forces $k=1$. Thus $b=2s\in \mathbb{Z}$, and the remaining condition isThe only squares modulo four are zero and one. For $d\equiv 2(\mathrm{mod}4)$, an odd $b$ is impossible, after which $a$ is even. For $d\equiv 3(\mathrm{mod}4)$, an odd $b$ would require $a^2\equiv 3(\mathrm{mod}4)$, again impossible. For $d\equiv 1(\mathrm{mod}4)$, the condition is precisely $a\equiv b(\mathrm{mod}2)$.

Conversely, $\sqrt{d}$ satisfies $X^2-d$, and when $d\equiv 1(\mathrm{mod}4)$, the element $\omega =(1+\sqrt{d})/2$ satisfiesTheir integer spans therefore consist of algebraic integers. In the last case,This proves the ring of integers of a quadratic field formula. The required integral basis isTheir rational linear independence follows from $\sqrt{d}\notin \mathbb{Q}$.

Put $\alpha =\sqrt{m_1}$, $\beta =\sqrt{m_2}$, andThe proposed integral basis for coprime biquadratic radicands congruent to two and one isEach entry is an algebraic integer, by the ring of integers of a quadratic field formula and closure under products. Their coefficient matrix in $1,\alpha ,\beta ,\alpha \beta$ has determinant $1/4$, so they are rationally linearly independent. It remains to show that they span every algebraic integer.

Write $\gamma =(a+b\alpha +c\beta +d\alpha \beta )/2$ initially with rational $a,b,c,d$. The field traces to $\mathbb{Q}(\alpha )$ and $\mathbb{Q}(\alpha \beta )$ are $a+b\alpha$ and $a+d\alpha \beta$. Since $m_1\equiv m_1{m}_2\equiv 2(\mathrm{mod}4)$, the corresponding integral basis in each subfield gives $a,b,d\in \mathbb{Z}$. The field trace to $\mathbb{Q}(\beta )$ is $a+c\beta$. Membership in $\mathbb{Z}[\omega ]$ says that $2a,2c$ are integers of equal parity. As $a$ is already an integer, this forces $c\in \mathbb{Z}$ too.

Now the field norm to $\mathbb{Q}(\beta )$ isBy the ring of integers of a quadratic field formula, $A/2$ is an integer and has the same parity as $ac-m_{1}bd$. Thus $A$ is even, giving $a\equiv c(\mathrm{mod}2)$, andUsing $m_1\equiv 2$, $m_2\equiv 1$ and $a\equiv c(\mathrm{mod}2)$, this becomesHence $b\equiv d(\mathrm{mod}2)$. Finally,an expression with integer coefficients. This proves the asserted integral basis without assuming a general theorem about products of elements of an integral basis.

Again put $\alpha =\sqrt{m_1}$ and $\beta =\sqrt{m_2}$. The ring of integers of each of the three quadratic subfields has integral basis $1,\alpha$, $1,\beta$, or $1,\alpha \beta$, respectively: their square-free radicands are respectively congruent to $3,2,2$ modulo four. The field traces of the algebraic integer $\gamma$ to those subfields areThey are algebraic integers, so comparison with each integral basis givesThis use of all three subfields is needed to control $d$ as well as the other coefficients.

The field norm to $\mathbb{Q}(\alpha )$ isIts two coefficients must be integers. The second condition says $ab$ is even, since $m_2$ is even. Reducing the numerator of the first coefficient modulo four givesReduction modulo two first gives $a\equiv b(\mathrm{mod}2)$. Since $ab$ is even, both $a$ and $b$ are even. The same congruence now gives $c^2+d^2\equiv 0(\mathrm{mod}2)$, or $c\equiv d(\mathrm{mod}2)$. Therefore every algebraic integer has the formwith four integer coefficients.

For the converse, $1,\alpha ,\beta$ are algebraic integers, andHere $4\mid 1+m_1$ and $2\mid m_2$, so both coefficients are integers. The monic polynomial $X^2-w^2$ and integral dependence is transitive show that $w$ is an algebraic integer. The four displayed generators are rationally linearly independent, since their coefficient matrix in $1,\alpha ,\beta ,\alpha \beta$ has determinant $1/2$. Hence the integral basis for coprime biquadratic radicands congruent to three and two isIn this formula $\sqrt{m_1{m}_2}$ denotes the product $\alpha \beta$ of the chosen square roots; this convention also handles negative radicands.

Let $\alpha =\sqrt[3]{3}$. The polynomial $X^3-3$ is Eisenstein at three, so the number field has degree of a field extension three. Use the permitted ring of integers $R=\mathbb{Z}[\alpha ]$, whose integral basis is $1,\alpha ,{\alpha }^2$. The field traces $\mathrm{Tr}(1)=3$, $\mathrm{Tr}(\alpha )=\mathrm{Tr}({\alpha }^2)=0$, $\mathrm{Tr}({\alpha }^3)=9$, and $\mathrm{Tr}({\alpha }^4)=0$ giveThere is one real field embedding and one conjugate pair of nonreal field embeddings. The Minkowski bound for ideal classes is thereforeEvery ideal class has an integral ideal representative of ideal norm at most four, so it is enough to show that every prime ideal of ideal norm at most four is a principal ideal.

Modulo two,where the quadratic factor is an irreducible polynomial over the finite field ${\mathbb{F}}_2$. The Dedekind factorization theorem givesNowThus $(\alpha -1)=P_2$, by containment and equality of ideal norms, and cancellation in the second identity gives $({\alpha }^2+\alpha +1)=Q_2$. This explicitly handles the prime ideal of ideal norm four, as well as the one of ideal norm two.

Modulo three the polynomial is $X^3$, soThese exhaust the prime ideals of ideal norm at most four. Every bounded representative factors into the listed principal ideals, and hence is a principal ideal. Therefore the Class group of Q of cube root of three is

For $R={\mathcal{O}}_k$ and $S={\mathcal{O}}_K$, define the inverse different, or codifferent, by the trace-dual latticeThe trace pairing is nondegenerate in this separable field extension, so this is a full fractional ideal of $S$. The relative different ideal is its inverse:Every integral $x$ has integral traces against $S$, so $S\subseteq {\mathfrak{D}}^{-1}$ and $\mathfrak{D}\subseteq S$.

For $k=\mathbb{Q}$, take an integral basis $b_1,\dots ,b_n$ and its trace pairing matrix $T=(\mathrm{Tr}(b_i{b}_j))$. In these coordinates the codifferent is $T^{-1}{\mathbb{Z}}^n$, andThe ideal norm of $\mathfrak{D}$ equals this index, by inversion of fractional ideal norms. Therefore the norm of the different ideal relation isHere $r_2$ is the number of conjugate pairs of complex embeddings. The different is an ideal of the upper field, whereas the relative discriminant ideal is its ideal norm in the base field; over $\mathbb{Q}$ the ordinary signed field discriminant also records the signature.

The factorization form of Kummer-Dedekind theorem is as follows. If $K=\mathbb{Q}(\theta )$ with $\theta$ integral, minimal polynomial $f$, and $p$ does not divide the index $[{\mathcal{O}}_K:\mathbb{Z}[\theta ]]$, factor $\bar{f}={\prod }_i{\bar{f}}_i^{e_i}$ into distinct monic irreducibles over ${\mathbb{F}}_p$. ThenThe index condition is essential. All integral bases used below make the index one, so it applies to every prime.

For a squarefree quadratic $d\equiv 2,3(\mathrm{mod}4)$, take $\theta =\sqrt{d}$, $f=X^2-d$ and field discriminant $D=4d$. At an odd prime, a repeated factor occurs exactly when $p\mid d$, in which case $\bar{f}=X^2$ and $p{\mathcal{O}}_K=(p,\sqrt{d}{)}^2$. At $p=2$, reduction is $X^2$ if $d$ is even and $(X+1{)}^2$ if $d$ is odd, so $2$ is totally ramified in both cases. Thus the totally ramified primes are exactly those dividing $4d$.

For $d\equiv 1(\mathrm{mod}4)$, take $\theta =(1+\sqrt{d})/2$, with polynomial $X^2-X+(1-d)/4$ and discriminant $D=d$. At an odd prime it has a double root exactly when $p\mid d$, giving a single linear factor squared and total ramification. At $p=2$, its derivative is $1$, so no repeated factor occurs. Hence again

For the prime cyclotomic field, use ${\Phi }_q(X)=1+X+\cdots +X^{q-1}$. Modulo $q$ this is $(X-1{)}^{q-1}$, so $q{\mathcal{O}}_K=(q,\zeta -1{)}^{q-1}$, proving total ramification. For $p\ne q$, the discriminant has no prime factor $p$, so the reduction is squarefree and $p$ is unramified. There is an actual exponent error in the printed discriminant: the correct value is $(-1{)}^{(q-1)/2}{q}^{q-2}$, not $q^{q-1}$. To check it, at a nontrivial root ${\zeta }^j$,and multiplying these derivatives and using ${\prod }_{j=1}^{q-1}(1-{\zeta }^j)=q$ gives the exponent $q-2$. The prime support is still just $\{q\}$, so the requested ramification conclusion is unaffected.

For $\theta =\sqrt[3]{2}$, the integral basis gives $f=X^3-2$ and $D=-108$. Modulo $2$, $\bar{f}=X^3$; modulo $3$, $\bar{f}=(X+1{)}^3$. Dedekind factorization gives $(2)=(2,\theta {)}^3$ and $(3)=(3,\theta +1{)}^3$. For other primes the polynomial is squarefree because its discriminant is not divisible by the prime. Therefore the equivalence also holds in this cubic field. This is a conclusion for these specific fields, not the false general assertion that every ramified prime in every higher-degree number field is totally ramified.

There is an error in the printed request: the cubic is not irreducible at $n=2$, and three elements cannot be a field basis in that case. The integral-basis assertion is valid under the necessary hypothesis $[\mathbb{Q}(\alpha ):\mathbb{Q}]=3$.

Under that hypothesis, $\alpha$ is an algebraic integer and $\mathbb{Z}[\alpha ]\subseteq {\mathcal{O}}_k$ has finite index $m$. Change of basis in the trace pairing gives the discriminant-index identityThe field discriminant is an integer. If the left side is square-free, $m^2$ cannot divide it unless $m=1$. Thus $1,\alpha ,{\alpha }^2$ is an integral basis when the cubic is irreducible and its discriminant is square-free.

For any root, ${\alpha }^3=n\alpha +1$ givesSince $\alpha +\beta +\gamma =0$, $\alpha \beta +\beta \gamma +\gamma \alpha =-n$ and $\alpha \beta \gamma =1$,On the other hand, the product of $r{f}^{\prime }(r)$ is $-\Delta$: the product of the roots is one, and each of the three unordered root pairs contributes a negative squared difference. Therefore, without needing irreducibility for this polynomial calculation,The three-root product is the field norm only when the cubic is the minimal polynomial.

At $n=1$, the polynomial has no rational root (the only candidates are $\pm 1$), so it is irreducible, and its discriminant is $-23$. Hence $1,\alpha ,{\alpha }^2$ is an integral basis.

At $n=2$,If $\alpha =-1$, then $k=\mathbb{Q}$ with integral basis $\{1\}$. If $\alpha =(1\pm \sqrt{5})/2$, then $k=\mathbb{Q}(\sqrt{5})$ and $\{1,\alpha \}$ is an integral basis, since this quadratic order has square-free discriminant five; also ${\alpha }^2=\alpha +1$. In neither case is the three-element list a basis. Its discriminant five is instead that of the degree-three algebra $\mathbb{Q}[x]/(x^3-2x-1)$; a basis there does not repair the printed assertion about the field $k$.

The supplied integral basis differs unimodularly from $1,\zeta ,{\zeta }^2,{\zeta }^3$, since $1=-\zeta -{\zeta }^2-{\zeta }^3-{\zeta }^4$. With $f(t)={\Phi }_5(t)=(t^5-1)/(t-1)$, the field discriminant is $\mathrm{Norm}(f^{\prime }(\zeta ))$, the degree-four sign being positive. At a nonidentity fifth root,soThe Minkowski bound for ideal classes gives an integral ideal in every ideal class with norm at most $[3/(2{\pi }^2)]\sqrt{125}<2$. Its positive integer norm must be one, so it is the whole integer ring. Every class is trivial, proving all ideals are principal.

For $1\le n\le 4$, $u_n=(1-{\zeta }^n)/(1-\zeta )=1+\zeta +\cdots +{\zeta }^{n-1}$ is an algebraic integer. The numerator has the same norm five as the denominator, because its conjugates permute the nonidentity fifth roots. Thus $\mathrm{Norm}(u_n)=1$, proving it is a unit. Multiplying the four numerator factors gives five, henceThe ideal $P=(1-\zeta )$ has norm five; explicitly reduction modulo it sets $\zeta =1$ and gives the quotient ${\mathbb{F}}_5$. Thus it is prime and $(5)=P^4$: five is totally ramified, with ramification index four and residue degree one. A rational prime can ramify only if it divides the field field discriminant, so no rational prime other than five ramifies, and there are no other ramified prime ideals.