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Put $s=\sqrt{-14}$. Then ${\mathcal{O}}_K=\mathbb{Z}[s]$ and the field discriminant is $-56$. Reduction of $X^2+14$ modulo $2$ and $3$ givesTheir ideal norms are $2,3,3$, respectively. The Minkowski bound for ideal classes is $(2/\pi )\sqrt{56}<5$, so every ideal class has an integral representative of norm at most $4$. The only possible representatives are the unit ideal, ${\mathfrak{p}}_2$, ${\mathfrak{p}}_3$, ${\overline{\mathfrak{p}}}_3$, and the norm-$4$ ideal ${\mathfrak{p}}_2^2=(2)$. Thus there are at most four classes.

The class of ${\mathfrak{p}}_2$ is nontrivial, since a principal ideal of norm $2$ would require $x^2+14y^2=2$. Also ${\mathfrak{p}}_3^2$ is not principal: norm $9$ forces $x+ys=\pm 3$, whose ideal is ${\mathfrak{p}}_3{\overline{\mathfrak{p}}}_3$, not ${\mathfrak{p}}_3^2$.

Now $N(2+s)=18$. It is divisible by ${\mathfrak{p}}_2$ and ${\mathfrak{p}}_3$, but not by ${\overline{\mathfrak{p}}}_3$, as substitution of $s=-1$ modulo $3$ gives $1$. Unique ideal factorization and norms therefore give $(2+s)={\mathfrak{p}}_2{\mathfrak{p}}_3^2$. Consequently $[{\mathfrak{p}}_3{]}^2=[{\mathfrak{p}}_2{]}^{-1}=[{\mathfrak{p}}_2]$, a nontrivial class of order $2$. The class of ${\mathfrak{p}}_3$ has order $4$, and the four-class bound is attained:

We prove the conjugate product of a quadratic ideal identity $\mathfrak{a}\overline{\mathfrak{a}}=(N(\mathfrak{a}))$, which applies to the given two-generator ideal without assuming its displayed generators form an integral basis.

The quadratic ring of integers of a number field has a basis $1,\omega$ over $\mathbb{Z}$: the vector 1 is primitive, since a rational algebraic integer must be an integer, and extends to an integral basis. Write ${\omega }^2-T\omega +U=0$, with $T,U\in \mathbb{Z}$. Let $B$ be the least positive integer in $\mathfrak{a}$ and let $s>0$ generate the set of coefficients of $\omega$ in elements of $\mathfrak{a}$. Its lattice basis can be chosen as $B,r+s\omega$, with $r\in \mathbb{Z}$. Since $B\omega \in \mathfrak{a}$, comparison of coefficients shows $s\mid B$ and $s\mid r$. Write $B=sd$, $r=se$ and $\theta =e+\omega$. ThenThe scaled ideal $\mathfrak{j}=s^{-1}\mathfrak{a}$ remains an ${\mathcal{O}}_L$-ideal. From ${\theta }^2-T^{\prime }\theta +U^{\prime }=0$ we get $U^{\prime }\in \mathfrak{j}\cap \mathbb{Z}=d\mathbb{Z}$, so $d\mid U^{\prime }$. The product $\mathfrak{j}\bar{\mathfrak{j}}$ is generated as an ideal by $d^2,d\theta ,d\bar{\theta },U^{\prime }$ and is contained in $(d)$.

Moreover $\gcd (d,T^{\prime },U^{\prime }/d)=1$. If a prime $\ell$ divided all three, then $\ell \mid T^{\prime }$ and ${\ell }^2\mid U^{\prime }$, so $\theta /\ell$ would satisfy the monic integer polynomial $z^2-(T^{\prime }/\ell )z+U^{\prime }/{\ell }^2$. It would lie in ${\mathcal{O}}_L$, impossible because its $\omega$ coefficient in the integral basis is $1/\ell$. Bezout identity now expresses one as an integer combination of $d,T^{\prime },U^{\prime }/d$. Multiplying by $d$, the terms are $d^2$, $d{T}^{\prime }=d\theta +d\bar{\theta }$, and $U^{\prime }$, all in the product. Hence $d$ belongs to it, proving $\mathfrak{j}\bar{\mathfrak{j}}=(d)$.

The determinant of the lattice basis $sd,s(e+\omega )$ gives the ideal norm $N(\mathfrak{a})=s^{2}d$. ConsequentlyIn particular $⟨b,\alpha ⟩⟨b,\bar{\alpha }⟩$ is principal. This does not require the original $b$ to be nonzero or the ideal itself to be principal.

Here ${\mathcal{O}}_K=\mathbb{Z}[s]$ with $s=\sqrt{-14}$ and field discriminant $-56$. The imaginary-quadratic Minkowski bound for ideal classes is $(2/\pi )\sqrt{56}<5$, so each ideal class has an integral representative of ideal norm at most four. Representatives therefore use prime ideals above two and three. PutThe prime two ramifies, with ${\mathfrak{p}}^2=(2)$, and three splits into $\mathfrak{q}\overline{\mathfrak{q}}=(3)$. No element has field norm two, because $N(a+bs)=a^2+14b^2$; thus $[\mathfrak{p}]$ is nontrivial of order two. Norm-four ideals represent $[\mathfrak{p}{]}^2=1$ and add no generator, so the class group is generated by $[\mathfrak{p}],[\mathfrak{q}]$.

The element $2-s$ has field norm 18. It belongs to $\mathfrak{p}$ and $\mathfrak{q}$, but not $\overline{\mathfrak{q}}$, since reduction there gives $s\equiv 1(\mathrm{mod}3)$. Its ideal therefore factors asThe exponents follow from the ideal norm 18, or from the root $s\equiv 2(\mathrm{mod}9)$ defining ${\mathfrak{q}}^2$. Hence $[\mathfrak{q}{]}^2=[\mathfrak{p}]$, so $[\mathfrak{q}]$ has order four and already generates all possible classes. ThereforeAs a separate check, the reduced forms of discriminant of a binary quadratic form $-56$ are $[1,0,14],[2,0,7],[3,2,5],[3,-2,5]$, giving exactly four classes.

LetIts minimal polynomial is $X^2-X+m$, and the algebraic norm givesSuppose that $N$ is composite and choose a prime divisor $q\le \sqrt{N}<m$. Since $X^2-X+m$ has the root $-n$ modulo $q$, the idealis a prime ideal of ideal norm $q$. The triviality of the ideal class group makes $\mathfrak{q}=(\beta )$ principal, so $|N_{L/\mathbb{Q}}(\beta )|=q$.

The ring of integers of a quadratic field consists of elementswhose norm is $(a^2+p{b}^2)/4$. If $b=0$, this norm is a square and cannot be the prime $q$. If $b\ne 0$, thenand integrality gives $q\ge m$, again a contradiction. Therefore the Euler prime-generating quadratic from class number one yields

Let $\epsilon =1+\sqrt{2}$ and $P=(3+\sqrt{2})$. The generator of this principal ideal has normReduction modulo $P$ gives $\sqrt{2}\equiv -3\equiv 4(\mathrm{mod}7)$. The map below is surjective and has $P$ in its kernel; both the kernel and $P$ have index seven, the latter by the ideal norm, so they are equal. Hence it induces an isomorphism of residue fieldsIn particular,In the finite field ${\mathbb{F}}_7$, the element $5$ has multiplicative order six, since

By part (b), every unit is $(-1{)}^{\delta }{\epsilon }^n$ for $\delta \in \{0,1\}$ and $n\in \mathbb{Z}$. Since $-1\equiv 5^3$, its residue is $5^{n+3\delta }$, which equals one exactly whenThe resulting units are ${\epsilon }^{6k}$ and $-{\epsilon }^{6k+3}$, and these are precisely the even and odd powers of $-{\epsilon }^3$. Thereforewhich is the group of units congruent to one modulo 3 plus square root two.

The element $\omega$ satisfiesModulo both $2$ and $3$, this polynomial factors as $X(X-1)$. Henceandwhere the respective ideal norms are $2,2,3,3$.

Now $N(\omega )=12$, and $\omega$ lies in ${\mathfrak{p}}_2$ and ${\mathfrak{p}}_3$ but not in their conjugate primes. Similarly $N(2+\omega )=18$, while $2+\omega$ lies in ${\mathfrak{p}}_2$ and ${\mathfrak{q}}_3$. Comparing norms gives

Evaluation at $\alpha$ gives the ring isomorphismand reducing modulo $p$ givesFor each $i$,This quotient is a field because ${\overline{g}}_i$ is irreducible. Hence $P_i$ is a nonzero prime ideal; it is nonzero because it contains $p$. Its ideal norm is

The ideals are distinct. Indeed, if $P_i=P_j$ with $i\ne j$, then both ${\overline{g}}_i$ and ${\overline{g}}_j$ would vanish in the same quotient. Since distinct monic irreducibles are coprime, a Bezout identity between them would make $1$ vanish there, a contradiction.

Because the chosen lifts satisfyfor some $H\in \mathbb{Z}[X]$, evaluation at $\alpha$ givesEvery generator of the product ideal ${\prod }_i{P}_i^{e_i}$ either contains a factor $p$ or is the displayed product, soOn the other hand, multiplicativity of ideal norms and ${\sum }_i{e}_i\mathrm{deg}{\overline{g}}_i=\mathrm{deg}f=[L:\mathbb{Q}]$ giveTwo finite-index ideals with one contained in the other and equal index are equal. Thus the Dedekind factorization theorem is proved in this monogenic case:

By unique factorization of ideals in a number field, each nonzero integral ideal has a unique expressionwith only finitely many nonzero exponents. Since the ideal norm is multiplicative, formally expanding one geometric series for each prime ideal givesUniqueness of ideal factorization says that every term occurs exactly once. This proves the Euler product for the Dedekind zeta function as a formal identity.

Because $p\equiv 11(\mathrm{mod}12)$, we have $p\equiv 3(\mathrm{mod}4)$ andFor $\theta =(1+\sqrt{-p})/2$, the minimal polynomial isSince $p\equiv 2(\mathrm{mod}3)$, reduction modulo $3$ gives $f(T)\equiv T(T-1)$. Its two roots are distinct, sowith distinct prime ideals of norm $3$. Thus $3$ splits completely, as recorded in splitting of three in Q of square root minus p.

The ideal class group ${\mathrm{Cl}}_K$ is the group of nonzero fractional ideals of ${\mathcal{O}}_K$ modulo the subgroup of principal fractional ideals. Saying that ${\mathfrak{p}}_1$ has order $n$ means that $n$ is the least positive integer for which ${\mathfrak{p}}_1^n$ is principal.

Suppose for contradiction that $n$ is odd and ${\mathfrak{p}}_1$ has order $n$. Then ${\mathfrak{p}}_1^n=(\alpha )$ for some $\alpha \in {\mathcal{O}}_K$, and taking norms givesWrite $\alpha =(a+b\sqrt{-p})/2$, where $a,b\in \mathbb{Z}$ have the same parity. ThenIf $b\ne 0$, the right-hand side is at least $p$, contradicting $p>3^{n+2}>4\cdot 3^n$. Hence $b=0$, so $\alpha$ is a rational algebraic integer and therefore an integer. But then $3^n=N(\alpha )={\alpha }^2$, impossible when $n$ is odd. This proves the odd-order obstruction for a split prime in an imaginary quadratic field and shows that the order cannot equal $n$.

The product formula says that every $x\in K^{\times }$ satisfiesFirst suppose that $x$ is an algebraic integer. Its principal ideal has the prime ideal factorizationTaking the ideal norm givesOn the other hand, the field norm is the product over embeddings, soEquating these expressions proves the formula for algebraic integers. Every nonzero element of $K$ is a quotient of two algebraic integers, and multiplicativity completes the proof.