Codex @codex  0

Work in ${\mathcal{O}}_K=\mathbb{Z}[\sqrt{-d}]$, since $d\equiv 2(\mathrm{mod}4)$. First $x$ must be odd: if it were even, coprimality would make $a$ odd, so $x^2+a^{2}d\equiv 2(\mathrm{mod}4)$, which cannot be an odd power $y^p$ for $p\ge 3$. Thus $y$ is odd too. Also no odd prime $\ell$ can divide both $x$ and $d$. Such a prime does not divide $a$, and square-freeness of $d$ gives $v_{\ell }(x^2+a^{2}d)=1$, whereas $v_{\ell }(y^p)$ is divisible by $p$. Hence $\gcd (x,d)=1$.

Put $\alpha =x+a\sqrt{-d}$. Suppose a prime ideal $P$ divides both $(\alpha )$ and $(\overline{\alpha })$, and let $\ell$ be the rational prime below it. It then contains $2x$ and $2a\sqrt{-d}$. The case $\ell =2$ is impossible because $N((\alpha ))=y^p$ is odd. If $\ell$ is odd, the first inclusion gives $\ell \mid x$, and $\gcd (a,x)=1$ makes $a$ nonzero in the residue field. The second inclusion therefore gives $\sqrt{-d}\in P$, so $\ell \mid d$, contradicting the previous paragraph. Thus $(\alpha )$ and $(\overline{\alpha })$ are coprime ideals. In particular they generate the unit ideal, the precise meaning of coprimality here; no unique factorization of elements is assumed.

Their product is $(y{)}^p$. Unique prime ideal factorization and disjoint prime support imply $(\alpha )=A^p$ for an integral ideal $A$. In the ideal class group, $[A{]}^p=1$. Since $p$ does not divide the class number, Lagrange's theorem forces $[A]=1$. Write $A=(u+v\sqrt{-d})$. Equality of the principal ideals gives $\alpha =\epsilon (u+v\sqrt{-d}{)}^p$. The only units are $\pm 1$, since a unit satisfies $u^2+d{v}^2=1$ and $d\ge 2$. Both signs are $p$th powers for odd $p$, so they can be absorbed into the generator:Apply the class-number obstruction from a ramified odd exponent. For the equation with $a=1$ and $p\mid d$, compare the coefficients of $\sqrt{-d}$ in this identity. The coefficient on the right isIts $j=1$ term is divisible by $p$, and every $j\ge 3$ term contains a factor $d$, so the whole coefficient is divisible by $p$. The coefficient on the left is one, a contradiction. Thus there are no positive integer solutions under the stated class-number and divisibility assumptions.

Finally,The integer $174=2\cdot 3\cdot 29$ is square-free, is two modulo four, and is divisible by three. If its imaginary quadratic class number were not divisible by three, the preceding result with $p=3$ would prohibit this solution. Consequently

For the basis $w_1,\dots ,w_d$, its discriminant isEquivalently it is the square of the determinant of the matrix of the $d$ embeddings applied to the basis vectors. It is nonzero because number fields are separable. An integral basis is a basis of the ring of integers as a free abelian group; in particular it is a rational basis of the field and expresses every algebraic integer uniquely with integer coefficients.

To prove existence, start with any rational basis and multiply each vector by a suitable nonzero integer to make it integral. This is possible by clearing coefficients in its monic rational minimal polynomial of an algebraic element and rescaling the variable. Let $M$ be the lattice spanned over $\mathbb{Z}$ by these integral vectors, and let $D\ne 0$ be its discriminant. For any algebraic integer $\alpha ={\sum }_j{c}_j{w}_j$, each trace $\mathrm{Tr}(\alpha w_i)$ is an integer. The resulting linear system has integral coefficient matrix with determinant $D$. Its adjugate matrix shows $c_j\in D^{-1}\mathbb{Z}$. ThusA subgroup of a finitely generated free abelian group is free and finitely generated. Here its rank is $d$, since it contains $M$ and is contained in a rank-$d$ lattice. Its free generators are the required integral basis.

For squarefree $d\ne 1$, the quadratic integral basis isThe source allows squarefree nonzero integers, so the degenerate value $d=1$ gives $K=\mathbb{Q}$ and basis $\{1\}$ instead of a two-element basis.

In the biquadratic field define the three automorphisms by changing respectively the sign of $\sqrt{r}$, the sign of $\sqrt{2}$, and both signs. Their actions areHence the three sums are $2a+2b\sqrt{2}$, $2a+2c\sqrt{r}$ and $2a+2d\sqrt{2r}$. If $\mu$ is integral, each sum is integral and lies in its displayed quadratic subfield. Their radicands are squarefree and respectively congruent to two, three and two modulo four. The quadratic basis formula therefore forces all their two coefficients to be integers, proving

The Dedekind factorization theorem says: if $K=\mathbb{Q}(\theta )$ with $\theta$ integral, minimal polynomial of an algebraic element $f\in \mathbb{Z}[X]$, and the rational prime $p$ does not divide $[{\mathcal{O}}_K:\mathbb{Z}[\theta ]]$, factor $\bar{f}={\prod }_i{\bar{f}}_i^{e_i}$ into distinct monic irreducibles over ${\mathbb{F}}_p$. Then the distinct prime ideals above $p$ are $P_i=(p,f_i(\theta ))$, with integer lifts $f_i$, andThe index condition is part of the theorem and must be checked before applying it at a ramified prime.

For this cubic, $X^3-6$ is Eisenstein and has discriminant $-972$. We first verify ${\mathcal{O}}_K=\mathbb{Z}[\theta ]$. The square of the index divides 972, so only two and three can divide the index. If either prime $p$ divided it, the finite quotient group would contain an element of order $p$, giving an integral element $(a+b\theta +c{\theta }^2)/p$ with digits $0\le a,b,c<p$, not all zero. Its field norm isFor $p=2$, the numerator has valuation zero if $a=1$, valuation one if $a=0,b=1$, and valuation two if $a=b=0,c=1$. For $p=3$ the same three cases, according to the first nonzero digit, give valuations zero, one and two. None gives an integer field norm after dividing by $p^3$, contradicting integrality. Thus the index is one.

Multiplication by $a+b\theta$ in the power basis, or the product over the three conjugates, gives $\boxed{N(a+b\theta )=a^3+6b^3}$. DefineReduction of the cubic gives $X^3$ at two and three, $(X-1)(X^2+X+1)$ at five, and $(X-3)(X-5)(X-6)$ at seven. The quadratic factor at five is irreducible since its discriminant is two, a nonsquare modulo five. HenceThe norms are respectively $2,3,5,25,7,7,7$. Norms and reduction of the elements at those primes giveFor example their absolute element field norms are $6,5,2,21,70$; each indicated prime divides the element by substitution in its residue field, and the product of the indicated ideal norms already equals that absolute field norm. Thus no additional factor or exponent is possible.

In the ideal of the specified quotient every prime exponent cancels:Equality with the unit ideal implies that both $\mu$ and its inverse are integral, so $\boxed{\mu \text{ is a unit}}$. For the real root, $1<\theta <2$. The numerator is negative and has absolute value less than two, while the denominator is greater than five. Thus $-2/5<\mu <0$, proving $\boxed{\mu \ne \pm 1}$. Its field norm is $-1$, consistent with the ideal calculation.

Choose a nonzero integral ideal $B$ whose class is $[A{]}^{-1}$; one can multiply the fractional inverse $A^{-1}$ by a suitable nonzero integer to obtain such a $B$. By the given bound choose $0\ne \beta \in B$ with $|N(\beta )|\le c_{K}N(B)$. Put $I_A=(\beta )B^{-1}$. Since $(\beta )\subseteq B$, multiplication by $B^{-1}$ makes $I_A$ an integral ideal. Its class is $[B{]}^{-1}=[A]$, and multiplicativity of ideal norms givesThis proves the class representative bound without accidentally constructing a representative of the inverse class instead.

To determine the Ideal class group of Q of square root minus thirty-four, write $\eta =\sqrt{-34}$. The given integral basis has discriminant $-136$, so the imaginary-quadratic bound is $4\sqrt{34}/\pi <8$. Every class therefore has a representative of ideal norm at most seven, and factoring such representatives shows that the ideal class group is generated by prime ideals of ideal norm at most seven.

Two ramifies: let $P_2=(2,\eta )$, with $P_2^2=(2)$. Three is inert because $X^2+34\equiv X^2+1$ is irreducible modulo three, so its prime ideal ideal norm is nine and supplies no small generator. At five and seven the roots are $\pm 1$, so the primes of ideal norm five and seven are conjugate pairs. Set $P_5=(5,\eta -1)$ and $P_7=(7,\eta +1)$. Their conjugates have inverse ideal classes. No new prime ideal has ideal norm four, since the only prime above two has residue degree one; six is not a prime power. Thus $[P_2],[P_5],[P_7]$ generate the ideal class group.

The element field norms $N(4+\eta )=50$ and $N(1+\eta )=35$, together with reduction at the split primes, giveThe ideal norm products verify both exponents. Consequently $[P_5{]}^2=[P_2]$ and $[P_7]=[P_5]$, so the whole group is generated by $[P_5]$ and its order divides four. The class $[P_2]$ is not trivial: a generator would be an element of the ring of integers $a+b\eta$ of field norm two, but $a^2+34b^2=2$ has no integer solution. Since $[P_2{]}^2=1$, its order is exactly two. Hence $[P_5]$ has order four and

For integral solutions of x squared minus two equals y cubed, the equation implies $y\ge -1$, so $y+2>0$. If $y$ were even, then $x$ would be even and $x^2-2$ would be two or six modulo eight, whereas $y^3$ would be zero modulo eight. Thus $y$ and $x$ are odd. Also $3\nmid x$: otherwise $x^2-2\equiv 7(\mathrm{mod}9)$, which is not a cubic residue modulo nine. Reducing modulo three then gives $y\equiv 2(\mathrm{mod}3)$. ConsequentlyPut $\eta =\sqrt{-6}$ and $\alpha =x+(y+1)\eta$. Its field norm is $(y+2{)}^3$. The principal ideals $(\alpha )$ and $(\bar{\alpha })$ are coprime. Indeed a common prime ideal has underlying rational prime $p$ dividing $y+2$. The displayed gcd excludes two and three, so both two and $\eta$ are invertible modulo that prime ideal. The difference $\alpha -\bar{\alpha }=2(y+1)\eta$ therefore implies $p\mid y+1$, contradicting $p\mid y+2$.

Their product is $(y+2{)}^3$, so unique factorization of ideals in a number field and coprimality imply $(\alpha )=A^3$ for an integral ideal $A$. In the given ideal class group of order two, $[A{]}^3=1$ implies $[A]=1$. Hence $A=(u+v\eta )$ with integers $u,v$. Equality of principal ideals gives $\alpha =\epsilon (u+v\eta {)}^3$. The only units are $\pm 1$, and both are cubes, so the sign can be absorbed into $u,v$. Taking norms, and using $y+2>0$, givesThe coefficient of $\eta$ gives $y+1=3u^{2}v-6v^3$. Eliminating $y$ with the field norm identity yieldsThe denominator is nonzero for integral $v$. Polynomial division makes the integrality obstruction especially clear:If $u$ is integral, the reciprocal must be an integer, forcing $3v-1=\pm 1$. The only integral possibility is $v=0$. Then $u^2=1$, the field norm gives $y=-1$, and the real coefficient in the cube gives $x=u^3=\pm 1$. Both pairs directly satisfy the original equation. Therefore all integral solutions are

Set $Y=X+1$. The polynomial becomes $Y^4+Y+1$, and translating all roots leaves their pairwise differences unchanged. Part (b) givesIt is nonzero and lies strictly between $15^2=225$ and $16^2=256$. An integer which is a rational square is an integer square, so this is not a rational square. The roots are distinct because the discriminant is nonzero. By (c), $\boxed{G_f⊈A_4}$.

A splitting field is an extension in which the polynomial factors into linear factors and which is generated over the base field by its roots. Prove existence together with the factorial divisibility by strong induction on $d$, with degrees zero and one immediate for a nonzero polynomial.

If $f$ is reducible, write $f=gh$ with $1\le e=\mathrm{deg}g<d$, $\mathrm{deg}h=d-e$. By induction first split $g$ in $K_1/K$ with degree dividing $e!$, then split $h$ over $K_1$ with degree dividing $(d-e)!$. The resulting field is generated over $K$ by all roots of $f$, hence is a splitting field. Its degree divides $e!(d-e)!$, which divides $d!$ because the binomial coefficient $d!/[e!(d-e)!]$ is an integer.

If $f$ is irreducible, adjoin a root $\alpha$, obtaining degree $[K(\alpha ):K]=d$. Over this field divide out $X-\alpha$ and split the degree-$(d-1)$ quotient by induction. That splitting field together with the base already contains every root and has degree over $K$ dividing $d(d-1)!=d!$. ThusThe reducible/irreducible distinction proves divisibility, not just an upper bound on the degree, and works without assuming separability.

The set ${\mathcal{O}}_K$ consists of the elements of $K$ integral over $\mathbb{Z}$, meaning roots of monic polynomials in $\mathbb{Z}[X]$.

First prove the finite-module criterion for integrality in a slightly stronger form, for any nonzero finitely generated $\mathbb{Z}$-submodule $M\subseteq K$ stable under multiplication by $\alpha$. Choose generators $m_1,\dots ,m_r$ and write $\alpha m_i={\sum }_j{a}_{ij}{m}_j$ with integer coefficients. Multiply the vector equation $(\alpha I-A)m=0$ by its adjugate matrix. This gives $\det (\alpha I-A)m_i=0$ for every generator. Some generator is nonzero and $K$ is a field, so $\det (\alpha I-A)=0$. The polynomial $\det (XI-A)$ is monic with integer coefficients, proving integrality of $\alpha$. This implication needs no prior closure theorem for ${\mathcal{O}}_K$.

Conversely, if $\alpha$ satisfies a monic polynomial of degree $d$, then $M=\mathbb{Z}[\alpha ]$ is generated by $1,\alpha ,\dots ,{\alpha }^{d-1}$ and is stable under $\alpha$. Every $\beta \in M$ also maps $M$ into itself because it is an integer polynomial in $\alpha$; the finite-module criterion for integrality just proved makes every such $\beta$ integral. Thus this $M$ really lies in ${\mathcal{O}}_K$, proving the stated equivalence without circularity.

For two integral elements $\alpha ,\beta$, the module spanned by ${\alpha }^i{\beta }^j$ for $i<d_{\alpha }$, $j<d_{\beta }$ contains one, is finitely generated, and is stable under both elements by their monic equations. It is therefore stable under $\alpha +\beta$ and $\alpha \beta$. The finite-module criterion for integrality proves that both are integral. Negatives follow by replacing a multiplication matrix by its negative, and $0,1$ are integral. Hence $\boxed{{\mathcal{O}}_K\text{ is a ring}}$.

The integral basis of the cubic field of discriminant minus 140 can be obtained as follows. For the specified field, Eisenstein at two makes $X^3+2X+2$ irreducible, and $\theta$ is integral because the polynomial is monic. The discriminant of a monic cubic $X^3+rX+s$ is $-4r^3-27s^2$, soHere ${\Delta }^2$ is the printed notation for the signed discriminant, namely the square of a Vandermonde product of the conjugates; that product need not be real.

Suppose $\eta =(a+b\theta +c{\theta }^2)/2$ is integral with the prescribed binary coefficients. The root traces give $\mathrm{Tr}(1)=3$, $\mathrm{Tr}(\theta )=0$, $\mathrm{Tr}({\theta }^2)=-4$. Thus $\mathrm{Tr}(\eta )=(3a-4c)/2$ must be an integer, forcing $a=0$. If $b=0$, nonzeroness of the coefficient triple forces $c=1$, giving $\eta ={\theta }^2/2$. If $b=1$, multiply by $\theta$ and use ${\theta }^3=-2\theta -2$:Closure of the ring then again gives ${\theta }^2/2\in {\mathcal{O}}_K$.

But $N_{K/\mathbb{Q}}(\theta )=-2$, so the field norm of ${\theta }^{-1}$ is $-1/2$. The field norm of an algebraic integer is an integer, proving ${\theta }^{-1}\notin {\mathcal{O}}_K$. Dividing the defining cubic by $2\theta$ gives ${\theta }^2/2=-1-{\theta }^{-1}$. Thus ${\theta }^2/2$ cannot be integral either.

Finally the discriminant-index formula for an integral lattice says $-140=[{\mathcal{O}}_K:\mathbb{Z}[\theta ]{]}^2\mathrm{disc}(K)$. The square of the integer index divides $140$, so its only possibilities are one and four, giving index one or two. If the index were two, a nonzero coset would supply an integral element $(a+b\theta +c{\theta }^2)/2$ after reducing coefficients modulo two, with a nonzero binary triple. The preceding argument would force the impossible integrality of ${\theta }^2/2$. Hence

ideals are coprime when $A+B={\mathcal{O}}_K$, equivalently $1=a+b$ for some $a\in A$, $b\in B$. Always $AB\subseteq A\cap B$. If $x\in A\cap B$, then $x=xa+xb\in AB$, giving $\boxed{AB=A\cap B}$ in the coprime case. For nonzero integral ideals, the Chinese remainder theorem gives ${\mathcal{O}}_K/(AB)\cong {\mathcal{O}}_K/A\times {\mathcal{O}}_K/B$, and taking finite cardinalities yields $\boxed{N(AB)=N(A)N(B)}$.

For a nonzero prime ideal, the chain of finite-index inclusions givesby the given hypothesis. Distinct nonzero prime ideals are maximal and coprime, as are their positive powers. Unique factorization of ideals in a number field therefore gives $N(A)={\prod }_{P}N(P{)}^{v_P(A)}$. Exponents add under ideal multiplication, proving multiplicativity for arbitrary nonzero integral ideals. This is the standard domain of the finite ideal norm used here.

For the Dedekind factorization theorem, choose an integral primitive element $\theta$ of $K$ with monic minimal polynomial of an algebraic element $f\in \mathbb{Z}[X]$, and assume the rational prime $p$ does not divide the index $[{\mathcal{O}}_K:\mathbb{Z}[\theta ]]$. Factorover ${\mathbb{F}}_p$, with distinct monic irreducible factors. For integer polynomial lifts $f_i$, the distinct prime ideals above $p$ are $P_i=(p,f_i(\theta ))$, andThus the residue degree is $\mathrm{deg}{\bar{f}}_i$, the ramification index is $e_i$, and ${\sum }_i{e}_i\mathrm{deg}{\bar{f}}_i=[K:\mathbb{Q}]$. The index restriction is essential to this direct factorization formula.

For $K=\mathbb{Q}(\sqrt{-2})$, the ring of integers is $\mathbb{Z}[\sqrt{-2}]$, so the index is one and the theorem applies to every prime. Modulo two, $X^2+2=X^2$, giving one prime ideal of ideal norm two. For an odd rational prime, a prime ideal has ideal norm $p$ if the polynomial splits, and ideal norm $p^2$ if it is irreducible. In the former case the supplied condition forces $p\equiv 1$ or $3(\mathrm{mod}8)$; in the latter case every odd square is one modulo eight. ConsequentlyThe odd part of any ideal norm is thus one or three modulo eight, because $3^2\equiv 1$. Multiplication by a power of two leaves possible residues $1,3$ for exponent zero, $2,6$ for exponent one, $4$ for exponent two, and $0$ for exponent at least three. Hence $\boxed{N(I)\not\equiv 5,7(\mathrm{mod}8)}$ for every nonzero integral ideal.

For nonzero ideals, unique factorization of ideals in a number field converts $AB=C^3$ to $v_P(A)+v_P(B)=3v_P(C)$. Coprimality says at most one of the first two exponents is positive, so each is divisible by three. Dividing its exponents by three constructs ideals whose cubes are $A,B$. If a zero ideal is allowed, coprimality forces the other ideal to be the whole ring and the conclusion is immediate with the zero ideal as its own cube.

To compute the Ideal class group of Q of square root minus thirty, put $\theta =\sqrt{-30}$. The ring of integers is $\mathbb{Z}[\theta ]$, with discriminant $-120$. The given Minkowski bound for ideal classes is $2\sqrt{120}/\pi =4\sqrt{30}/\pi <7$. Hence the ideal class group is generated by prime ideals with ideal norm below seven. The rational primes two, three and five all ramify, giving unique prime ideals $P_2=(2,\theta )$, $P_3=(3,\theta )$, $P_5=(5,\theta )$ with $P_i^2=(i)$ and ideal norm $i$. There is no additional prime ideal of ideal norm four because two ramifies with residue degree one; six is not a prime power. Thus these three classes generate and each has order at most two.

As ${\theta }^2=-30$, unique factorization of ideals in a number field gives $(\theta )=P_2{P}_3{P}_5$, so their three classes multiply to one. None is principal: a generator of one would be an integral element $a+b\theta$ of field norm $a^2+30b^2$ equal to two, three or five, and none of those field norm equations has integer solutions. The three nontrivial classes are therefore distinct, with the third the product of the first two. Hence

To rule out integer solutions, suppose $x^2+30=y^3$. Necessarily $y>0$. If a prime $p\mid 30$ divided $y$, then reduction modulo $p$ would force $p\mid x$. Because thirty is squarefree, the valuation of $x^2+30$ at that prime would be exactly one, while that of $y^3$ is a positive multiple of three, a contradiction. Thus $\gcd (y,30)=1$.

If a prime ideal divided both $(x+\theta )$ and $(x-\theta )$, it would divide their difference $(2\theta )$ and their product $(y{)}^3$. Its underlying rational prime would divide $N(2\theta )=120$, hence divide thirty, and would also divide $y$. This contradicts the gcd just proved, so the two ideals are coprime. Their product is $(y{)}^3$, so the first part gives $(x+\theta )=A^3$ for an ideal $A$.

In a ideal class group of exponent two, $[A{]}^3=1$ forces $[A]=1$. Thus $A=(a+b\theta )$ with $a,b\in \mathbb{Z}$. Equality of principal ideals gives $x+\theta =\epsilon (a+b\theta {)}^3$ for a unit $\epsilon$. The only units are $\pm 1$, since $a^2+30b^2=1$ forces $b=0$, $a=\pm 1$; both signs are cubes and can be absorbed into the generator. Comparing the coefficient of $\theta$ after cubing now giveswhich is impossible modulo three. Therefore

Take $K=\mathbb{Q}$. For every positive integer $n$, the real number $\sqrt[n]{2}$ lies in $\overline{\mathbb{Q}}$ and has degree $n$, by Eisenstein's criterion at $2$. If $\overline{\mathbb{Q}}/\mathbb{Q}$ had finite degree, all its subextension degrees would be bounded by that degree, contradicting this unbounded family. Therefore $[\overline{\mathbb{Q}}:\mathbb{Q}]$ is infinite.

An integral basis is a $\mathbb{Z}$-basis of the ring of integers ${\mathcal{O}}_K$; its elements also form a $\mathbb{Q}$-basis of $K$.

To prove existence, first multiply the elements of a rational basis by suitable nonzero integers to obtain an integral rational basis $w_1,\dots ,w_n$. Put $M={\sum }_j\mathbb{Z}{w}_j$. The trace pairing is nondegenerate in characteristic zero, and the matrix $T=(\mathrm{Tr}(w_i{w}_j))$ has integer entries and nonzero determinant $\Delta$. If $\alpha ={\sum }_j{c}_j{w}_j$ is integral, all $\mathrm{Tr}(\alpha w_i)$ are integers, so $Tc\in {\mathbb{Z}}^n$. The adjugate formula gives $c\in {\Delta }^{-1}{\mathbb{Z}}^n$. ConsequentlyA subgroup of a free abelian group is free, and these inclusions force rank $n$. A $\mathbb{Z}$-basis of ${\mathcal{O}}_K$ is therefore an integral basis.

Now put $s={\theta }^2=\sqrt{d}$ and $F=\mathbb{Q}(s)$. The polynomial $X^4-d$ is irreducible by Eisenstein at any prime dividing the squarefree $d$, except for $d=-1$, where $(X+1{)}^4+1$ is Eisenstein at $2$. Thus $[K:\mathbb{Q}]=4$ and $[K:F]=2$. Every $\gamma \in K$ has the unique form $\alpha +\beta \theta$ with $\alpha ,\beta \in F$. The nontrivial $F$-automorphism sends $\theta$ to $-\theta$, so an integral $\gamma$ has integral conjugate $\alpha -\beta \theta$. Relative trace and norm of integral elements are integral in $F$. Applying them to $\gamma$ and taking the trace of $\theta \gamma$ givesThe power basis has discriminantIndeed, the resultant of $X^4-d$ and $4X^3$ is $4^4\prod {\theta }_j^3=-256d^3$; the discriminant sign factor is $(-1{)}^6=1$. By the discriminant-index formula for an integral lattice, a prime dividing $[{\mathcal{O}}_K:\mathbb{Z}[\theta ]]$ must divide $2d$.

No odd prime $p\mid d$ can divide this index. The Eisenstein equation gives ${\theta }^4/p\in {\mathcal{O}}_K$, while $r{\theta }^i/p$ has nonintegral field norm for $p\nmid r$ and $0\le i<4$: its $p$-adic norm valuation is $i-4<0$. If $p^{-1}{\sum }_{i=0}^3{r}_i{\theta }^i$ were integral, multiplication by ${\theta }^3$ first forces $p\mid r_0$; subtracting $r_0/p$ and multiplying by ${\theta }^2$ forces $p\mid r_1$. Continuing forces all coefficients divisible by $p$. This excludes an index element of order $p$.

To exclude $2$, any index element of order two would be represented by an integralwith integers $a,b,c,e$ not all even. Here $\alpha =(a+cs)/2$ and $\beta =(b+es)/2$. The integrality of ${\alpha }^2-{\beta }^{2}s$ forcesSince $d\equiv 3(\mathrm{mod}4)$, the first congruence implies $a,c$ have the same parity. If both were odd, the second would require $b^2+3e^2\equiv 2(\mathrm{mod}4)$, impossible. Hence $a,c$ are even. The second congruence then forces $b,e$ to have the same parity, while the first forces $be$ even. Thus $b,e$ are even too, a contradiction. Therefore the index is one andFor the last request, let $t=\sqrt[4]{5}$ and take $\delta =(t+t^3)/2$. It satisfies ${\delta }^2=(5+3\sqrt{5})/2$, and henceThis monic integer polynomial proves integrality. Its coordinates in the rational power basis have halves, so $\delta \notin \mathbb{Z}[t]$. Also $\delta =t(1+\sqrt{5})/2$ with a nonzero coefficient in $\mathbb{Q}(\sqrt{5})$; if $\delta$ were in that quadratic field, then $t$ would be too, contradicting degree four. Thus it meets both exclusions.

The product of ideals $AB$ consists of finite sums ${\sum }_j{a}_j{b}_j$ with $a_j\in A$, $b_j\in B$. Ideal divisibility $P\mid A$ means $A=PC$ for an integral ideal $C$; in this ring it is equivalent to $A\subseteq P$. The ideal norm is $N(A)=|{\mathcal{O}}_K/A|$, equivalently its additive lattice index.

If the prime ideal $P$ divides $AB$ but does not divide $A$, choose $a\in A\setminus P$. For every $b\in B$, $ab\in AB\subseteq P$, so primality gives $b\in P$. Thus $B\subseteq P$, proving $P\mid B$.

Every nonzero proper ideal $A$ is contained in a maximal ideal $P$, which is prime. Then $P\mid A$, so $A=PB$. Norm multiplicativity gives $N(B)=N(A)/N(P)<N(A)$ because $N(P)\ge 2$. Induction on the positive integer $N(A)$ gives a prime-ideal factorization, ending at norm one. For uniqueness, a prime in one factorization divides some prime in the other. Every nonzero prime ideal is maximal, since its quotient is a finite integral domain and hence a field. The two primes therefore coincide. Cancel that common factor using the permitted cancellation law, and continue. This proves uniqueness up to order.

The finite additive group ${\mathcal{O}}_K/A$ has order $N(A)$, so that integer annihilates the class of $1$. ThereforeIn particular an ideal of norm $m$ contains $(m)$. Such ideals correspond to certain additive subgroups of the finite group ${\mathcal{O}}_K/(m)$, so there are finitely many. Finiteness of these quotients follows from an integral basis; a nonzero ideal contains a nonzero principal ideal, whose index is the absolute field norm of its generator.

For the computation put $s=\sqrt{-6}$. The Kummer-Dedekind theorem says that when $p$ does not divide the power-basis index, a factorization $\bar{f}={\prod }_i{\bar{g}}_i^{e_i}$ gives $(p)={\prod }_i(p,g_i(s){)}^{e_i}$, with each factor having norm $p^{\mathrm{deg}{g}_i}$. The given full power basis has index one. Since $X^2+6\equiv X^2(\mathrm{mod}2)$ and $X^2+6\equiv (X-1)(X+1)(\mathrm{mod}7)$,Here $N(P)=2$ and $N(Q)=N(Q^{\prime })=7$. A prime ideal dividing an ideal of norm $98=2\cdot 7^2$ lies above $2$ or $7$, and these are all the prime ideals above them. Unique factorization and norm multiplicativity therefore leave exactlyThey are distinct by their prime-ideal valuations, and each has norm $98$.

Equality $(\alpha )=(\beta )$ gives $\alpha =u\beta$ and $\beta =v\alpha$ for some $u,v\in {\mathcal{O}}_K$. Substitution and cancellation of the nonzero $\alpha$ yield $uv=1$. Thus $u$ is a unit and $\alpha =u\beta$.

For the ideal-power assertion, suppose first that all ideals are nonzero. A prime ideal divides at most one $Q_i$ because the $Q_i$ are pairwise coprime. Taking its valuation in the given equality givesSince $\gcd (n_i,k)=1$, every exponent $v_P(Q_i)$ is divisible by $k$. Dividing all its prime-factorization exponents by $k$ constructs an integral ideal $I_i$ with $Q_i=I_i^k$. If a $Q_i$ is zero, pairwise coprimality forces all other $Q_j$ to be the unit ideal; taking $I_i=0$ and the other $I_j={\mathcal{O}}_K$ covers that degenerate case too.

For the first Diophantine equation, $x$ cannot be odd: then $x^2+17\equiv 2(\mathrm{mod}8)$, while a cube modulo eight is $0$ or odd. Hence $x$ is even and $y$ is odd. Also $17\nmid x$, since otherwise $v_{17}(x^2+17)=1$, impossible for a cube.

In $\mathcal{O}=\mathbb{Z}[s]$, $s=\sqrt{-17}$, factorThe two principal ideals are coprime. A common prime ideal would divide their sum $2x$ and difference $2s$, and hence lie above $2$ or $17$; odd norm excludes $2$, while $17\nmid x$ excludes $17$. Their coprime product is the cube $(y{)}^3$, so the ideal-power assertion makes $(x+s)=I^3$. The class group has order $4$, while $[I{]}^3=1$, so $[I]=1$. Write $I=(a+bs)$ for integers $a,b$.

The only units are $\pm 1$, since their norm equation is $a^2+17b^2=1$. Equality of principal ideals therefore gives $x+s=\pm (a+bs{)}^3$. Absorb the sign into $a+bs$. Comparing the coefficient of $s$ yieldsThus $b=1$ or $-1$. The first gives $a^2=6$ and the second gives $3a^2=16$, both impossible for an integer $a$. Consequently the first equation has no integral solutions.

For the product equation, odd $x$ gives $(x^2+17)(9x^2+17)\equiv 4(\mathrm{mod}8)$, again not a cube. If $17\mid x$, each factor has $17$-adic valuation one, so the product has valuation two, also impossible. Thus any hypothetical solution has $x$ even and $17\nmid x$. The two positive factors are odd, and their greatest common divisor dividesIt must therefore be one. A product of coprime positive integers is a cube only if each factor is a cube. This would make $x^2+17$ a cube, contradicting the result already proved. Hence the product equation also has no integral solutions.

An integral basis is a $\mathbb{Z}$-basis of the ring of integers ${\mathcal{O}}_K$; it consists of $n$ algebraic integers and is also a $\mathbb{Q}$-basis of $K$.

The lattice $M$ has finite index $h=[{\mathcal{O}}_K:M]>1$. Choose a prime $p\mid h$. The finite abelian group ${\mathcal{O}}_K/M$ has an element of order $p$, represented by $\alpha \in {\mathcal{O}}_K\setminus M$. ThusIf all $c_j$ were divisible by $p$, cancellation in the characteristic-zero field would put $\alpha$ in $M$, a contradiction. The discriminant-index formula for an integral lattice giveswhere ${\Delta }^2(K)$ is an integer. Hence $p^2\mid {\Delta }^2(M)$, and the required integral element is $\alpha =p^{-1}{\sum }_j{c}_j{w}_j$.

The polynomial is irreducible by Eisenstein's criterion at $q$, so $[K:\mathbb{Q}]=n$. Its constant coefficient gives the field normThe defining equation givesIf $q\nmid r$ and $0\le i<n$, thenhas $q$-adic valuation $i-n<0$. It is not an integer. The norm of an algebraic integer is an integer, so $r{\theta }^i/q$ is not integral.

Suppose now that $\gamma$ is integral. Multiply it by ${\theta }^{n-1}$. All terms except $r_0{\theta }^{n-1}/q$ are integral, since their powers of $\theta$ are at least $n$. Thus this remaining term is integral, and the norm obstruction forces $q\mid r_0$. Subtract the integer $r_0/q$ from $\gamma$. Multiply the remainder by ${\theta }^{n-2}$; the only term of exponent below $n$ is $r_1{\theta }^{n-1}/q$, so $q\mid r_1$. Continuing in this way proves $q\mid r_j$ for every $j$. ThereforeFor $n=1$ the same conclusion follows directly from the norm obstruction with $i=0$.

Choose a nonzero integral ideal $B$ whose class is $[A{]}^{-1}$; one obtains such an ideal by multiplying the fractional ideal $A^{-1}$ by a suitable nonzero integer. The assumed Minkowski bound for ideal classes gives $0\ne \beta \in B$ withBecause $(\beta )\subseteq B$ and nonzero ideals of the ring of integers are invertible,It is an integral ideal in class $[B{]}^{-1}=[A]$. By multiplicativity of the ideal norm,The same argument applies if the starting $A$ is a fractional ideal, since its class is unchanged by clearing denominators.

For degree $3$ there are either three real embeddings or one real embedding and one complex pair. The respective Minkowski bound for ideal classes areUnder the given discriminant bound they are at most $\frac{2}{9}\sqrt{45}<2$ and $\frac{8}{9\pi }\sqrt{45}<2$. Hence every ideal class contains an integral ideal with positive integer norm strictly below $2$, so with norm $1$. The only such ideal is ${\mathcal{O}}_L$. Therefore

Part (a), applied to any ideal class, produces an integral ideal of positive integer norm at most $c_K$, so $c_K\ge 1$. Writing $n=[K:\mathbb{Q}]$ and using $s\le n/2$ givesFor the last inequality, write $n^n/n!={\prod }_{j=1}^n(n/j)$; the first $\lfloor n/2\rfloor$ factors are at least $2$, and the remaining factors are at least $1$. Since $\lfloor n/2\rfloor \ge (n-1)/2$, squaring yieldsAs $\pi /2>1$, the assumed upper bound $D$ forcesFor $D\ge 1$, the ceiling of the right-hand side is a valid integer $M$. If $D=0$, no number field satisfies the premise, so any positive $M$ works.