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For a number field $K$, let ${\mathcal{O}}_K$ be its ring of integers of a number field, $r_1$ the number of real embeddings and $r_2$ the number of conjugate pairs of nonreal complex embeddings, so $[K:\mathbb{Q}]=r_1+2r_2$. The Dirichlet unit theorem sayswhere $\mu (K)$ is the finite group of roots of unity in $K$. Thus an imaginary quadratic number field has only roots of unity as units: usually $\{\pm 1\}$, with four for $d=-1$ and six for $d=-3$. A real quadratic number field has units $\pm {ϵ}^n$ for a fundamental unit $ϵ>1$ and integers $n$.

For $K=\mathbb{Q}(\sqrt{26})$, ${\mathcal{O}}_K=\mathbb{Z}[\sqrt{26}]$ and the field norm is $x^2-26y^2$. The element $ϵ=5+\sqrt{26}$ has norm $-1$ and is a unit. It is the smallest positive unit greater than one. Indeed for $1<u<ϵ$, its conjugate is $\pm 1/u$, so the coefficient $y=(u\mp 1/u)/(2\sqrt{26})$ is a positive integer less than two. Therefore $y=1$, and $x^2=26\pm 1$ forces $x=5$, giving $u=ϵ$, a contradiction. The Dirichlet unit theorem now gives $\boxed{{\mathcal{O}}_K^{\times }=\{\pm (5+\sqrt{26}{)}^n:n\in \mathbb{Z}\}}$. Its least positive norm-one generator is ${ϵ}^2=51+10\sqrt{26}$.

For the generalized Pell equation, write $\alpha =x+y\sqrt{26}$ with norm $N=\pm 10$. Change the sign if necessary and multiply by a power of $ϵ$ to place it in $\sqrt{10}\le \alpha <ϵ\sqrt{10}$. These operations preserve integrality and the absolute norm, though the latter operation can change the norm's sign. Since ${\alpha }^{\prime }=N/\alpha$, we have $x=(\alpha +N/\alpha )/2\ge 0$ and $y=(\alpha -N/\alpha )/(2\sqrt{26})\ge 0$. The upper bound gives $y<4$. Checking $y=0,1,2,3$ in $x^2=26y^2\pm 10$ leaves only $(x,y)=(4,1),(6,1)$. Both reduced elements lie in the chosen interval. Therefore all integral solutions of either sign areNegative powers are integral because ${ϵ}^{-1}=\sqrt{26}-5$. In the first family the norm is $-10(-1{)}^n$ and in the second it is $10(-1{)}^n$, so this also separates the two equations. The reduction proves exhaustion rather than merely generating examples.

The multiplicative field norm on this ring isIf an element is a unit, its positive integer norm has a multiplicative inverse in the positive integers, hence is one. The only integer solutions of $a^2+5b^2=1$ are $(a,b)=(1,0),(-1,0)$. Therefore the units are exactly $1$ and $-1$.

The standard failure of unique factorization isThere is no element of norm two or three: $b\ne 0$ gives norm at least five, while $b=0$ gives a square. An alleged nontrivial factorization of $2$ (norm four), $3$ (norm nine), or $1\pm \sqrt{-5}$ (norm six) would require a factor of norm two or three. Thus all four displayed factors are irreducible elements. No factor in the first factorization is associated to a factor in the second, since associates have the same norm and the norms are respectively four, nine and six. Hence $\mathbb{Z}[\sqrt{-5}]$ is not a unique factorization domain.

For a number field with $r$ real embeddings ${\sigma }_i$ and $s$ conjugate pairs of complex embeddings ${\tau }_j$, use the logarithmic embedding of number field unitsIt is a group homomorphism from multiplication to addition. A unit has field norm $\pm 1$, so the coordinates sum to zero. Deleting one coordinate identifies this hyperplane with ${\mathbb{R}}^{r+s-1}$ and gives the desired map.

If its image coordinates remain in a bounded set, the omitted coordinate is bounded too, and every conjugate of $u$ has bounded absolute value. The coefficients of its monic minimal polynomial are integers and elementary symmetric expressions in at most $[K:\mathbb{Q}]$ bounded conjugates. They are therefore bounded integers. Only finitely many such polynomials, and finitely many roots of them, exist. Thus the inverse image of every bounded set is finite. In particular the kernel is finite and the image is discrete. This argument includes the zero-dimensional target case; it does not yet assert the full-rank part of Dirichlet unit theorem.

For a square-free integer $\delta \equiv 2$ or $3(\mathrm{mod}4)$, the ring of integers of the quadratic field is ${\mathcal{O}}_{\mathbb{Q}(\sqrt{\delta })}=\mathbb{Z}[\sqrt{\delta }]$. Square-freeness is the usual convention underlying the first request: congruence alone would not suffice, for example $\delta =27$ gives the same field as $\delta =3$.

Now take the stipulated square-free $\delta \equiv 3(\mathrm{mod}4)$ and put $r=\sqrt{2}$, $s=\sqrt{\delta }$. For any integral $\alpha =a+br+cs+drs$, relative traces to the three quadratic subfields show $2a,2b,2c,2d\in \mathbb{Z}$. Those subfields all have integral bases $1$ and their square root, including $\mathbb{Q}(\sqrt{2\delta })$. Write $\alpha =(A+Br+Cs+Drs)/2$ with integer coefficients.

Its relative field norm to $\mathbb{Q}(s)$ isThus $AC$ is even, and the constant numerator is divisible by four. Since $\delta \equiv 3(\mathrm{mod}4)$, these conditions force $A,C$ both even: one odd makes the numerator odd, and both odd contradicts $AC$ even. They then force $B,D$ to have the same integer parity. Every algebraic integer consequently belongs to the lattice generated byConversely the last generator $w$ satisfies $w^2=(1+\delta )/2+\sqrt{\delta }$ and the monic integer equationAll four generators are integral and linearly independent over $\mathbb{Q}$, so this lattice is exactly ${\mathcal{O}}_L$, proving this integral basis of an odd biquadratic field containing square root of two without an unproved index assumption.

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.

The Dirichlet unit theorem states that for a number field with $r$ real embeddings and $s$ conjugate pairs of complex embeddings,where $\mu (K)$ is its finite group of roots of unity. Here $K$ is real quadratic, so $(r,s)=(2,0)$ and $\mu (K)=\{\pm 1\}$. Thus its unit group has rank one.

The elementis a unit because its field norm isIt is the smallest unit greater than one in the chosen real embedding. Indeed, write such a unit as $u=a+b\sqrt{2}$ and let $u^{\prime }=a-b\sqrt{2}=N(u)/u$ be its algebraic conjugate. Since $|u^{\prime }|<1<u$, both $a=(u+u^{\prime })/2$ and $b=(u-u^{\prime })/(2\sqrt{2})$ are positive integers. Hence $a,b\ge 1$ and $u\ge 1+\sqrt{2}$. The rank-one logarithmic lattice in Dirichlet's theorem is therefore generated by $\log \epsilon$. Consequently the units of Q of square root two are

The Minkowski convex body theorem states that if $\Lambda \subset {\mathbb{R}}^n$ is a full lattice and $C$ is a measurable, convex, centrally symmetric set withthen $C$ contains a nonzero point of $\Lambda$.

Here $d\not\equiv 3(\mathrm{mod}4)$, so the ring of integers of a quadratic field isUnder its complex embedding it has real-lattice covolume $\sqrt{d}$. An integral ideal $I$ has index $N(I)$ in ${\mathcal{O}}_K$, so its covolume is $\sqrt{d}N(I)$. Apply Minkowski's theorem to the disk $|z|\le R$, whose area is $\pi R^2$. Taking $R^2$ down to $4\sqrt{d}N(I)/\pi$ gives a nonzero $\alpha \in I$ withSince the field norm is $N_{K/\mathbb{Q}}(\alpha )=|\alpha {|}^2$ in this imaginary quadratic field, this proves the imaginary quadratic Minkowski disk bound

Because $(\alpha )\subseteq I$, the idealis integral, represents the inverse class of $I$, and hasThus every ideal class has a bounded-norm representative. Only finitely many integral ideals have bounded norm, proving the finiteness of the ideal class group.

Now take $K=\mathbb{Q}(\sqrt{-22})$. Its ring of integers is $\mathbb{Z}[\sqrt{-22}]$, its discriminant is $-88$, andEvery class therefore has an integral representative of norm at most five. The polynomial $X^2+22$ is irreducible modulo three and modulo five, so both primes are inert and there are no ideals of norm three or five. At two it reduces to $X^2$, and the splitting of rational primes in a quadratic field givesAn ideal of norm four is ${\mathfrak{p}}_2^2=(2)$ and is principal. The ideal ${\mathfrak{p}}_2$ is not principal, since a generator would have norm two, whereashas no integer solution. Consequently the Ideal class group of Q of square root of minus twenty-two is

Suppose finally that $y^3=x^2+22$. If $x$ were even, the right-hand side would be $2$ modulo four, which is not a cube modulo four. Hence $x$ and $y$ are odd. In ${\mathcal{O}}_K$,The two ideals on the left are coprime. Indeed, a common prime ideal would divide both $2x$ and $2\sqrt{-22}$. It cannot lie above two because their norms are odd. It would therefore lie above eleven and force $11\mid x$; but then $x^2+22$ has $11$-adic valuation exactly one, impossible for the cube $y^3$.

By unique factorization of ideals in a number field, coprimality impliesfor some ideal $\mathfrak{a}$. The class group has order two, so it has no element of order three; $[\mathfrak{a}{]}^3=1$ therefore forces $\mathfrak{a}$ to be principal. The only units are $\pm 1$, both of which can be absorbed into a cube, and hencefor integers $a,b$. Comparing the coefficient of $\sqrt{-22}$ givesThus $b=\pm 1$. For $b=1$ this requires $3a^2=23$, and for $b=-1$ it requires $3a^2=21$; neither is possible. Therefore the result on no integral solutions to y cubed equals x squared plus twenty-two is

Letbe a primitive polynomial. The Eisenstein criterion states that if a prime number $p$ satisfiesthen $F$ is an irreducible polynomial in $\mathbb{Z}[X]$, equivalently in $\mathbb{Q}[X]$ by Gauss lemma for polynomials.

To prove it, suppose that $F=GH$ with $G,H\in \mathbb{Z}[X]$ of positive degree. Primitivity and Gauss's lemma let us take such an integral factorisation if a factorisation over $\mathbb{Q}$ exists. Reducing modulo $p$ givesBecause $p$ does not divide the leading coefficient of $F$, neither factor loses degree on reduction. The polynomial ring ${\mathbb{F}}_p[X]$ is a unique factorization domain, so both reductions are monomials of positive degree. In particular, $p$ divides both constant terms $G(0)$ and $H(0)$. It follows that $p^2$ dividesa contradiction. This proves the criterion.

For a prime $p$, translate the geometric sum by one:Its leading coefficient is one, every other coefficient is divisible by $p$, and its constant coefficient is $p$, which is not divisible by $p^2$. It is therefore Eisenstein at $p$. Translation $X\mapsto X+1$ is an automorphism of $\mathbb{Z}[X]$, soThis is the geometric-sum irreducibility criterion.

The evaluation homomorphismhas image $\mathbb{Z}[\zeta ]$ and contains $(f)$ in its kernel. Since $f$ is monic, division by $f$ in $\mathbb{Z}[X]$ writes every $G$ as $G=Qf+R$ with $\mathrm{deg}R<\mathrm{deg}f$. If $G(\zeta )=0$, then $R(\zeta )=0$; the irreducibility of $f$ says that $f$ is the minimal polynomial of $\zeta$, so $R=0$. Thus $\ker ({\mathrm{ev}}_{\zeta })=(f)$, and the first isomorphism theorem for rings gives

Now take $p=3$. Then ${\zeta }^2+\zeta +1=0$, and the Eisenstein integers areComplex conjugation sends $\zeta$ to ${\zeta }^2$, so the field norm isGiven $z=x+y\zeta \in \mathbb{C}$, choose integers $m,n$ with $|x-m|,|y-n|\le \frac{1}{2}$. For $q=m+n\zeta$,For $\alpha ,\beta \in \mathbb{Z}[\zeta ]$ with $\beta \ne 0$, apply this to $z=\alpha /\beta$ and put $r=\alpha -q\beta$. ThenHence the norm is a Euclidean function, proving that $\mathbb{Z}[\zeta ]$ is a Euclidean domain. This is the Euclidean norm on the Eisenstein integers.

Finally suppose $A\in {\mathrm{GL}}_n(\mathbb{Z})$ satisfies $A^2+A+I=0$. Make the free abelian group $M={\mathbb{Z}}^n$ into a $\mathbb{Z}[\zeta ]$-module by definingThis is well-defined precisely because $A$ obeys the same polynomial relation as $\zeta$. The module is finitely generated. It is also torsion-free module: if $0\ne \alpha \in \mathbb{Z}[\zeta ]$ and $\alpha v=0$, multiplication by the conjugate of $\alpha$ gives $N(\alpha )v=0$, and the additive group ${\mathbb{Z}}^n$ has no nonzero integer torsion.

A Euclidean domain is a principal ideal domain, and the structure theorem for finitely generated modules over a principal ideal domain says that a finitely generated torsion-free module over one is free. Thusfor some $m$. Since $\mathbb{Z}[\zeta ]$ has free module basis $1,\zeta$ over $\mathbb{Z}$, comparison of abelian ranks gives $n=2m$. There can consequently be no such matrix when $n$ is odd.

If $n=2m$, choose a $\mathbb{Z}[\zeta ]$-basis $v_1,\dots ,v_m$. Thenis a $\mathbb{Z}$-basis, and $A^2{v}_i=-v_i-A{v}_i$. In this basis the matrix of $A$ is a direct sum of $m$ copies ofEvery admissible matrix is therefore conjugate in ${\mathrm{GL}}_n(\mathbb{Z})$ to $C^{\oplus m}$. Hence there is exactly one conjugacy class for even $n$, and none for odd $n$. This is the classification of integral matrices satisfying the third cyclotomic polynomial.

The main theorem of local class field theory gives a continuous Local Artin mapwith dense image, normalized by sending a uniformizer to a chosen Frobenius. For every finite abelian extension $L/K$, it induces the Local Artin reciprocity isomorphismThe existence theorem of local class field theory says that the finite-index open subgroups of $K^{\times }$ are exactly the norm subgroups $N_{L/K}(L^{\times })$ for finite abelian extensions $L/K$, and that the extension is uniquely determined inside $K^{\mathrm{ab}}$.

Write $e=e(L/K)$ and $f=f(L/K)$. The valuation of a field norm satisfiesso the valuation image of the norm subgroup is $f\mathbb{Z}$. The exact sequence obtained from $v_K:K^{\times }\to \mathbb{Z}$ therefore givesThe left side is $[L:K]=ef$ by Local Artin reciprocity. Cancelling $f$ proves

Minkowski's lemma states that if $\Lambda$ is a full lattice in ${\mathbb{R}}^n$ and $C\subset {\mathbb{R}}^n$ is measurable, convex, and centrally symmetric, thenimplies that $C$ contains a nonzero point of $\Lambda$.

Let ${\sigma }_1,{\sigma }_2:K\to \mathbb{R}$ be the two real embeddings. The Minkowski embedding of a real quadratic field identifies ${\mathcal{O}}_K$ with a latticeof covolume $\sqrt{|D_K|}$. Choose a constant $c$ satisfyingFor $R>0$, consider the closed diamondIt is convex and centrally symmetric, and its area isMinkowski's lemma supplies a nonzero ${\alpha }_R\in {\mathcal{O}}_K$ whose two embeddings lie in $C_R$.

For $(x,y)\in C_R$, the arithmetic-geometric mean inequality givesSince the coordinate product is the field norm,Finally, $|{\sigma }_2({\alpha }_R)|\le c/R\to 0$ as $R\to \infty$. No fixed nonzero algebraic integer can occur for arbitrarily large $R$, because its second embedding is nonzero. Consequently the elements ${\alpha }_R$ obtained along an unbounded sequence of $R$ contain infinitely many distinct values. Here $N(\alpha )$ denotes the absolute norm, as usual in this inequality.

For $K=\mathbb{Q}(\sqrt{10})$, the ring of integers of a quadratic field isThe elementhas field norm $9-10=-1$, so it is a unit. We claim that

Let $u$ be any unit. Its norm is $\pm 1$. After changing its sign and replacing it by its inverse if necessary, its first real embedding satisfies $u\ge 1$. Choose $n\in \mathbb{Z}$ so thatWrite $v=a+b\sqrt{10}$ with $a,b\in \mathbb{Z}$. Its conjugate is $v^{\prime }=\pm v^{-1}$, so $|v^{\prime }|\le 1$. ThereforeThus $|a|\le 3$ and $|b|\le 1$. The equation $a^2-10b^2=\pm 1$ now shows directly that the only value in $[1,\epsilon )$ is $v=1$: if $b=0$ then $a=\pm 1$, while if $|b|=1$ then $a=\pm 3$ and the positive possibilities are either below $1$ or equal to $\epsilon$. Hence $u={\epsilon }^n$ after normalization, proving the claimed description of the units of Q of square root ten.

Put $d=[L:K]$. If $\alpha =0$ the result is immediate, so suppose $\alpha =x^n\ne 0$ with $x\in L$. The field norm is multiplicative, sends an element of $K$ to its $d$th power, and commutes with powers. ThereforeChoose integers $r,s$ with $rd+sn=1$. ThenThe expression in parentheses belongs to $K$, so $\alpha$ is an $n$th power in $K$. This is coprime-degree descent for powers.

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.