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In a real quadratic number field, a fundamental unit $ϵ>1$ generates the infinite cyclic factor of its unit group, so every unit is $\pm {ϵ}^n$. Existence follows from the Dirichlet unit theorem; choosing the positive generator of the rank-one logarithmic lattice gives this normalization.

Classical Kummer theory relates extraction of $n$th roots to Galois cohomology. In characteristic zero the exact sequenceand Hilbert theorem 90 identify $H^1(K,{\mu }_n)$ with $K^{\times }/K^{\times n}$. For an elliptic curve, the corresponding Kummer exact sequence of an elliptic curve isMultiplication by $n$ is surjective over the algebraic closure. If $nQ=P\in E(K)$, the cocycle $\sigma \mapsto \sigma Q-Q$ takes values in $E[n]$. Changing the choice of $Q$ changes it by a coboundary, and changing $P$ by an element of $nE(K)$ does not change its class. Conversely, a trivial cocycle class allows $Q$ to be adjusted by an $n$-torsion point to become $K$-rational. Thus the Kummer map of an elliptic curve is an injectionThis is the elliptic form of Kummer theory. The full cohomology group need not be finite; the arithmetic restriction on these classes is essential.

Choose a finite set $S$ of places containing the archimedean places, the primes over $n$, and all primes of bad reduction of an elliptic curve. At a finite place outside $S$, $E$ has good reduction and $n$ is a unit. A division point of the reduction of $P$ exists over the algebraic closure of the residue field. Smooth lifting gives a point over a finite unramified extension whose multiple differs from $P$ by an element of the kernel of reduction of an elliptic curve. In the formal group of an elliptic curve, $[n](T)=nT+O(T^2)$ is an isomorphism by the invertible morphism criterion for formal group laws, and its integral inverse converges on the maximal ideal. Correcting that difference produces a division point in the maximal unramified extension. Hence the Kummer map of an elliptic curve class is unramified outside $S$.

To prove finiteness explicitly, choose a finite Galois extension $L/K$ containing all $E[n]$ and all $n$th roots of unity, and enlarge $S$ to include its ramified places. A basis of $E[n]\cong (\mathbb{Z}/n\mathbb{Z}{)}^2$ identifies it over $L$ with ${\mu }_n^2$. Classical Kummer theory then identifiesThe restriction of every class in the image of $\delta$ belongs to $L(S_L,n{)}^2$, where the S-unramified power class group isIndeed an unramified local Kummer extension at residue characteristic prime to $n$ has valuation divisible by $n$: in an unramified field containing a root, $nv(a^{1/n})=v(a)$ with integral valuations.

The finiteness of S-unramified Kummer classes follows from the exact sequenceTo see the final map, write the ideal of $a$ away from $S_L$ as ${\mathfrak{a}}^n$ and take the ideal class of $\mathfrak{a}$. Its kernel is represented by an S-unit, after division by an $n$th power; conversely an $n$-torsion ideal class yields such an $a$. The S-unit group is finitely generated by the Dirichlet unit theorem together with the finitely many inverted primes. The ideal class group of the localized ring is a quotient of the finite ordinary ideal class group. Both outer groups are therefore finite.

Finally restriction has finite kernel: inflation-restriction puts it in the finite group $H^1(\mathrm{Gal}(L/K),E[n])$. Thus the image of $\delta$ has finite restriction image and finite kernel, andThis proves the weak Mordell-Weil theorem. Combining it with the height descent lemma proves the full Mordell-Weil theorem. Local restrictions at every place refine the finite group used here to the Selmer group of an elliptic curve, which is useful for explicit descent calculations.

The idele group is the multiplicative restricted productwith respect to ${\mathcal{O}}_v^{\times }$ at the finite places of a number field; $K_v$ is the completion of a valued field at the place $v$. Thus each tuple has nonzero components, and all but finitely many finite components are units. Its restricted product topology on the idele group has basic open sets ${\prod }_{v\in S}{W}_v\times {\prod }_{v\notin S}{\mathcal{O}}_v^{\times }$, where $S$ is finite and contains the infinite places, and each $W_v$ is open in $K_v^{\times }$. In particular $U_K$ is an open subgroup.

Embed $K^{\times }$ diagonally. Take a neighbourhood of one whose finite components all lie in ${\mathcal{O}}_v^{\times }$ and whose infinite components satisfy $|x_v-1|<1/2$ in the usual real or complex modulus. A diagonal element there is an algebraic unit $a$. If $a\ne 1$, then $a-1$ is a nonzero algebraic integer, so its field norm is a nonzero integer. Buta contradiction. Thus $K^{\times }$ is discrete. It is also closed: in a topological group, a subgroup with an isolated identity cannot have an external accumulation point, since quotients of two nearby subgroup elements would approach the identity.

Send an idele to its associated fractional ideal byOnly finitely many exponents are nonzero. This homomorphism is onto, by choosing powers of local uniformizers, and its kernel is $U_K$. Diagonal elements map to principal fractional ideals. The resulting quotient givesIt is a topological isomorphism when the ideal class group is given the discrete topology, since $U_K$ is open.

Use normalized local moduli: real modulus, squared complex modulus, and $|{\pi }_v{|}_v=(N{\mathfrak{p}}_v{)}^{-1}$ at a finite place. The idelic modulus $\Vert x\Vert ={\prod }_v|x_v{|}_v$ defines $J_K^1=\ker \Vert \cdot \Vert$, the norm-one idele group. The product formula puts $K^{\times }$ inside this kernel. Every ideal class has a representative in $J_K^1$, because an infinite component can be rescaled to correct the modulus without altering its fractional ideal. The compact space $J_K^1/K^{\times }$ therefore maps continuously onto the discrete ideal class group. Its image must be finite, proving $\mathrm{Cl}(K)$ is finite.

For the Dirichlet unit theorem, put $U^1=U_K\cap J_K^1$ and $E=K^{\times }\cap U^1={\mathcal{O}}_K^{\times }$. Let $r_1$ count real embeddings and $r_2$ count conjugate complex pairs. Infinite logarithms define a continuous surjectionusing $\log |x_v|$ at real places and $2\log |x_v|$ at complex places. Its kernel is compact: it consists of real signs, complex unit circles, and the product of compact finite-place unit groups. More generally the inverse image of a bounded closed subset of $H$ is compact. Since $E$ is closed and discrete, its intersection with each such inverse image is finite. Hence $\Lambda =\ell (E)$ is discrete in $H$, and the kernel $E\cap \ker \ell$ is a finite group. It is exactly the roots of unity $\mu (K)$, since every element of a finite multiplicative group has finite order and every root of unity has all local moduli one.

The image of $U^1$ in $J_K^1/K^{\times }$ is an open subgroup, hence also closed, and is homeomorphic to $U^1/E$. The assumed compactness therefore makes $U^1/E$ compact, and its continuous quotient $H/\Lambda$ is compact. A discrete cocompact subgroup of a real vector space is a full Euclidean lattice, of rank $\dim H=r_1+r_2-1$. Thus $E/\mu (K)\cong {\mathbb{Z}}^{r_1+r_2-1}$, and lifting a lattice basis splits off the free factor:This derives both finiteness and the unit rank from the stated compactness assumption, rather than assuming either conclusion to prove compactness.

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.

Galois cohomology records the obstruction to choosing Galois-invariant division points. Let $G_K=\mathrm{Gal}(\overline{K}/K)$ be the absolute Galois group and give its algebraic-point modules the discrete topology. A continuous $1$-cocycle with values in a Galois module $M$ is a function $c:G_K\to M$ satisfyingA group coboundary has the form $c_{\sigma }=\sigma a-a$. Quotienting cocycles by coboundaries defines $H^1(K,M)$. For the finite module $E[m]$ these cocycles have finite image and factor through finite data; continuity is essential because $G_K$ is a profinite group.

In characteristic zero, multiplication by $m\ge 2$ on the algebraic points of an elliptic curve is surjective, with kernel $E[m]\cong (\mathbb{Z}/m\mathbb{Z}{)}^2$. The short exact sequence of Galois modulesgives the Kummer exact sequence of an elliptic curveExplicitly, choose $Q$ with $mQ=P$ and set $\delta (P{)}_{\sigma }=\sigma Q-Q$. Changing $Q$ by an $m$-torsion point changes the cocycle by a coboundary. The class vanishes exactly when a suitable choice of $Q$ is Galois-fixed, that is, when $P\in mE(K)$. This proves the injection directly and identifies its arithmetic meaning.

The entire group $H^1(K,E[m])$ need not be finite. For example, Kummer theory gives $H^1(K,{\mu }_m)=K^{\ast }/K^{\ast m}$, which has classes supported on arbitrarily many different primes. The finite part needed for the Weak Mordell-Weil theorem comes from a ramification restriction on the image of $\delta$.

Choose a finite set $S$ of places containing the archimedean places, the primes dividing $m$ and all primes of bad reduction of an elliptic curve. At a finite place $v\notin S$, the curve has good reduction and $m$ is invertible in its valuation ring. The elliptic curve extends to a smooth proper group scheme, and $[m]$ on that model is finite etale. A point $P\in E(K_v)$ extends to an integral section by properness. Its division-point fibre is consequently finite etale over the valuation ring. Over the maximal unramified extension it has a point, so the Kummer cocycle restricts trivially to the inertia group. The module $E[m]$ itself is unramified there for the same reason. Thuswhere $H_S^1$ denotes classes unramified outside $S$.

Here is a proof that this restricted group is finite. Take a finite Galois extension $L/K$ containing all coordinates of $E[m]$ and the $m$th roots of unity, and enlarge $S$ by its ramified primes. Over $L$ the module is trivial and, after choosing a basis, is isomorphic to ${\mu }_m^2$. Hilbert theorem 90 and the multiplicative Kummer sequence therefore giveFor unramified Kummer classes with bounded prime support, a class unramified outside $S_L$ has valuations divisible by $m$ at every prime outside $S_L$: the valuation of an $m$th root in an unramified extension is integral. Consequently both coordinates lie in

To show this set finite, write the outside-$S_L$ divisor of $a$ as $mD$. The ideal class of $D$ belongs to the $m$-torsion of the ideal class group of the ring of $S_L$-integers. This gives the exact sequenceThe last map is onto: if $mD$ is principal in the $S_L$-ideal group, a generator represents a class with outside valuations divisible by $m$. The kernel consists exactly of S-units modulo $m$th powers. The Dirichlet unit theorem, with the finitely many inverted primes adjoined, makes the S-unit group finitely generated; the ideal class group is finite, and localization only quotients it. Both ends of the sequence are therefore finite.

Finally, the inflation-restriction exact sequence bounds the kernel of restriction from $H^1(K,E[m])$ to $H^1(L,E[m])$ by the finite group $H^1(\mathrm{Gal}(L/K),E[m])$. The unramified subgroup has finite image, contained in $L(S_L,m{)}^2$, and finite kernel, so it is finite. The Kummer injection now provesThis is the Weak Mordell-Weil theorem, proved without first assuming finite generation of $E(K)$.

For computation one refines the unramified group by local solvability. The Selmer group of an elliptic curve consists of classes whose restriction at every completion lies in the corresponding local Kummer image. It is finite and sits inwhere the Tate–Shafarevich group $\mathrm{Sha}$ measures classes in $H^1(K,E)$ that become trivial at every completion. Thus locally soluble descent equations can give an upper bound without every class coming from a rational point. No finiteness assumption on the whole Tate-Shafarevich group is needed for the weak theorem. Combining the finite quotient with the height descent lemma from the height essay yields the full Mordell-Weil theorem. Replacing multiplication by $m$ with a smaller isogeny gives the same cohomological framework for two-isogeny descent.

Galois cohomology and the finite division quotient. Let $K$ be a number field, $G_K=\mathrm{Gal}(\overline{K}/K)$ with its profinite topology, and $M$ a discrete continuous $G_K$-module. Its Galois cohomology is continuous group cohomology. In degrees zero and one,Thus a crossed homomorphism measures the failure of a choice to be Galois-invariant, and changing the choice changes the cocycle by a coboundary. A short exact sequence of such modules gives a long exact sequence in group cohomology. The connecting map sends an invariant element to the cocycle obtained by lifting it and comparing its Galois translates.

For an elliptic curve $E/K$ and $n\ge 2$, multiplication by $n$ is surjective on $E(\overline{K})$ and has kernel $E[n]$. The Kummer exact sequence of an elliptic curve therefore givesExplicitly, choose $nQ=P$ and set $\kappa (P)(\sigma )=\sigma Q-Q$. Replacing $Q$ changes this by a coboundary; replacing $P$ by $P+nR$, $R\in E(K)$, does not change it. If its class is zero, some torsion translate of $Q$ is Galois-invariant, so $P\in nE(K)$. This proves injectivity of the Kummer map of an elliptic curve.

The whole group $H^1(K,E[n])$ need not be finite. The crucial restriction is ramification. Let $S$ contain the finite primes of bad reduction of an elliptic curve and those dividing $n$. At $v\notin S$, good reduction and prime-to-residue-characteristic multiplication imply that every division point of a local point is in an unramified extension, by the formal-group lifting argument. Thus every global Kummer class restricts trivially to inertia there.

Choose a finite Galois extension $L/K$ containing $E[n]$ and the $n$th roots of unity, and enlarge $S$ to include its ramified primes. Over $L$, the torsion module is constant and can be identified with ${\mu }_n^2$. By Kummer theory,For classes unramified outside the primes $S_L$ above $S$, valuations outside $S_L$ must be divisible by $n$. Their restriction images therefore lie in $L(S_L,n{)}^2$, the square of an S-unramified power class group. Its finiteness can be seen without assuming the Mordell-Weil theorem: the finiteness of S-unramified Kummer classes follows fromThe S-unit group is finitely generated by the Dirichlet unit theorem and its valuations at $S_L$. The localized ideal class group is a quotient of the ordinary ideal class group of $L$, which is finite by the finiteness of the ideal class group. The Minkowski bound for ideal classes supplies an integral ideal of bounded norm in each class, and only finitely many integral ideals have bounded norm.

The kernel of restriction from $H^1(K,E[n])$ to $H^1(L,E[n])$ is also finite: a cocycle trivial after restriction can, after subtracting a coboundary, be made to factor through the finite group $\mathrm{Gal}(L/K)$. There are only finitely many maps from that group to the finite module $E[n]$. This is the finite-extension kernel of a Kummer map mechanism. Consequently the possible Kummer classes form a finite set, provingThis is the Weak Mordell-Weil theorem. Imposing membership in the local Kummer image at every place refines the finite unramified collection to the n-Selmer group. Its relation with the obstruction to a globally rational point iswhere the last group is the $n$-torsion of the Tate–Shafarevich group. Thus local solubility gives a finite computable upper bound but can leave a genuine global obstruction.

Heights and finite generation. For the other essay, normalize the absolute values on a field $K$ so that the product formula holds. The Absolute logarithmic Weil height of $[x_0:x_1]\in {\mathbb{P}}^1(K)$ isAt the infinite places the local degrees are one or two and the usual real or complex modulus is used. The product formula makes this independent of scaling; it is also unchanged by extending $K$. Over $\mathbb{Q}$, using coprime integral coordinates gives $h([u:v])=\log \max (|u|,|v|)$, the naive height on the projective line. The Northcott theorem says that points of bounded height and bounded field degree form a finite set; in particular this holds over the fixed field $K$. One way to see the finiteness is the height-Mahler measure formula: for an algebraic number of degree $d$, the coefficients of its primitive minimal integer polynomial are bounded by binomial factors times $\exp (dh)$. Bounds on $d$ and $h$ therefore leave only finitely many integer polynomials and hence only finitely many algebraic numbers.

Set $h_E(P)=h(x(P))$ for $P\ne O$, and $h_E(O)=0$. Since $x:E\to {\mathbb{P}}^1$ has degree two, bounded $h_E$ gives only finitely many points of $E(K)$. The height growth under a morphism of the projective line states that a degree-$d$ morphism gives $h(f(t))=dh(t)+O(1)$. The upper bound follows from its homogeneous coordinate polynomials; the lower bound follows from their having no common zero, using a resultant identity at each place. Applied to the degree-four duplication map, it gives a uniform constant $C$ withConsequently $4^{-j}{h}_E(2^{j}P)$ is Cauchy: the absolute difference of consecutive terms is at most $C{4}^{-(j+1)}$. Define the canonical height of an elliptic curve byIt is nonnegative and satisfies $\hat{h}(2P)=4\hat{h}(P)$. The factor one-half is the conventional normalization for the divisor $(O)$, since the first-coordinate height belongs to $2(O)$.

The addition-divisor identity also gives the bounded-error height identityuniformly in $P,Q$. This is the height form of the identity for the sum and difference pullbacks of the line bundle associated to $2(O)$; it includes points where the affine addition formulas have a zero denominator. Apply it to $2^{j}P,2^{j}Q$, divide by $2\cdot 4^j$, and take the limit. Thus $\hat{h}$ satisfies the parallelogram law, and its polarization is a bilinear form, the height pairing. Nonnegativity implies the Cauchy-Schwarz inequality for that pairing, by evaluating its nonnegative quadratic polynomial on $mP+nQ$ for all integers $m,n$ and approximating real ratios by rational numbers. In particularThese properties do not presuppose finite generation. Bounded canonical height gives a finite set by the bounded difference and the Northcott theorem. Torsion points have canonical height zero. Conversely, if $\hat{h}(P)=0$, all $2^{j}P$ have bounded naive height, so two coincide; their difference makes $P$ torsion. Thus the height detects precisely the free part, once finite generation is proved.

Use the Weak Mordell-Weil theorem with $n=2$ and choose finitely many coset representatives $R_i$. Write $P=2Q+R_i$, and put $H={\max }_i\hat{h}(R_i)$. The height inequality yieldsIteration enters the finite set $\{P:\hat{h}(P)\le H+1\}$, since after $j$ steps the height is at most $H+2^{-j}\max (\hat{h}(P)-H,0)$. Unwinding $P=2Q+R_i$ expresses every point using that finite set and the $R_i$. This proves the canonical-height proof of Mordell-Weil finite generation and henceThe Fundamental theorem of finitely generated abelian groups supplies this decomposition. The Mordell-Weil theorem combines a finite quotient from arithmetic Galois cohomology with a contracting height descent; either ingredient alone would not establish finite generation.

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 the real quadratic field $K=\mathbb{Q}(\sqrt{d})$ with squarefree $d>1$, Dirichlet unit theorem givesfor a fundamental unit $ϵ>1$, since the roots of unity in a real field are $\pm 1$.

If $p\mid d$ and $p\equiv 3(\mathrm{mod}4)$, a unit of norm $-1$ would have the form $(a+b\sqrt{d})/2$ with integers $a,b$ and satisfy $a^2-d{b}^2=-4$. Reducing modulo the odd prime $p$ gives $(a/2{)}^2\equiv -1(\mathrm{mod}p)$, impossible because $-1$ is not a quadratic residue for such a prime. Hence every unit has norm $+1$ in this case. In contrast, $\boxed{d=2,u=1+\sqrt{2}}$ has norm $-1$.

Kummer theory begins with the exact sequenceGalois cohomology and Hilbert theorem 90 identifyso cyclic extensions of exponent dividing $n$ are described by adjoining $n$th roots when $K$ contains ${\mu }_n$.

For an elliptic curve $E/K$, multiplication by $n$ gives the Kummer exact sequence of an elliptic curveIts connecting homomorphism is the injective Kummer map of an elliptic curvewhere $nQ=P$. Passing to the finite division field of an elliptic curve $K(E[n])$ makes $E[n]$ constant. Rational functions whose divisors are $n(T)-n(O)$ then express the classes through finitely many elements of $K(E[n]{)}^{\times }/K(E[n]{)}^{\times n}$.

Let $S$ contain the primes above $n$, the primes of bad reduction of an elliptic curve and the finitely many primes introduced by these functions. The local theory of good reduction shows that every Kummer class coming from $E(K)$ is unramified outside $S$, so it lies in an S-unramified power class group. Such a group is finite: valuations outside $S$ vanish modulo $n$, the ideal class group is finite, and the Dirichlet unit theorem makes the group of $S$-units modulo $n$th powers finite. The kernel created by passing to $K(E[n])$ is finite by finite-group Galois cohomology. HenceThis is the Kummer-theoretic proof of the weak Mordell-Weil theorem. Combining it with the height descent lemma proves the Mordell-Weil theorem: $E(K)$ is a finitely generated abelian group.

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

First suppose $|{\sigma }_i(\alpha )|=1$ for every complex embedding. The coefficients of the monic polynomialare rational algebraic integers and hence integers. They are uniformly bounded in $m$, because each is an elementary symmetric sum of numbers of modulus one. Only finitely many such integer polynomials can occur, so only finitely many algebraic integers ${\alpha }^m$ occur. Two powers coincide; since $\alpha \ne 0$, this gives ${\alpha }^{r-s}=1$. Thus $\alpha$ is a root of unity. This is Kronecker theorem on algebraic integers in the unit disk.

The second printed assertion is literally false for $\alpha =0$. For the intended statement with $\alpha \ne 0$, use the Minkowski embedding: algebraic integers of $K$ form a lattice, so only finitely many have all conjugates of modulus at most $2$. A nonzero algebraic integer with every conjugate of modulus at most one hasso every conjugate has modulus one and the first part applies. Consequently every non-root in that finite box has ${\max }_i|{\sigma }_i(\alpha )|>1$. Choose $c>1$ no larger than the least of these finitely many maxima, taking $c=2$ if there are no non-roots. Then every nonzero $\alpha \in {\mathcal{O}}_K$ satisfying $|{\sigma }_i(\alpha )|<c$ for all $i$ is a root of unity.

Dirichlet unit theorem states that for signature $(r_1,r_2)$,and the logarithmic embedding of number field units maps the free part to a full lattice in the hyperplane where the weighted coordinate sum is zero.

For a real quadratic field, the roots of unity are $\{\pm 1\}$ and the logarithms of positive units form a nonzero discrete subgroup $a\mathbb{Z}\subset \mathbb{R}$. Its smallest positive element is $\log ϵ$ for a unit $ϵ>1$. Given any positive unit $u>1$, choose $n$ so that $1\le u{ϵ}^{-n}<ϵ$; minimality forces the quotient to be $1$. Applying signs and inverses shows

Since $11\equiv 3(\mathrm{mod}4)$, the ring of integers of a quadratic field is $\mathbb{Z}[\sqrt{11}]$. A unit $a+b\sqrt{11}$ satisfies the Pell equation $a^2-11b^2=\pm 1$. The elementhas norm $1$. To prove it is the smallest unit above one without invoking the continued-fraction algorithm, suppose $1<a+b\sqrt{11}<10+3\sqrt{11}$. Replacing by its inverse or negative if needed makes $b>0$, and the conjugate relation implies $b\sqrt{11}<(u+u^{-1})/2<10$, so $b\le 3$. For $b=1,2,3$, the numbers $11b^2\pm 1$ are respectivelyOnly the last list contains a square, namely $100=10^2$, which yields the endpoint itself. Hence no smaller unit exists and

Assume that $\zeta$ is a primitive fifth root of unity. The cyclotomic field $K=\mathbb{Q}(\zeta )$ has degree four and signature $(r_1,r_2)=(0,2)$. The Dirichlet unit theorem therefore gives free rankThe roots of unity in a rational cyclotomic field are the ten elements $\pm {\zeta }^j$, so

For explicit units, $-\zeta$ has order ten, whileHence $1+\zeta$ is a unit. It has infinite order because in the standard complex embeddingwhereas every root of unity has modulus one. Its image in the free $\mathbb{Z}$ factor is therefore nonzero and generates a finite-index subgroup of that factor. Consequentlyhas finite index in ${\mathcal{O}}_K^{\times }$, as described by the unit group of the fifth cyclotomic field.

The Dirichlet unit theorem states that for a number field of signature $(r_1,r_2)$,

For $K=\mathbb{Q}(\sqrt{5})$,The field is real quadratic, so the unit rank is one and its only roots of unity are $\pm 1$. To see that $\varphi$ is fundamental, suppose a positive unit satisfies $1<u<\varphi$. If its norm is $1$, its integral trace $u+u^{-1}$ lies strictly between $2$ and $\sqrt{5}$; if its norm is $-1$, its trace $u-u^{-1}$ lies strictly between $0$ and $1$. Neither interval contains an integer. Reducing any positive unit by a suitable power of $\varphi$ now proves the units of the quadratic field Q square root of five formula

If ${\mathcal{O}}_L^{\times }/{\mathcal{O}}_K^{\times }$ is finite, the two unit groups have the same rank, namely one. Put $d=[L:K]$ and let $(r_1,r_2)$ be the signature of $L$. ThenTherefore $r_1=4-2d$ and $r_2=2d-2$. For a proper extension, nonnegativity forceswith signature $(0,2)$.

This degree occurs: take $L=K(i)$. It is a totally imaginary quadratic extension of $K$, so the unit ranks agree and the quotient is finite. It is nontrivial because $i\in {\mathcal{O}}_L^{\times }$ but $i\notin {\mathcal{O}}_K^{\times }$ (and its coset has order two). This is the finite relative unit quotient over a real quadratic field example.

Define the S-unramified power class group byThere is an exact sequence from the $S$-unit group modulo $n$th powers into $K(S,n)$ and then into the $n$-torsion of the ideal class group. By the Dirichlet unit theorem,after absorbing the fixed roots of unity into the exponent. The ideal class group has fixed finite order $h_K$. Since $n\ge 2$, choose a constant $c$ depending only on $K$ with $h_K$ and the unit contribution bounded by $n^c$. Then