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Let $W$ be the subgroup fixed by the Coleman norm operator. The Coleman norm-fixed lift gives the unique factorization $f=w(\overline{f})v$, with $v\in 1+pR$. Multiplication consequently gives the displayed direct product of groups, while reduction identifies $W$ with ${\mathbb{F}}_p[[T]{]}^{\times }$. This is a decomposition of unit groups, not an identification of a power-series ring with a product of groups.

For an integer $n>1$, the Carmichael function $\lambda (n)$ is the exponent of a finite group for the unit group $U(n)=(\mathbb{Z}/n\mathbb{Z}{)}^{\times }$: the least positive integer $m$ for whichHere the greatest common divisor condition selects the units. Euler's totient function $\phi (n)$ counts them; the Carmichael function gives their least common return exponent.^[ref]

Modulo eight, the units are $1,3,5,7$, and their squares are all congruent to one. Their multiplicative orders are $1,2,2,2$, so $\lambda (8)=2$ although $\phi (8)=4$. Modulo fifteen, the units $1,2,4,7,8,11,13,14$ have multiplicative orders $1,4,2,4,4,2,4,2$. For example, the powers of two return through $1,2,4,8,1$. Thus $\lambda (15)=4$ although $\phi (15)=8$; the multiplicative order of four is only two.

The three quantities are related byA common return exponent must be a multiple of every multiplicative order, hence of their least common multiple. Conversely, that least common multiple makes every unit return to one. Lagrange's theorem makes each multiplicative order divide the group order $\phi (n)$, proving the final divisibility. The Carmichael function applies to all units at once, while the multiplicative order concerns a chosen unit.

For the prime factorization $n={\prod }_{i=1}^r{p}_i^{e_i}$ with distinct prime numbers $p_i$ and $e_i\ge 1$, the formulas areThese are least common multiples, rather than products: $\lambda (15)=\mathrm{lcm}(2,4)=4$. The convention $\lambda (1)=1$ handles the trivial case separately.^[ref]

The Chinese remainder theorem for unit groups explains the least common multiple: a power returns to one in every prime power component exactly when it is a multiple of each component's exponent of a finite group. For an odd prime power, the unit group is a cyclic group of group order $p^{e-1}(p-1)$. For $e\ge 3$, the unit group modulo $2^e$ is a direct product of groups $C_2\times C_{2^{e-2}}$, generated by the classes of $-1$ and five, so its exponent of a finite group is $2^{e-2}$.^[ref]

This gives a uniform reduction for modular exponentiation: for a unit $a$ and a nonnegative integer $k$, reduce $k$ modulo $\lambda (n)$. For example, $2^{1001}\equiv 2(\mathrm{mod}15)$ because $1001\equiv 1(\mathrm{mod}4)$. The unit hypothesis matters: $3^2\equiv 1(\mathrm{mod}8)$, but $2^2\not\equiv 1(\mathrm{mod}8)$. Computing the formula assumes the prime factorization is available. The same common-exponent viewpoint explains Carmichael numbers: a composite number $n$ passes the Fermat primality test for every unit base exactly when $\lambda (n)\mid n-1$. For example, $\lambda (561)=\mathrm{lcm}(2,10,16)=80\mid 560$.

For an integer modulus $n\ge 2$ and an integer $a$ with greatest common divisor $\gcd (a,n)=1$, Euler's theorem, also called the Fermat-Euler theorem, statesHere Euler's totient function $\phi (n)$ counts the units in the unit group $U(n)=(\mathbb{Z}/n\mathbb{Z}{)}^{\times }$. These congruence classes have modular multiplicative inverses, equivalently their representatives are coprime to $n$. The guaranteed return to one permits exponent reduction for units.^[ref]

For a prime number $p$, $\phi (p)=p-1$, recovering Fermat's little theorem in the form $a^{p-1}\equiv 1(\mathrm{mod}p)$ when $p\nmid a$. Multiplying by $a$ and separately handling $p\mid a$ gives $a^p\equiv a(\mathrm{mod}p)$ for every integer $a$. For a composite number used as the integer modulus, a nonzero congruence class can be a nonunit.

Modulo twelve, the units have representatives $1,5,7,11$, so Euler's totient function gives $\phi (12)=4$. Multiplication by five producesThus this permutation has two permutation cycles, each of length two. In particular $5^2\equiv 1(\mathrm{mod}12)$, and the guaranteed return $5^4\equiv 1$ follows. The number of units need not be the length of one permutation cycle. Modulo fifteen, multiplication by seven similarly gives the two permutation cycles $(1,7,4,13)$ and $(2,14,8,11)$, covering all eight units. The powers of seven follow the permutation cycle containing one; the other permutation cycle shows that these powers do not visit every unit.

There is a stronger obstruction for a nonunit than just an occasional failed calculation. If a positive integer $d>1$ divides both $a$ and $n$, then $a^k\equiv 0(\mathrm{mod}d)$ for every $k\ge 1$, so $a^k$ cannot be congruent to one modulo $n$. For example, $\gcd (3,15)=3$ and $3^{\phi (15)}=3^8\equiv 6(\mathrm{mod}15)$, rather than one. Such a base has no multiplicative order modulo that integer modulus.^[ref]

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.

Choose a reduced residue system $u_1,\dots ,u_{\phi (n)}$. For a unit modulo n $a$, the bijection $u\mapsto au$ on the unit group has inverse $u\mapsto a^{-1}u$. Comparing products before and after this permutation givesThe product $P$ is a unit; its modular inverse cancels it, proving Euler's theorem. Merely being nonzero would not justify cancellation.^[ref]

In the modulo-twelve example, $P=1\cdot 5\cdot 7\cdot 11=385\equiv 1(\mathrm{mod}12)$. The permuted representatives $5,1,11,7$ have the same product, while the unreduced products $(5\cdot 1)(5\cdot 5)(5\cdot 7)(5\cdot 11)$ equal $5^{4}P$. Equating these products modulo twelve exposes the fourth power without calculating it directly. Using all nonzero congruence classes instead would include nonunits, whose product need not permit cancellation.

For coprime integers $a,n$ with $n>1$, the multiplicative order is the least positive integer $d$ such thatIt counts the distinct powers in the cyclic subgroup generated by the congruence class of $a$. Thus it is an order of a group element in the unit group $(\mathbb{Z}/n\mathbb{Z}{)}^{\times }$.^[ref]

For example, the modular exponentiation $2^k\mathrm{mod}7$, starting at $k=0$, produces $1,2,4,1,2,4,\dots$. The first return to one takes three steps, so ${\mathrm{ord}}_7(2)=3$. Starting at the congruence class three and repeatedly multiplying by two instead gives $3,6,5,3,\dots$: these are the other three units modulo seven, outside the cyclic subgroup generated by two.

Euler's theorem guarantees a return to one at exponent $\phi (n)$, where $\phi$ is Euler's totient function. Thus the multiplicative order exists for every unit.^[ref] In fact,By Lagrange's theorem, the size of the cyclic subgroup divides the size of the unit group, which is Euler's totient function. For two modulo seven, this is $3\mid 6$; the multiplicative order need not equal Euler's totient function. If $a\equiv 1(\mathrm{mod}n)$, its multiplicative order is one.

The coprime integers hypothesis is necessary: if $a^d=1+kn$, any common divisor of $a$ and $n$ divides one. For instance, powers of two modulo eight give $1,2,4,0,0,\dots$, with no positive-exponent return to one. This sequence eventually settles, but two has no multiplicative order modulo eight.

In a nonzero unital ring with finite odd unit group, the unit $-1$ has order dividing two. By Lagrange's theorem, that order also divides $|U(R)|$, so $-1=1$ and the characteristic of a ring is two. Consequently $\overline{a}\mapsto a{1}_R$ defines a unital ring homomorphism ${\mathbb{F}}_2\to R$. This map still exists for the zero ring, whose characteristic is one under the least-positive-integer convention.

Among the units $U(N)$ modulo $N$, the passing residue classes form the kernel of a group homomorphism $a\mapsto a^{N-1}$. If each trial samples independently and uniformly from those units, with replacement, the probability that $r$ trials all pass is ${\alpha }_N^r$. This statement assumes that sampling rule; it does not describe arbitrary or dependent choices of bases.

For $341=11\cdot 31$, the Chinese remainder theorem identifies the unit group with cyclic groups of orders $10$ and $30$. The number of solutions of $x^{340}=1$ is $\gcd (340,10)\gcd (340,30)=100$, among $\phi (341)=300$ units. Thus ${\alpha }_{341}=1/3$. The Carmichael numbers $561$ and $1105$ have passing fractions $320/320$ and $768/768$, so their conditional false-pass probability remains one regardless of the number of independent trials.

The Fermat test example uses selected bases; this diagram exhausts the specified finite base sets. Both make the same logical distinction: an exponent witness can prove compositeness, whereas a passing result needs further justification before it can support a claim of primality.

Fix fundamental units ${\epsilon }_1,\dots ,{\epsilon }_r$ and write $x=\zeta \prod {\epsilon }_i^{b_i}$. Rank zero already gives a finite unit group. Otherwise put $B=\max (1,|b_i|)$. The logarithmic embedding in Dirichlet's theorem has a weighted sum-zero image and is injective on the free exponent lattice. Norm equivalence and that sum-zero condition give an archimedean embedding $\sigma$ withThis is the small archimedean value of a large unit. From the equation, $\sigma (\beta y)=1-\sigma (\alpha x)$ is exponentially close to one. Also $h(y)\le h(x)+O(1)$ by the arithmetic height inequalities for $y=(1-\alpha x)/\beta$, so the fundamental-unit exponents of $y$ have size $O(B)$.

For large $B$, take the small-branch logarithm of $\sigma (\beta y)$. With $y=\eta \prod {\epsilon }_i^{c_i}$ it is the nonzero formusing fixed logarithms and $\log (-1)=\pi i$. The integer $k$ corrects the branch, and nonzero follows since $\beta y=1$ would force $\alpha x=0$. Its upper bound is $|\Lambda |\le C^{\prime }{e}^{-cB}$. The assumed effective logarithmic-form estimate for the fixed algebraic numbers gives $\log |\Lambda |\ge -C^{\prime \prime }\log (2B)$. Thus $cB\le C^{\prime \prime }\log (2B)+O(1)$, effectively bounding $B$. The finitely many embeddings and torsion choices give uniform constants. Therefore there are only finitely many unit solutions, and their exponents are effectively bounded.

For the Thue application, take an irreducible integral binary form $F$ of degree $d\ge 3$ and $m\ne 0$. In a splitting field, write $F(X,Y)=a{\prod }_i(X-{\rho }_{i}Y)$ and put ${\theta }_i=a{\rho }_i$, which are algebraic integers. The integral factors $L_i=aX-{\theta }_{i}Y$ satisfy ${\prod }_i{L}_i=a^{d-1}m$. Each ideal $(L_i)$ therefore divides one fixed ideal, giving finitely many possibilities. Choose a generator ${\delta }_i$ for each principal possibility, so $L_i={\delta }_i{u}_i$ with units $u_i$.

For three distinct roots,Division by the third term gives a fixed-coefficient unit equation in $u_1/u_3,u_2/u_3$. Its solutions are finite, hence there are finitely many values of $L_1/L_2$. This ratio determines the rational direction $X:Y$, since the roots are distinct. In a fixed direction, homogeneity and $F(X,Y)=m\ne 0$ leave only finitely many integer scales. This proves Thue theorem, and the argument also works for a form with at least three distinct projective roots even if reducible.

The hyperelliptic alternative uses a related factorization for the usual nondegenerate case with squarefree $f$ of degree at least three. For $y^2=f(x)$, ideals of two different root factors can share primes only above the fixed root differences and the leading coefficient. At other primes their valuations are even. Finite exceptional-prime and ideal-class choices, together with units modulo squares, reduce the factors to finitely many expressions ${\delta }_i{z}_i^2$. Relations between distinct root factors then give simultaneous norm or Pell-type equations. Writing their solutions in fundamental units and applying the same small-value/logarithmic-form estimates bounds the unit exponents, hence $x,y$. The common-factor and square-class reductions are essential; a single quadratic norm equation alone can have infinitely many Pell solutions. This outlines the effective hyperelliptic treatment as well as the explicit Thue reduction.

A modulus of a number field is $\mathfrak{m}={\mathfrak{m}}_0{\mathfrak{m}}_{\infty }$, with ${\mathfrak{m}}_0$ a nonzero integral ideal and ${\mathfrak{m}}_{\infty }$ a subset of the real places. Let $I_{\mathfrak{m}}$ be the group of fractional ideals prime to ${\mathfrak{m}}_0$, and let $P_{\mathfrak{m}}$ consist of principal ideals $(a)$ with $a\equiv 1$ at each prime power in ${\mathfrak{m}}_0$ and $a>0$ at each selected real place. The ray class group isThe ray class field $K_{\mathfrak{m}}$ is the finite abelian extension for which the Artin map has kernel $P_{\mathfrak{m}}$ and induces an isomorphism ${\mathrm{Cl}}_{\mathfrak{m}}(K)\cong \mathrm{Gal}(K_{\mathfrak{m}}/K)$. Equivalently it is the maximal abelian extension whose conductor divides $\mathfrak{m}$.

To compute its order use the ray class exact sequence, including signatures:The middle map lifts a residue and sign tuple to an element of $K$ by weak approximation theorem, then takes its principal ideal as a ray class. Different lifts differ by a principal ray factor. Its kernel consists exactly of tuples represented by global units. Every ordinary ideal class has a representative prime to the finite modulus, again by weak approximation theorem; this gives surjectivity of the last map. If $U_{\mathfrak{m}}$ denotes the ray units, the kernel of the unit map is $U_{\mathfrak{m}}$. The Chinese remainder theorem and counting units in each prime-power quotient give the ray class number formula

First take $K=\mathbb{Q}(\sqrt{-3})$. Its ring of integers is $\mathbb{Z}[\omega ]$, where $\omega =(-1+\sqrt{-3})/2$ and ${\omega }^2+\omega +1=0$. The Minkowski bound for ideal classes is $(2/\pi )\sqrt{3}<2$, so every ideal class has a representative of norm one and $h_K=1$. Its units are the six roots of unity. Modulo seven, the polynomial $X^2+X+1$ has distinct roots $2,4$, and thereforeThe unit group of this quotient has order $36$. The unit $1+\omega$ maps to $(3,5)$, which has order six; consequently unit reduction is injective and $U_{(7)}=\{1\}$. The ray class number formula gives $36/6=6$.

We also identify the quotient explicitly. Conjugation interchanges the two factors, so the residue norm is $(x,y)\mapsto xy$. Its kernel has order six and equals the image of the six units, since $3\cdot 5=1(\mathrm{mod}7)$. HenceThe field $M=K({\zeta }_7)=\mathbb{Q}({\zeta }_{21})$ has degree six over $K$, since the degrees over $\mathbb{Q}$ are twelve and two. It is unramified outside seven by the cyclotomic computation and the base-change argument above. At primes over seven, $K$ has completion ${\mathbb{Q}}_7$, and $M/K$ is the tame degree-six cyclotomic extension; its conductor exponent is one. Its relative arithmetic Artin symbol at an ideal $\mathfrak{a}$ prime to seven acts byIf $\mathfrak{a}=(a)$ with $a\equiv 1(\mathrm{mod}7{\mathcal{O}}_K)$, then $N_{K/\mathbb{Q}}(a)\equiv 1(\mathrm{mod}7)$. Thus the Artin map kills the principal ray subgroup. It factors through the order-six ray class group; the residue norm quotient computed above shows that it is onto the degree-six Galois group, so the fields coincide. This proves the explicit ray class field of Q of square root minus three modulo seven:

Now take $K=\mathbb{Q}(\sqrt{-6})$, and write $s=\sqrt{-6}$. Its ring of integers is $\mathbb{Z}[s]$, its discriminant is $-24$, and its units are $\pm 1$. The quotient by $(s)$ iswhose two units are both images of $\pm 1$. The ray class exact sequence therefore gives ${\mathrm{Cl}}_{(s)}(K)\cong \mathrm{Cl}(K)$.

For its class number, the Minkowski bound for ideal classes is $(2/\pi )\sqrt{24}<4$, so each ideal class contains an integral ideal of norm at most three. The primes of norms two and three areThese are the only nontrivial possibilities within the bound, and their ideal classes coincide and have order at most two. The ideal ${\mathfrak{p}}_2$ is not principal: its being principal would give an integer solution to $x^2+6y^2=2$, which is impossible. Thus $h_K=2$, the ideal class group is $C_2$, and the ray field is the Hilbert class field.

We verify its actual identity rather than stopping at the degree. PutThe two quadratic fields $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{-3})$ are distinct, so $[H:\mathbb{Q}]=4$ and $[H:K]=2$. The integral elements $1,\sqrt{2},b,\sqrt{2}b$ are a rational basis. Their trace matrix is the Kronecker product of the two quadratic trace matrices, with discriminants $8$ and $-3$. Its determinant is therefore $8^2(-3{)}^2=576$. The discriminant-index formula for an integral lattice shows that the field discriminant divides $576$. On the other hand the relative discriminant ideal formula givesThe norm on the right is a positive integer, so equality is forced. Thus the relative discriminant ideal is the unit ideal, and $H/K$ is unramified at every finite prime. There are no real places of $K$. The extension is quadratic and therefore Abelian; its degree equals $h_K$, identifying it with the Hilbert class field of Q of square root minus six. Hence the ray class field of Q of square root minus six modulo square root minus six is

Write $\tau =(1+\sqrt{-3})/2$. The parity condition gives $R=\mathbb{Z}[\tau ]$, since $(a+b\sqrt{-3})/2=(a-b)/2+b\tau$. For any such element,If $a,b$ are even the numerator is divisible by four; if they are odd it is congruent to $1+3=0$ modulo four. Therefore $N(z)$ is a nonnegative integer. It is zero only at zero, and the Eisenstein-integer norm is multiplicative because complex conjugation respects multiplication.

If $z$ is a unit, $N(z)N(z^{-1})=1$, so $N(z)=1$. Conversely $N(z)=1$ makes $\overline{z}$ its inverse in $R$. The equation $a^2+3b^2=4$ permits exactly $(a,b)=(\pm 2,0)$ or $(\pm 1,\pm 1)$. HenceIndeed ${\tau }^2=\tau -1$ and ${\tau }^3=-1$, so this unit group is cyclic of order six.

For the prime-ideal assertion, ${\tau }^2-\tau +1=0$ and the polynomial $t^2-t+1$ has distinct roots $3,5$ modulo seven. Consequently evaluation at either root defines a surjective ring map $R\to {\mathbb{F}}_7$. Its kernelsare distinct maximal ideals, hence prime ideals. If $A+B\tau$ belongs to both, then $A+3B\equiv A+5B\equiv 0(\mathrm{mod}7)$. Subtraction gives $2B\equiv 0$, so both $A$ and $B$ are divisible by seven. The converse inclusion is immediate, provingEquivalently, the two distinct factors of the reduced polynomial give the Chinese remainder theorem decomposition $R/7R\cong {\mathbb{F}}_7\times {\mathbb{F}}_7$. This proof does not require the permitted unique-factorization assumption.

Put $\mathcal{D}={\mathcal{K}}_X^{\ast }/{\mathcal{O}}_X^{\ast }$, the quotient sheaf of abelian groups. At a closed point $P$, the discrete valuation gives an isomorphismIndeed, a local parameter on a smooth algebraic curve has valuation one, and the valuation kernel is precisely the unit group of the discrete valuation ring. Thus a local section represented by $f$ determines its local principal divisor, and changing $f$ by a regular function which is a unit does not change those coefficients.

These local valuations identify $\Gamma (X,\mathcal{D})$ with $\mathrm{Div}(X)$. To see that they give finite support, represent a section locally by finitely many rational functions on an algebraic variety, using Noetherian Zariski topology to take a finite cover. Each representative has only finitely many zeros and poles, so their combined support is finite. Conversely, for $D=\sum n_{P}P$, choose at each support point a local parameter $t_P$ and shrink its neighbourhood to exclude every other zero or pole of $t_P$ and every other support point. The local rational equation $t_P^{n_P}$ represents the desired valuations there; on the complement of the support use $1$. On overlaps these equations differ by units, so their classes glue in the quotient sheaf. Finally, a section with every valuation zero has zero germ everywhere and is the identity section. Hence the identification is both surjective and injective.

The map on global sections sends $f\in k(X{)}^{\ast }$ to $\mathrm{div}(f)$. Therefore its cokernel is

Use the following fixed-number form of a logarithmic estimate: for chosen logarithms of fixed nonzero algebraic numbers $a_1,\dots ,a_s$, a nonzero form $\Lambda =\sum b_j\log a_j$ with integer coefficients $|b_j|\le B$ satisfies $|\Lambda |\ge c(2B{)}^{-C}$, for fixed positive constants $c,C$. This follows from the Baker lower bound for a homogeneous linear form in logarithms.

By the Dirichlet unit theorem, write units as $x=\zeta \prod {ϵ}_j^{m_j}$ and $y=\xi \prod {ϵ}_j^{n_j}$, with finitely many roots of unity and fixed fundamental units. If the unit rank is zero, there are already only finitely many choices. Otherwise put $B=\max (1,|m_j|,|n_j|)$. The real logarithmic embedding of the free unit group has full rank and lies in the product-formula hyperplane. Equivalence of norms in that finite-dimensional space shows that a unit with exponent size $B$ has some archimedean conjugate of modulus at most $e^{-c_{0}B}$: a large positive logarithmic coordinate forces a comparably large negative one, and a large negative coordinate itself suffices.

Interchanging the two terms if necessary, choose such a conjugate for $y$. At this embedding the equation makes $\sigma (\alpha x)=1-\sigma (\beta y)$ with $|\sigma (\beta y)|\le C_0{e}^{-c_{0}B}$. For large $B$ define the logarithm near one. It is nonzero because $\beta y\ne 0$, andOn fixed branches it isfor an integer $k=O(B)$. The bound on $k$ follows from summing the fixed imaginary parts and then reducing to the near-one branch. There are only finitely many possible embeddings and roots of unity, so the stated lower bound applies uniformly with coefficients $O(B)$. It contradicts the exponential upper bound when $B$ is large. Hence the exponents are bounded and $\boxed{\alpha x+\beta y=1\text{ has finitely many unit solutions}}$ in ${\mathcal{O}}_K^{\times }$. This is the unit equation over a number field proof; “unit” here means an algebraic-integer unit, not an arbitrary nonzero field element.

For the Thue theorem, let $F\in \mathbb{Z}[X,Y]$ be irreducible homogeneous of degree $d\ge 3$ and let $m\ne 0$ be fixed. In a splitting field $L$, write $F(X,Y)=a{\prod }_i(X-{\theta }_{i}Y)$ with distinct roots. Choose a fixed integer $c$ so that each $c{\theta }_i$ is integral and $c^{d}m/a$ is a nonzero integer. For a solution $(u,v)$ the nonzero algebraic integers $t_i=c(u-{\theta }_{i}v)$ satisfy ${\prod }_i{t}_i=c^{d}m/a$. Their principal ideals therefore divide a fixed ideal and have only finitely many possibilities. For each occurring principal ideal choose a generator ${\gamma }_i$, so $t_i={\gamma }_i{\epsilon }_i$ with ${\epsilon }_i$ a unit.

Any three distinct roots give the identityDivide by the third term and absorb the finitely many generator ratios into the coefficients. This is a fixed-coefficient unit equation in ${\epsilon }_1/{\epsilon }_3$ and ${\epsilon }_2/{\epsilon }_3$, so it has finitely many solutions for each generator choice. Thus $\rho =t_1/t_3$ has finitely many values. The relation $(1-\rho )u=({\theta }_1-\rho {\theta }_3)v$ fixes a rational direction whenever an integer solution exists; $\rho =1$ simply gives $v=0$. On each such direction, $(u,v)=k(u_0,v_0)$ for a primitive integer pair, and $k^{d}F(u_0,v_0)=m$ allows finitely many $k$. This proves the desired finiteness. When $m=0$, irreducibility excludes a nonzero rational root direction, leaving only $(0,0)$.

The argument also proves finiteness for any binary form with at least three distinct projective roots: irreducibility was used only to guarantee distinct roots and to dispose of the zero-value case. A rational invertible change of variables can move every projective root away from infinity; after clearing its fixed denominators, the resulting pairs lie in an integer lattice and the same proof applies. This stronger nonzero-value version will apply to the binomial forms in question 6.

Let $M_f$ be the matrix of multiplication by $f$ in the basis from Question 1. Define the Coleman norm operator by the norm for a finite free ring extension:The entries lie in $\varphi (R)$, and $\varphi :R\to \varphi (R)$ is an isomorphism, so this determines $Nf$ uniquely. If $f$ is a unit, $M_f$ is invertible and its determinant is a unit. Also $M_{fg}=M_f{M}_g$, giving $N(fg)=NfNg$. Splitting the conjugates, or diagonalizing multiplication after adjoining the roots of unity and inverting $p$, gives the equivalent formulaIn particular, this definition has the required target $R^{\times }$, rather than merely a ring with extra roots of unity in its coefficients.

For the requested ring congruence, observe that the reduction of $\varphi$ is the injective map $\overline{g}(T)\mapsto \overline{g}(T^p)$. If $\varphi (g)\in p^{k}R$, its reduction first gives $g=p{g}_1$. Then $\varphi (g_1)\in p^{k-1}R$; repeat to obtain $g\in p^{k}R$. Applying this to $g=f-1$ provesThis congruence reflection for Frobenius substitution is valid for all $f\in R$, without assuming $f$ is a unit.

A further property needed for the fixed points is Coleman norm contraction. If $g=1+p^{k}h$ with $k\ge 1$, expand the determinant:Here $e_j(M_h)\in \varphi (R)$ is the coefficient of degree $j$ in $\det (I+t{M}_h)$. The trace formula from Question 1 makes the linear term divisible by $p^{k+1}$; every later term is divisible by $p^{2k}$ and $2k\ge k+1$. Reflecting the resulting ring congruence through $\varphi$ givesIf $Nf=f$ and $f\in 1+pR$, iteration gives $f\in 1+p^{k}R$ for every $k$. Since ${\bigcap }_k{p}^{k}R=0$, $f=1$.

To identify all fixed points, we also need $Nf\equiv f(\mathrm{mod}pR)$ for every unit $f$. In characteristic $p$, the extension ${\mathbb{F}}_p[[T]]/{\mathbb{F}}_p[[T^p]]$ has degree $p$ and its norm for a finite free ring extension is $\overline{f}\mapsto {\overline{f}}^p$. Indeed, after passing to fraction fields it is a purely inseparable field extension of degree $p$, so multiplication has only the eigenvalue $\overline{f}$, with multiplicity $p$, after splitting. Its determinant is ${\overline{f}}^p$. Thereforeand congruence reflection for Frobenius substitution gives $Nf\equiv f(\mathrm{mod}pR)$.

Now choose any lift $f_0\in R^{\times }$ of a given $a\in A^{\times }$ and put $f_n=N^n{f}_0$. Multiplicativity givesThus $(f_n)$ is a Cauchy sequence for the finer of the topologies on integral formal power series and converges to a unit $w(a)$ reducing to $a$. The Coleman norm operator is continuous: multiplication matrices and their determinants are continuous, and the finite-free coordinate decomposition and its inverse preserve the $p$-adic filtration defining the finer of the topologies on integral formal power series. Hence $Nw(a)=w(a)$. If two fixed units reduce to $a$, their quotient is fixed and belongs to $1+pR$, so it is $1$ by the preceding argument. This proves uniqueness, independence of the initial lift, and multiplicativity of the Coleman norm-fixed lift. In particular,Every unit $f\in R$ consequently has a unique factorization $f=w(\overline{f})v$ with $v\in 1+pR$, giving the canonical Coleman decomposition of power-series units:The final assertion as printed is false. The original PDF identifies a ring with a product of unit groups. With multiplication, $A$ has a zero and is not a group. With addition, it has exponent $p$, whereas $W\times A^{\times }$ has a nonidentity element of order $2$: $-1\in W$ because $p$ is odd and $N(-1)=(-1{)}^p=-1$. Thus neither interpretation makes the printed assertion true. The two boxed isomorphisms above are the valid natural conclusions; they do not require guessing which symbols the author intended to replace.

We give the detailed norm-index argument for a cyclic extension $L/K$ of p-adic fields; thus $K$ is a finite extension of ${\mathbb{Q}}_p$. Write $G=⟨\sigma ⟩$, $|G|=n$. For an additive $G$-module $M$, put $D=\sigma -1$ and $N=1+\sigma +\cdots +{\sigma }^{n-1}$. The two periodic Tate cohomology of a cyclic group groups and the Herbrand quotient arewhen these groups are finite. For multiplicative modules, $N$ is the product of conjugates and $D(z)=\sigma (z)/z$.

The periodic complex alternating $D$ and $N$ gives a six-term exact sequence for every short exact sequence of modules. Taking the alternating product of the finite group orders in that exact sequence provesFor a finite module, the counting fibers gives $|NM|=|M|/|\ker N|$ and $|DM|=|M|/|\ker D|$, so the two Tate groups have equal order. This proves the Herbrand quotient of a finite module is one. It follows that replacing a module by a finite-index submodule does not change its quotient, whenever the cohomology groups are finite.

Two basic calculations drive the norm computation. For the trivial module $\mathbb{Z}$, $D=0$ and $N$ is multiplication by $n$, so $h_G(\mathbb{Z})=n$. For a regular lattice $R[G]$, with $R={\mathbb{Z}}_p$, invariants are constant coefficient vectors and every such vector is a norm: put its constant coefficient in just one coordinate before applying $N$. The group kernel of $N$ consists of the coefficient vectors of sum zero. Successively taking partial sums solves the cyclic difference equation $Dx=y$ for every such vector $y$. Thus both Tate groups vanish and $h_G(R[G])=1$. This is cyclic cohomology of a regular lattice.

Let $U_L={\mathcal{O}}_L^{\times }$ and $U_L^r=1+{\mathfrak{P}}_L^r$. For sufficiently large $r$, the p-adic logarithm and p-adic exponential are inverse $G$-equivariant isomorphismsTo make the range explicit, $r>v_L(p)/(p-1)$ suffices: the higher terms in the logarithm and exponential have greater valuation than their leading terms, their series converge there, and the identities $\log (\exp x)=x$ and $\exp (\log (1+x))=1+x$ remain valid. Thus a deep multiplicative unit group becomes an additive ${\mathbb{Z}}_p$-lattice in $L$.

By the normal basis theorem, $L\cong K[G]$ as a $K[G]$-module. As a ${\mathbb{Q}}_p[G]$-module it is therefore ${\mathbb{Q}}_p[G{]}^{[K:{\mathbb{Q}}_p]}$. Transport the regular lattice $M_0={\mathbb{Z}}_p[G{]}^{[K:{\mathbb{Q}}_p]}$ into $L$. Both $M_0$ and ${\mathfrak{P}}_L^r$ are full $G$-stable ${\mathbb{Z}}_p$-lattices, so their intersection has finite index in each. The exact sequences with finite quotients show that the Tate groups of ${\mathfrak{P}}_L^r$ are finite and thatSince $U_L/U_L^r$ is finite, the exact-sequence formula for the Herbrand quotient gives $h_G(U_L)=1$. Finally, the valuation sequencehas trivial $G$-action on $\mathbb{Z}$. Hence the Herbrand quotient of the local multiplicative group is

For completeness, the denominator vanishes by Hilbert theorem 90. If $N_{L/K}a=1$, put $c_0=1$ and $c_j=a\sigma (a)\cdots {\sigma }^{j-1}(a)$. Independence of distinct field automorphisms lets us choose $t\in L$ with $b={\sum }_{j=0}^{n-1}{c}_j{\sigma }^j(t)\ne 0$. Since $c_n=1$, shifting the sum gives $\sigma (b)=b/a$. Thus $a=b/\sigma (b)$, a multiplicative coboundary. Therefore ${\hat{H}}^{-1}(G,L^{\times })=1$. The invariant subgroup of $L^{\times }$ is $K^{\times }$, soThis proves the cyclic local norm index without assuming local reciprocity. If the extension has ramification index $e$ and residue degree $f$, the valuation formula $v_K(Nz)=f{v}_L(z)$ shows that the valuation contribution to the index is $f$. A norm is a unit only when $z$ is a unit, so the remaining contribution isIn particular, unit norms are surjective in an unramified cyclic extension. The norm subgroup is open: on deep units its logarithm is ${\mathrm{Tr}}_{L/K}({\mathfrak{P}}_L^r)$, a nonzero full lattice in $K$, and therefore contains a sufficiently deep principal-unit group. This connects the index calculation to the open finite-index norm subgroups classified by local class field theory.

To finish, the Hasse norm theorem for a cyclic extension of number fields states that a nonzero element is a global norm if and only if it is a norm at every finite and infinite completion. Here is its Brauer-theoretic proof outline. Associate to $a\in K^{\times }$ the cyclic algebra $A=(L/K,\sigma ,a)$, with relations $z\ell =\sigma (\ell )z$ and $z^n=a$. The splitting criterion for a cyclic algebra says that it splits exactly when $a=Nc$: if $a=Nc$, multiplication by $L$ and the operator $c\sigma$ realize it as ${\mathrm{End}}_K(L)$; conversely, in a split matrix representation, the underlying $n$-dimensional space is one-dimensional over $L$, so $z$ must act as $c\sigma$ and $z^n=Nc$.

Apply this criterion at every completion, using the restricted cyclic character and any one completion of $L$ above that place. If $a$ is a local norm everywhere, every localized Brauer class is zero. Injectivity in the Albert-Brauer-Hasse-Noether theorem forces $[A]=0$ in $\mathrm{Br}(K)$; the global splitting criterion gives $a=Nc$. Conversely, a global norm is a local norm after completion, since the norm on $L{\otimes }_K{K}_v$ is the product of the component norms, whose images coincide in the Galois case. The cyclic hypothesis is essential; the stated local-global principle does not hold for arbitrary extensions.

Every nonzero P-adic number has a unique decomposition $z=p^{m}u$ with $m\in \mathbb{Z}$ and $u\in {\mathbb{Z}}_p^{\times }$. The group homomorphism kernel of reduction from the unit group onto ${\mathbb{F}}_p^{\times }$ is $1+p{\mathbb{Z}}_p$, the group of principal units. Each nonzero residue class has a unique lift $\omega$ satisfying ${\omega }^{p-1}=1$, by the simple-root Hensel lemma proved in question 3. Uniqueness makes these lifts multiplicative. They form the cyclic group ${\mu }_{p-1}$, and every unit has a unique factorizationThus ${\mathbb{Q}}_p^{\times }=p^{\mathbb{Z}}\times {\mu }_{p-1}\times (1+p{\mathbb{Z}}_p)$.

For odd $p$, the p-adic logarithm and p-adic exponential function give inverse group isomorphisms $1+p{\mathbb{Z}}_p\leftrightarrow p{\mathbb{Z}}_p$. To verify the convergence domain, for $t\in p{\mathbb{Z}}_p$ the p-adic logarithm terms have valuations $n{v}_p(t)-v_p(n)\to \infty$, and the exponential terms have valuations $n{v}_p(t)-v_p(n!)\to \infty$, since $v_p(n!)\le (n-1)/(p-1)$ and $p>2$. In both series every term after the linear one has strictly larger valuation than the linear term. Consequently $\log (1+t)\in p{\mathbb{Z}}_p$ and $\exp (t)\in 1+p{\mathbb{Z}}_p$. The formal identities $\log (vw)=\log v+\log w$ and $\exp (\log v)=v$, $\log (\exp t)=t$ hold on these convergent domains, as follows by multiplying the convergent series or passing to their formal identities termwise.

Choose a generator $\zeta$ of ${\mu }_{p-1}$. An explicit isomorphism isThe inverse reads off the valuation, the root-of-unity unit factor, and $p^{-1}\log (v)$. Both directions are continuous with the standard product topology, so this is also a topological group isomorphism. The odd-prime condition is essential for using all of $1+p{\mathbb{Z}}_p$ as the p-adic logarithm-isomorphism domain.

An automorphism sends the primitive root of unity ${\xi }_n$ to another primitive root of unity, hence uniquely to ${\xi }_n^a$ with $a\in (\mathbb{Z}/n\mathbb{Z}{)}^{\ast }$. Define $\chi (\sigma )=a$. Composition multiplies exponents, and an automorphism with exponent one fixes the generator and thus all of $L$. Therefore $\chi$ is an injective group homomorphism and $G$ is abelian.

The extension is finite Galois: it is the splitting field of $X^n-1$, whose roots are all present and distinct. Existence of a root of order $n$ in positive characteristic implies that the characteristic does not divide $n$. The Fundamental theorem of Galois theory identifies an intermediate field with $L^H$. Since $G$ is abelian, $H$ is normal; the normal-subgroup criterion gives $L^H/K$ Galois, with group $G/H$.

For a non-surjective nontrivial example, take $K=\mathbb{Q}(i)$ and $n=8$. Its cyclotomic extension has degree two, and its image is $\{1,5\}$ inside the four-element group $(\mathbb{Z}/8\mathbb{Z}{)}^{\ast }$. For $K=\mathbb{Q}$ and prime $p$, ${\Phi }_p(X)=1+\cdots +X^{p-1}$ is irreducible: ${\Phi }_p(X+1)$ is Eisenstein at $p$. Thus the extension has degree $p-1$, equal to the unit group's order, proving surjectivity.

For $n=7$, let $\zeta ={\xi }_7$. The cyclic Galois group of order six has exactly one subgroup of each order dividing six, so there are exactly four fields:Here ${\sigma }_a(\zeta )={\zeta }^a$. To identify the quadratic field without guessing, put $A=\zeta +{\zeta }^2+{\zeta }^4$ and $B={\zeta }^3+{\zeta }^5+{\zeta }^6$. Multiplication gives $A+B=-1$ and $AB=2$, so $(A-B{)}^2=-7$. The order-three subgroup fixes $A$, and $A-B\notin \mathbb{Q}$, proving the quadratic identification. The real field is cubic because complex conjugation is ${\sigma }_6$, with fixed subgroup of order two.