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Fix integers $q\ge 1$ and $a$ with $(a,q)=1$. We prove the Dirichlet theorem on primes in arithmetic progressions by showing that the prime Dirichlet sum in this residue class diverges as its real exponent decreases to one.

First justify the required Dirichlet L-function facts. If $\chi$ is a nonprincipal Dirichlet character modulo $q$, its sum over a full period is zero: choose a unit $b$ with $\chi (b)\ne 1$, and multiplication by $b$ permutes the residue classes, forcing the sum to equal $\chi (b)$ times itself. Therefore the partial sums $C_{\chi }(u)={\sum }_{n\le u}\chi (n)$ are bounded by $q$. Partial summation giveswhich supplies holomorphic continuation near one, with locally uniform convergence. For the principal Dirichlet character ${\chi }_0$,has a simple pole at one, with residue $\phi (q)/q$.

The assumption in the question gives nonvanishing at one for real nonprincipal characters. We must also prove it for nonreal characters. For a nonreal $\chi$, its square ${\chi }^2$ is nonprincipal. For real $\sigma >1$, logarithms of the Euler products giveFor a prime not dividing $q$, its logarithmic coefficient is $3+4\mathrm{Re}\chi (p{)}^k+\mathrm{Re}\chi (p{)}^{2k}=2(1+\cos \theta {)}^2\ge 0$, where $\chi (p{)}^k=e^{i\theta }$. Primes dividing $q$ contribute only the positive zeta term. If $L(1,\chi )$ vanished to order $m\ge 1$, the product would be $O((\sigma -1{)}^{4m-3})$ and tend to zero: the squared-character factor is bounded because ${\chi }^2$ is nonprincipal. This is a contradiction. HenceThis step uses the given real-character hypothesis and the Euler product positivity for L-function nonvanishing for all remaining characters.

In $\sigma >1$ choose the logarithm furnished by the absolutely convergent Euler product. Its prime-power expansion givesThe error from powers $k\ge 2$ is uniformly bounded as $\sigma \downarrow 1$, by ${\sum }_p{\sum }_{k\ge 2}1/(k{p}^{k\sigma })<\infty$. For nonprincipal $\chi$, nonvanishing and holomorphic continuation at one provide a bounded logarithm in a neighborhood of one. The Euler-product branch differs from that branch by a constant integral multiple of $2\pi i$ on the final real interval, so it too is bounded. ThusFor the principal character, its simple pole instead gives

The Orthogonality of Dirichlet characters gives, for primes not dividing $q$,Multiply by $p^{-\sigma }$ and sum. Absolute convergence allows the interchange for $\sigma >1$, soThe right side tends to infinity. If there were only finitely many primes in the progression, the left side would have a finite limit at one. This contradiction proves infinitely many primes occur in every reduced residue class. For $q=1$ or $2$ the character argument simply has no nonprincipal terms and gives the same conclusion.

For the final deduction the modulus $q$ is fixed. Let $A(t)=\pi (t;q,a)$. The given consequence of the Siegel–Walfisz theorem is $A(t)\sim t/(\phi (q)\log t)$. Partial summation, including the prime $2$ through the lower endpoint $2^{-}$ when appropriate, givesThe boundary is $O_q(1/\log x)$. Write $A(t)=t(1+\epsilon (t))/(\phi (q)\log t)$ with $\epsilon (t)\to 0$. The main integral is $(\log \log x-\log \log 2)/\phi (q)$. The error is $o(\log \log x)$: given $\eta >0$, choose $T$ such that $|\epsilon (t)|\le \eta$ for $t\ge T$. Its integral from $2$ to $T$ is a fixed constant, and its absolute value from $T$ to $x$ is at most $\eta \log \log x/\phi (q)$ plus a fixed constant. Divide by $\log \log x$ and then let $\eta \downarrow 0$. We obtain the reciprocal primes in a fixed arithmetic progression asymptoticThe argument uses the supplied prime-counting asymptotic in this last part; the preceding infinitude proof did not assume the Prime number theorem in arithmetic progressions.

The Selberg upper-bound sieve replaces a difficult coprimality indicator by a nonnegative square whose coefficients can be optimized. Let $P(z)={\prod }_{p\le z}p$, and suppose a finite sequence has divisibility countswhere $g$ is multiplicative on squarefree integers and $0<g(p)<1$. Choose real Selberg sieve weights ${\lambda }_1=1$ supported on $d\mid P(z)$ with $d\le R$. If an entry has no prime factor in $P(z)$, its only weighted divisor is $1$, so its indicator is bounded above by the square of the divisor sum. Summing giveswhere $Q(\lambda )={\sum }_{d,e}{\lambda }_d{\lambda }_{e}g([d,e])$. This is an upper bound regardless of the signs of the weights.

Here is the optimization. On the permitted squarefree integers putThe identitycan be checked one prime at a time: a prime in both $d$ and $e$ gives the factor $1+h(p)=1/g(p)$. ThusFinite Möbius inversion gives ${\lambda }_{d}g(d)={\sum }_{d\mid r}\mu (r/d)Y_r$, and in particular the condition ${\lambda }_1=1$ becomes ${\sum }_r\mu (r)Y_r=1$. The Cauchy-Schwarz inequality now yieldsEquality is attained by $Y_r=\mu (r)k(r)/J$. Inverting again gives the explicit optimizing weightsThey satisfy $|{\lambda }_d|\le 1$. To see this, multiply out ${\prod }_{p\mid d}(1-g(p){)}^{-1}={\sum }_{e\mid d}k(e)$. The numerator becomes the sum of $k(ev)$ over distinct permitted integers $ev\le R$ with $e\mid d$ and $(v,d)=1$, a subset of the terms in $J$. We therefore have a main term $X/J$ and a controllable remainder. This proves the basic Selberg sieve diagonalization and the useful bound on the weights, rather than merely asserting an optimized density.

Apply it to the polynomial $F(n)=n(n+2)$ for $1\le n\le N$. If $\omega (d)$ is the number of roots of $F$ modulo squarefree $d$, the Chinese remainder theorem givesUse $g(d)=\omega (d)/d$, $R=z=N^{1/8}$, and sieve the indices for which $F(n)$ has no prime factor at most $z$. The local weights areWe claim $J(z,z)\gg (\log z{)}^2$. The following truncated-product argument supplies the necessary lower bound. Put $w=z^{1/8}$, and consider all squarefree integers on the primes at most $w$, with total weightGive such an integer $d$ probability $k(d)/W$. Prime inclusion is independent with probability $g(p)$, soby the Mertens first theorem, absorbing the prime $2$ separately. For sufficiently large $z$, this is at most $\frac{1}{2}\log z$. The Markov inequality shows that at least half of $W$ lies on $d\le z$, all of which are terms in $J(z,z)$.

Moreover, partial summation applied to the Mertens first theorem givesIndeed it is $A(w)/\log w+{\int }_2^{w}A(t)/(t(\log t{)}^2)dt$, and the contribution of $A(t)-\log t=O(1)$ has a finite limiting constant with tail $O(1/\log w)$. Expanding the logarithm of $W$ now givesThe error is bounded since ${\sum }_p{p}^{-2}<\infty$. Consequently $W\asymp (\log w{)}^2\asymp (\log z{)}^2$, proving the claim. The exponent two reflects the two forbidden classes at almost every prime.

For the remainder use $|{\lambda }_d|\le 1$ and the simple bound $\omega ([d,e])\le [d,e]\le de$. HenceThe number of surviving indices is thereforeEvery pair of primes $p,p+2$ with $p>z$ survives, since both its prime factors exceed $z$. The exceptional pairs with $p\le z$ number at most $z$. With $z=N^{1/8}$ we obtainThis is the twin-prime upper bound from a quadratic sieve. The method constructs a majorant and bounds the sifted set; it does not give a positive lower bound for twin primes.

Use the meromorphic completionThe functional equation of the Riemann zeta function isIts entire version is $\xi (s)=\frac{1}{2}s(s-1)Z(s)$ with $\xi (s)=\xi (1-s)$. The completion $Z$ itself has poles at zero and one; it should not be confused with the entire Riemann xi function.

We prove the equation by a theta integral. SetThe Gaussian Fourier transform, with kernel $e^{-2\pi ixy}$, isFor example, differentiation in $y$ and integration by parts give the differential equation $I^{\prime }(y)=-(2\pi y/t)I(y)$; the Gaussian integral gives $I(0)=t^{-1/2}$. Periodize the Gaussian by summing $e^{-\pi t(x+n{)}^2}$ over integers $n$. Its $k$th Fourier coefficient is the displayed transform at $k$. Both the periodized function and its Fourier series converge absolutely, so evaluation at $x=0$ gives the Poisson summation formula in this case:

For $\mathrm{Re}s>1$, integration term by term in the Gamma function integral givesIndeed, the two terms for $n$ and $-n$ cancel the factor $1/2$, and substitution $v=\pi n^{2}t$ gives ${\pi }^{-s/2}\Gamma (s/2)n^{-s}$. Split at one, and use the transformation above in the integral from zero to one. The elementary contribution isSubstitution $t=1/u$ in the remaining term therefore givesThe theta tail decays exponentially. The integral is consequently an entire function of $s$, with uniform convergence and termwise differentiation on compact subsets. The formula continues $Z$ meromorphically to the whole plane. Both its rational part and its integral are unchanged by $s\mapsto 1-s$, proving the equation and the entire xi formulation. Using the reflection and duplication identities of the Gamma function, one may equivalently writeunderstood as an identity of meromorphic functions, including points where individual factors need continuation.

Now let $N(T)$ count the Nontrivial zeros of the Riemann zeta function $\rho =\beta +i\gamma$ with $0<\gamma \le T$, including multiplicity. The Riemann–von Mangoldt formula isFor $T$ not a zero ordinate, the more precise usual form isHere the argument is continued from $\zeta (2)>0$ along the path $2\to 2+iT\to \frac{1}{2}+iT$. The first formula remains valid at zero ordinates with the stated endpoint convention. Only its leading asymptotic is needed below.

List the positive ordinates as ${\gamma }_1\le {\gamma }_2\le \cdots$, with multiplicity, consistent with this counting function. Put $b_n=2\pi n/\log n$. For every fixed $\epsilon \in (0,1)$,For all sufficiently large $n$, the lower count is less than $n$ and the upper count is greater than $n$. Consequently $(1-\epsilon )b_n<{\gamma }_n\le (1+\epsilon )b_n$. Letting $\epsilon \downarrow 0$ proves the asymptotic inversion of the zeta zero countThis argument works even when several zeros have the same ordinate; “increasing” is interpreted in the standard nondecreasing, multiplicity-counted sense. A count listing only distinct ordinates is a different object and is not determined by this formula alone.

In this notation the Riemann hypothesis says that every Nontrivial zero of the Riemann zeta function has real part one-half:It concerns their horizontal location, not merely the growth of the ordinates, and does not assert that the zeros are simple. The functional equation and complex conjugation make the zero set symmetric about both the real axis and the critical line; the hypothesis puts all the Nontrivial zeros of the Riemann zeta function on that line.

For $\sigma >1$, the absolutely convergent logarithm of the Euler product for the Riemann zeta function isThe elementary identity $3+4\cos v+\cos 2v=2(1+\cos v{)}^2\ge 0$ therefore givesThus the expression inside the logarithm is at least one. The Meromorphic continuation of the Riemann zeta function to the right half-plane has only a simple pole at one, of residue one. It follows, for example, fromwhose integral is holomorphic there.

Suppose $t\ne 0$ and $\zeta$ has a zero of order $m\ge 1$ at $1+it$. As $\sigma \downarrow 1$, the real factor is $O((\sigma -1{)}^{-3})$, the middle factor is $O((\sigma -1{)}^{4m})$, and the factor at $1+2it$ is bounded. Their product would tend to zero, contradicting its lower bound one. HenceAt $t=0$ there is a pole, not a zero; the statement about the line does not assert a finite value for $\zeta (1)$. This proves the three-four-one product proof of zeta boundary nonvanishing.

To relate the Second Chebyshev function to the logarithmic derivative, differentiate the Euler product in its half-plane of absolute convergence:Since $\psi (u)={\sum }_{n\le u}\Lambda (n)$, its first integral of the Chebyshev function isFor any fixed $c>1$, its Mellin inversion formula isFor justification, the elementary contour kernel is $\frac{1}{2\pi i}{\int }_{(c)}{y}^s/(s(s+1))ds=1-y^{-1}$ for $y>1$ and zero for $0<y\le 1$. Close left or right and take the residues at zero and minus one; at $y=1$ the continuous value is zero. Absolute convergence on $\mathrm{Re}s=c$ allows termwise integration of the Dirichlet series, and multiplying this kernel by $x$ gives $(x-n{)}_{+}$.

Here is how the relation yields the asymptotic. Move the contour to the left, using the functional equation of the Riemann zeta function to control the left-hand side. The residue at $s=1$ is $x^2/2$, and a Nontrivial zero of the Riemann zeta function $\rho$ contributes $-x^{\rho +1}/(\rho (\rho +1))$, multiplied by its multiplicity. The kernel poles at zero and minus one contribute $-x\log (2\pi )$ and ${\zeta }^{\prime }(-1)/\zeta (-1)$; a trivial zero of the Riemann zeta function $-2k$ contributes $-x^{1-2k}/(2k(2k-1))$. Thus the smoothed explicit formula for the Chebyshev function isFor the contour argument, use heights avoiding the zero ordinates and then let those heights increase. The local zero bound from the Riemann–von Mangoldt formula, together with the Local partial-fraction expansion of the Riemann zeta logarithmic derivative, permits heights with logarithmic-derivative bound $O({\log }^{2}T)$ in each fixed vertical strip. The horizontal integrals then vanish because the denominator is of size $T^2$. After that, send the left edge through negative odd integers to minus infinity; the functional equation of the Riemann zeta function bounds the logarithmic derivative there by a logarithm, and $x^{s+1}$ makes the left integral vanish for fixed $x>1$. This explains why this smoothing allows a convergent contour calculation without a quantitative zero-free region.

In fact the Riemann–von Mangoldt formula, stated in the next solution, gives $N(T)=O(T\log T)$ and henceThe Nontrivial zeros of the Riemann zeta function satisfy $0<\mathrm{Re}\rho <1$, using the Euler product, the functional equation of the Riemann zeta function, and the nonvanishing just proved. For each such zero, $x^{\rho -1}\to 0$ as $x\to \infty$, whereas $|x^{\rho -1}|\le 1$ for $x\ge 1$. The dominated convergence theorem therefore makes the absolutely convergent zero sum, divided by $x^2$, tend to zero. All the other displayed terms are $o(x^2)$. We obtain

It remains to unsmooth; differentiating an asymptotic without justification would not suffice. Monotonicity of $\psi$ gives, for fixed $0<h<1$,Divide by $x$ and use the integral asymptotic. The lower and upper limits lie between $1-h/2$ and $1+h/2$. Letting $h\downarrow 0$ proves $\psi (x)\sim x$. Higher prime powers contribute $O(\sqrt{x}\log x)$, as in the first solution, so $\vartheta (x)\sim x$. Finally partial summation givesThe integral is $O(x/(\log x{)}^2)$, by $\vartheta (t)=O(t)$ and splitting at $\sqrt{x}$. Hencewhich is the Prime number theorem.

Write $\vartheta (x)={\sum }_{p\le x}\log p$ and $\psi (x)={\sum }_{n\le x}\Lambda (n)$ for the two Chebyshev functions, and $\pi (x)$ for the prime-counting function. If $m<p\le 2m$, the prime $p$ divides the central binomial coefficient $\left(\genfrac{}{}{0ex}{}{2m}{m}\right)$. ConsequentlySum this inequality over dyadic integers $m=1,2,4,\dots ,2^{k-1}$. It gives $\vartheta (2^k)\le 2^{k+1}\log 2$. Monotonicity, with $2^{k-1}<x\le 2^k$, gives $\vartheta (x)=O(x)$ for real $x\ge 2$. This is the Chebyshev estimate from central binomial coefficients; no information about the distribution of primes beyond unique factorization has been used.

Split the primes at $\sqrt{x}$. For $p>\sqrt{x}$, $\log p>\frac{1}{2}\log x$, soSince $\sqrt{x}=O(x/\log x)$,We also need $\psi (x)=O(x)$. The contributions of higher prime powers satisfybecause only $k\le \log x/\log 2$ contribute, and $\vartheta (x^{1/k})\ll \sqrt{x}$ for $k\ge 2$.

The Von Mangoldt divisor identity is $\log n={\sum }_{d\mid n}\Lambda (d)$, which follows by factoring $n$ into prime powers. Summing it for $n\le M$, with $M$ a positive integer, givesThe bound already proved makes the last error $O(M)$. Integral comparison for ${\sum }_{n\le M}\log n$, or the Stirling formula, gives $\log (M!)=M\log M-M+O(\log M)$. ThereforeThe contribution from higher prime powers is uniformly bounded:Subtract it, and replace $M$ by $\lfloor x\rfloor$, to obtain the Mertens first theorem

Now fix $\delta >0$. Apply partial summation to $A(t)$ and $f(t)=(\log t{)}^{-1-\delta }$. With the lower endpoint interpreted as $2^{-}$, so that the prime $2$ is included, this givesThe boundary term tends to zero, and the integral converges, since $A(t)\ll \log t$ and ${\int }_2^{\infty }dt/(t(\log t{)}^{1+\delta })<\infty$. Thus the series converges for every positive $\delta$. Keeping the main term of $A(t)$ also gives the useful logarithmically weighted reciprocal-prime tailIndeed, subtract the finite partial sum from the limiting integral: the boundary contributes $-(\log x{)}^{-\delta }$ and the main integral contributes $(1+\delta )(\log x{)}^{-\delta }/\delta$.

We first prove the simple-root Hensel lemma directly. Put $x_1=x$. Suppose $x_n\equiv x(\mathrm{mod}\pi )$ and $f(x_n)\equiv 0(\mathrm{mod}{\pi }^n)$. The derivative remains a unit since its reduction agrees with $f^{\prime }(x)$. Choose $t_n\in {\mathcal{O}}_K$ whose residue solvesand set $x_{n+1}=x_n+{\pi }^n{t}_n$. Polynomial expansion gives $f(x_{n+1})=f(x_n)+{\pi }^n{t}_n{f}^{\prime }(x_n)+{\pi }^{2n}B$ with $B\in {\mathcal{O}}_K$. Since $2n\ge n+1$, the new value vanishes modulo ${\pi }^{n+1}$. The sequence is Cauchy because $v(x_{n+1}-x_n)\ge n$, and completeness supplies $y\in {\mathcal{O}}_K$ with $y\equiv x(\mathrm{mod}\pi )$. Continuity gives $f(y)=0$.

For uniqueness, if $y,z$ are two roots in that residue class, thenThe factor in parentheses is a unit, since $f^{\prime }(z)$ is a unit and $y-z\in \pi {\mathcal{O}}_K$. Thus $y=z$. This proves existence and uniqueness of the specified lifted root.

Now suppose the residue field has $q=p^r$ elements and put $e=v(p)$. Every element of $k_K^{\times }$ is a simple root of $T^{q-1}-1$. The Hensel lemma therefore lifts its $q-1$ roots uniquely to a subgroup of the roots of unity in $K$. These are the Teichmuller representatives and give all the prime-to-$p$ torsion. To verify the latter assertion, if a root of unity $u\equiv 1$ has order $m$ prime to $p$, write $u=1+t$. In $u^m-1=mt+\left(\genfrac{}{}{0ex}{}{m}{2}\right)t^2+\cdots$, the first term has valuation $v(t)$ and all later terms have greater valuation, so the sum cannot vanish unless $t=0$. Reduction is therefore injective on prime-to-$p$ torsion, whose order divides $q-1$.

Any root of unity has a unique decomposition into a $p$-power-order factor and a prime-to-$p$ factor, by splitting its finite order using coprime powers. For a primitive $p^a$th root $\zeta$, $a\ge 1$, reduction in characteristic $p$ gives $\bar{\zeta }=1$. The cyclotomic identityshows that all factors have the same valuation: $({\zeta }^j-1)/(\zeta -1)=1+\zeta +\cdots +{\zeta }^{j-1}$ reduces to $j\ne 0$. Consequently the valuation of a primitive p-power root of unity isThe left side is a positive integer. Hence $p^{a-1}(p-1)\le e$, so the possible $p$-power orders are bounded. Combining this bound with the prime-to-$p$ result bounds all root orders by one integer, proving finiteness. A finite multiplicative subgroup of a field is cyclic, and it contains exactly the full prime-to-$p$ subgroup of order $q-1$. ThusIf $e<p-1$, even a primitive $p$th root would violate the valuation formula, so $s=0$.

At the boundary $e=p-1$, assume first that $p$ is odd. Take $K_{+}={\mathbb{Q}}_p({\zeta }_p)$. The polynomial ${\Phi }_p(1+T)=p+\left(\genfrac{}{}{0ex}{}{p}{2}\right)T+\cdots +T^{p-1}$ is Eisenstein, so $K_{+}/{\mathbb{Q}}_p$ is totally ramified of degree $p-1$ and $v_{K_{+}}(p)=p-1$. It contains a primitive $p$th root, while the valuation formula excludes primitive $p^2$th roots. Thus $s=1$.

For the other example, take $K_0={\mathbb{Q}}_p(\pi )$ with ${\pi }^{p-1}=p$. The polynomial $T^{p-1}-p$ is Eisenstein, giving the same ramification index and residue field ${\mathbb{F}}_p$. If a primitive $p$th root belonged to $K_0$, the valuation formula would give ${\zeta }_p-1=u\pi$ with $u$ a unit. Substitute this into ${\Phi }_p(1+({\zeta }_p-1))=0$ and divide by $p$. All intermediate terms reduce to zero, leaving $1+{\bar{u}}^{p-1}=0$ in ${\mathbb{F}}_p$. But ${\bar{u}}^{p-1}=1$, contradicting $2\ne 0$. Thus $s=0$ in $K_0$. This proves both cases of roots of unity at the tame ramification boundary for odd primes.

The printed boundary claim needs the qualification $p>2$. When $p=2$, every characteristic-zero field already contains the root $-1$ of order two, so $s=0$ is impossible. If $v(2)=1$, the valuation formula excludes order four and higher, forcing $s=1$. For example $K={\mathbb{Q}}_2$ has exactly this behavior. This is a genuine exception to the final request as printed, rather than an omitted construction.

Write $F_n=K_n^{+}$, $F_{\infty }={\bigcup }_n{F}_n$, $G=\mathrm{Gal}(F_{\infty }/\mathbb{Q})={\Delta }^{+}\times \Gamma$, where $|{\Delta }^{+}|=(p-1)/2$ and $\Gamma \simeq {\mathbb{Z}}_p$. Set $A={\mathbb{Z}}_p[[G]]$. Choosing a topological generator $\gamma$ of $\Gamma$ identifies its Iwasawa algebra with $\Lambda ={\mathbb{Z}}_p[[S]]$, $S=\gamma -1$; this group variable is distinct from the local-unit interpolation variable $T$. The Teichmüller character decomposes $A$ into a product of copies of $\Lambda$, one for each character of ${\Delta }^{+}$, equivalently each even Teichmüller character of $\mathrm{Gal}(K_0/\mathbb{Q})$. All characteristic-ideal statements below are componentwise in this product.

Let $U_n^1$ be the local principal units at the unique prime above $p$. Let $E_n^1$ and $C_n^1$ be respectively the p-adic closures of global units and cyclotomic units inside $U_n^1$, and take their inverse limits under the field norms. Define $X_{\infty }$ as the Galois group of the maximal abelian pro-p extension of $F_{\infty }$ unramified away from $p$, and $Y_{\infty }$ as that of the maximal everywhere unramified extension. This is the real tower: its prime-ramified module is torsion, by the Leopoldt theorem for abelian number fields. It must not be confused with the positive-rank prime-ramified module of the full complex cyclotomic tower. The unramified Iwasawa torsion theorem gives torsion for $Y_{\infty }$.

A convenient formulation of the Iwasawa main conjecture uses the p-adic zeta pseudomeasure ${\zeta }_p$ on $G$. With the sign conventionits product with the augmentation ideal $I(G)$ is integral and principal. The pole on the trivial-character component is cancelled by that ideal; simply writing an integral quotient by ${\zeta }_p$ would be incorrect. The main assertion in this convention isWe explain the local calculation, the remaining global obstruction, and how the cyclotomic Euler system removes it.

The local calculation is the Iwasawa theorem on local cyclotomic units:One can see why this theorem has exactly this analytic term. For a norm-fixed Coleman power series $f$, its corrected logarithm isThe ratio lies in $1+pR$, so the expression is integral. The norm identity puts its Amice transform in the kernel of the Coleman trace operator, hence gives a measure on ${\mathbb{Z}}_p^{\times }$. Its kth moment is $(1-p^{k-1}){\delta }_k(f)$. The full local exact sequence has p-power roots of unity at its kernel and cokernel. Passing to the real, or even, part removes both, since complex conjugation acts as minus one on their Tate module and $p$ is odd. The corrected logarithm therefore identifies $U_{\infty }^1$ with $A$.

Question 2 now shows that the image of $c(a,b)$ is $([b]-[a]){\zeta }_p$: its even moments are $(1-p^{k-1})(a^k-b^k)B_k/k$. A choice of integer $e$ generating ${\mathbb{Z}}_p^{\times }$ topologically gives a norm-compatible generator $c(e,1)$, after multiplication by its constant Teichmuller lift to make it principal. That constant has zero logarithmic derivatives. Its conjugates generate the closed cyclotomic-unit module, and $[e]-[1]$ generates $I(G)$. This yields the displayed local quotient and its characteristic ideal. The generation statement is an important ingredient; the mere calculation of a few moments would not prove the theorem.

The class-field unit sequence gives the global comparisonMultiplicativity of characteristic ideals in exact sequences givesThus the local theorem proves the main conjecture precisely when one proves $\mathrm{char}(B)=\mathrm{char}(Y_{\infty })$. These terms measure the global-unit and ideal-class obstructions. Discarding them would silently impose a much stronger arithmetic hypothesis.

The first global step is the Euler-system divisibility for real cyclotomic class modules: a generator $f_Y$ of $\mathrm{char}(Y_{\infty })$ divides a generator $f_B$ of $\mathrm{char}(B)$ on every character component. Here is the mechanism. Choose auxiliary primes $q$ splitting completely in a finite layer and satisfying $q\equiv 1(\mathrm{mod}{p}^m)$. For a generator ${\sigma }_q$ of its auxiliary cyclic group, putApply products of these operators to the unit Cyclotomic Euler system of Question 3. The norm axiom and this group-ring identity make the derived elements invariant modulo $p^m$-th powers, so Kummer descent gives classes in $F_n^{\times }/(F_n^{\times }{)}^{p^m}$. The congruence axiom identifies the valuation at each newly introduced prime with a residue symbol of the preceding class; valuations away from the chosen auxiliary primes vanish modulo $p^m$. The Chebotarev density theorem supplies primes in prescribed ideal classes with the required residue-symbol behavior. Adding these primes successively forces the elementary divisors of the class group to divide the available cyclotomic-unit index. Taking the norm limit yields $f_Y\mid f_B$. Finite control defects do not survive localization at height-one primes; using two coprime annihilators of such a finite defect also shows that no hidden p-power factor is left. This is the substantive global argument furnished by the Cyclotomic Euler system, not a consequence of Iwasawa's local theorem alone.

For the reverse comparison use the cyclotomic unit index formula and norm descent, keeping invariants as well as coinvariants. At the base field put $A_0$ for its p-primary ideal class group, and let $N_{\infty }(E_0^1)$ be the universal norm of units in a Zp-extension. Class-field and unit descent, using total ramification and the principal prime over $p$, giveThese are the required norm-defect control statements. The cyclotomic unit index formula, with the p-adic closure justified by Leopoldt, gives $\#A_0=\#(E_0^1/C_0^1)$. Also $B^{\Gamma }=0$: its coinvariants are finite by the displayed control, so its invariants are finite by the Iwasawa module structure theorem, whereas $B$ is a submodule of the cyclic local quotient $H$, which has no nonzero finite submodule. One must retain $Y_{\infty }^{\Gamma }$; dropping it would incorrectly identify all global units with universal norms.

Define the finite Euler characteristic of a one-variable Iwasawa module by ${\chi }_{\Gamma }(M)=\#M_{\Gamma }/\#M^{\Gamma }$. The preceding identities now give an exact cancellation:For a torsion $\Lambda$-module with finite invariant and coinvariant groups, the structure theorem gives ${\chi }_{\Gamma }(M)=|f_M(0){|}_p^{-1}$. Write $f_{B,\chi }=f_{Y,\chi }{h}_{\chi }$ using the Euler-system divisibility. After forgetting ${\Delta }^{+}$, the characteristic series are the products of their character-component series. Equality of these Euler characteristic of a one-variable Iwasawa module values givesEvery summand is nonnegative, since $h_{\chi }\in {\mathbb{Z}}_p[[S]]$. Each is therefore zero. A formal power series with p-adic-unit constant term is a unit, so every $h_{\chi }$ is a unit. Hence $\mathrm{char}(B)=\mathrm{char}(Y_{\infty })$, and cancellation in the class-field exact sequence completes the boxed main conjecture.