Iwasawa’s theorem on class numbers

Fix an odd prime $p$ and consider the cyclotomic fields $K_r=\mathbb Q(\zeta_{p^r})$ , where $r=1,2,3,\ldots$ . Let $h_r$ denote the class number of $K_r$ . We want to understand how many factors of $p$ divide these class numbers as the conductor increases. More precisely, write $e_r=v_p(h_r)$ , so that $p^{e_r}$ is the largest power of $p$ dividing $h_r$ . At first, there is no obvious reason why the numbers $e_r$ should follow any simple pattern. The degree of $K_r$ is $(p-1)p^{r-1}$ , so the fields become much larger at every step. Their discriminants grow rapidly, and the corresponding ideal class groups can be complicated. Nevertheless, Iwasawa proved that there are integers $\mu,\lambda,\nu$ , independent of $r$ , such that for all sufficiently large $r$ ,

$\displaystyle \boxed{v_p(h_r)=\mu p^{r-1}+\lambda(r-1)+\nu.}$

This is a remarkable regularity statement. Although the fields become much larger, the exponent of $p$ in their class numbers eventually follows a very simple formula. The terms have different meanings: $\mu p^{r-1}$ grows exponentially with the level, $\lambda(r-1)$ grows linearly, and $\nu$ accounts for the initial arithmetic. We will try to understand where these terms come from using finite calculations. The natural starting point is the classical class number formula, which expresses the class number through special values of zeta functions. In cyclotomic fields, the relevant zeta functions factor into Dirichlet $L$ -functions, and some of their special values can be calculated using finite character sums. We will use these sums to study divisibility by $p$ , connect them to ordinary Bernoulli numbers, and then examine what happens when we increase the conductor from $p^r$ to $p^{r+1}$ .

There is one distinction we should make at the beginning. Let $K_r^+=\mathbb Q(\zeta_{p^r}+\zeta_{p^r}^{-1})$ be the maximal real subfield, and let $h_r^+$ be its class number. The relative class number $h_r^-=h_r/h_r^+$ has a particularly simple expression involving finite character sums. We will study this first, because it lets us see the arithmetic very explicitly. The full theorem for $h_r$ also involves the real class numbers, and we will distinguish what our calculations establish from the deeper results needed to control them.

Class Number Formula

Let $K$ be a number field of degree $d$ , with ring of integers $\mathcal O_K$ . Every nonzero integral ideal $\mathfrak a$ has a norm $N\mathfrak a=\#(\mathcal O_K/\mathfrak a)$ , which is an ordinary positive integer. We can therefore count ideals of bounded norm, say those satisfying $N\mathfrak a\le X$ . The class number enters because these ideals are divided into $h_K$ ideal classes. There is a geometric way to approach this count. Under the real and complex embeddings of $K$ , the ring of integers becomes a lattice in a Euclidean space of dimension $d$ . An integral ideal is also a lattice, with covolume $2^{-r_2}\sqrt{|D_K|},N\mathfrak a$ , where $D_K$ is the discriminant and $r_2$ is the number of pairs of complex embeddings. Thus the discriminant measures the size of the lattice cells, while the norm of the ideal gives the additional scaling. To count ideals in a fixed class, we choose an ideal representing its inverse class and describe the ideals using algebraic integers in that lattice. There is one complication: multiplying an algebraic integer by a unit gives the same principal ideal. So we must count elements modulo the unit group. Under logarithms of the absolute values of the embeddings, the units form a lattice of rank $r_1+r_2-1$ . Its fundamental volume is the regulator $R_K$ . The roots of unity give a further finite repetition, accounted for by their number $w_K$ . This explains the different quantities appearing in the analytic class number formula. The discriminant comes from lattice covolumes, the regulator comes from units, the roots of unity correct the finite repetitions, and the class number counts the ideal classes. The resulting formula is

$\displaystyle \boxed{{\text{Res}}_{s=1}\zeta_K(s)=\frac{2^{r_1}(2\pi)^{r_2}h_KR_K}{w_K\sqrt{|D_K|}}.}$

Here $\zeta_K(s)=\sum_{\mathfrak a}(N\mathfrak a)^{-s}$ is the Dedekind zeta function, with the sum over nonzero integral ideals. Its pole at $s=1$ records the leading term in the ideal-counting problem. This is a useful way to remember the formula: it is ultimately a statement about the number of ideals of bounded norm. For cyclotomic fields, it will be more convenient to use the corresponding class number formula at $s=0$ . The functional equation relates the two versions, giving the leading term $\zeta_K(s)\sim-(h_KR_K/w_K)s^{r_1+r_2-1}$ near zero. We will not prove the analytic class number formula here; that is our main analytic input. Once we use it to obtain the cyclotomic class number formula, the calculations that follow will involve finite sums and congruences.

Relative Class Number Formula

Write $q=p^r$ , and put $K=\mathbb Q(\zeta_q)$ . The degree of $K$ is $\varphi(q)=2g$ , where $g=(p-1)p^{r-1}/2$ . Its maximal real subfield $K^+$ has degree $g$ . For cyclotomic fields, the Dedekind zeta function factors into Dirichlet $L$ -functions. The reason is that splitting of primes in $K$ is determined by their residue classes modulo $q$ . Characters of $(\mathbb Z/q\mathbb Z)^\times$ provide a natural way to separate these residue classes. Comparing the Euler factors prime by prime gives $\zeta_K(s)=\prod_{\chi\bmod q}L(s,\chi)$ , where an imprimitive character is represented by the $L$ -function of its primitive character. The even characters, those satisfying $\chi(-1)=1$ , correspond to the real subfield. Thus $\zeta_{K^+}(s)=\prod_{\chi\text{ even}}L(s,\chi)$ . Dividing the two formulas leaves only the odd characters:

$\displaystyle \frac{\zeta_K(s)}{\zeta_{K^+}(s)}=\prod_{\chi(-1)=-1}L(s,\chi).$

This is already a useful simplification. We have replaced one large zeta function by a product of special values associated with individual characters. Now apply the class number formula at zero to both fields. Since $K$ is totally imaginary with $g$ pairs of complex embeddings, and $K^+$ is totally real of degree $g$ , both zeta functions vanish to order $g-1$ . Their quotient therefore has a finite nonzero value at zero. The class number formulas give

$\displaystyle \prod_{\chi\text{ odd}}L(0,\chi)=2h^-R_K/(w_KR_{K^+})$ .

For prime-power cyclotomic fields, the regulator ratio is $R_K/R_{K^+}=2^{g-1}$ . Roughly, the units of $K$ are roots of unity times units of the real subfield, and the logarithmic coordinates for each complex conjugate pair carry an extra factor of two. Since there are $g-1$ independent unit directions, the regulator acquires a factor $2^{g-1}$ . The absence of an additional unit-index factor can be checked using the fact that $u/\bar u$ is a root of unity for every unit $u$ , and then adjusting $u$ by a suitable power of $\zeta_q$ to make it real. Finally, $K$ contains exactly $2q$ roots of unity. Substituting the regulator ratio and $w_K=2q$ , we obtain

$\displaystyle \boxed{h^-=\frac{2q}{2^g}\prod_{\chi\text{ odd}}L(0,\chi).}$

This is the relative class number formula. Notice that the regulator, which is usually one of the difficult parts of the class number formula, has disappeared. We are left with special values of Dirichlet $L$ -functions. The next step is to replace those values by finite sums.

Character Sums

Let $\chi$ be a nontrivial Dirichlet character of conductor $f$ . Define its first generalized Bernoulli number by the finite sum

$\displaystyle B_{1,\chi}=\frac{1}{f}\sum_{a=1}^{f}a\chi(a)$ .

The classical special-value formula says that $L(0,\chi)=-B_{1,\chi}$ . The appearance of this sum is not difficult to motivate. Consider the rational function $R_\chi(X)=(\sum_{a=1}^{f}\chi(a)X^a)/(1-X^f)$ . Since $\sum_a\chi(a)=0$ , both the numerator and denominator vanish at $X=1$ . After cancelling their common factor and evaluating, we obtain $R_\chi(1)=-f^{-1}\sum_a a\chi(a)$ . The analytic relation between this rational function and the $L$ -function identifies this finite value with $L(0,\chi)$ . For an even nontrivial character, the sum vanishes. Indeed, replacing $a$ by $f-a$ and using $\chi(-1)=1$ shows that the sum is its own negative. This is another way to see why odd characters naturally appear in our calculation. Substituting the finite sums into the relative class number formula gives

$\displaystyle \boxed{h_r^-=2p^r\prod_{\substack{\chi\bmod p^r\\chi(-1)=-1}}\left(-\frac{B_{1,\chi}}2\right).}$

If a character modulo $p^r$ is induced from a smaller conductor $p^s$ , its generalized Bernoulli number can still be calculated using modulus $p^r$ . This follows by grouping the terms according to their residues modulo $p^s$ . Each residue has $p^{r-s}$ lifts, and the extra terms disappear because the sum of a nontrivial character is zero. Thus every factor in the formula is obtained from a finite sum over residues. The number of summands may be large, but there is nothing mysterious about the quantities being calculated. Now we want only the exponent of $p$ in the class number. The factors of two can be ignored, since $p$ is odd. Taking valuations gives the particularly convenient formula

$\displaystyle \boxed{v_p(h_r^-)=r+\sum_{\chi\text{ odd}}v_p(B_{1,\chi}).}$

The values $B_{1,\chi}$ may belong to a cyclotomic field rather than to $\mathbb Q$ . To speak of their divisibility by $p$ , we choose a prime above $p$ in a field containing the character values. The final product is rational, and its total valuation is an ordinary integer. This is the first major reduction. Instead of calculating the entire class group, or even its order, we can calculate the valuations of explicit finite character sums. We should now see a small example before trying to understand the general pattern.

Take $p=3$ . At conductor $3$ , there is only one odd character, with $\chi(1)=1$ and $\chi(2)=-1$ . Its Bernoulli sum is $B_{1,\chi}=(1-2)/3=-1/3$ . The relative class number formula therefore gives $h_1^-=6(1/6)=1$ . At conductor $9$ , the unit group has order six and is generated by $2$ . Its elements, in the order of successive powers of $2$ , are $1,2,4,8,7,5$ . Put $z=e^{2\pi i/6}$ and define characters $\chi_j$ by $\chi_j(2)=z^j$ . Because $-1\equiv2^3\pmod9$ , the odd characters are $\chi_1,\chi_3,\chi_5$ . For $\chi_1$ , the numerator of $B_{1,\chi_1}$ is $1+2z+4z^2+8z^3+7z^4+5z^5$ . Using $z^3=-1$ , $z^4=-z$ , $z^5=-z^2$ , and $z^2=z-1$ , this reduces to $-6(1+z)$ . Hence $B_{1,\chi_1}=-(2/3)(1+z)$ . The character $\chi_5$ is the complex conjugate of $\chi_1$ , so $B_{1,\chi_5}=-(2/3)(1+\bar z)$ . Their product is $(4/9)(1+z)(1+\bar z)=4/3$ , since $z+\bar z=1$ . The remaining odd character $\chi_3$ comes from conductor three, giving $B_{1,\chi_3}=-1/3$ . Substituting these three values,

$\displaystyle h_2^-=18\left(\frac13\right)\left(\frac16\right)=1.$

The calculation also shows why we need to track denominators carefully. The prefactor $18$ contributes two factors of $3$ . The factor coming from $\chi_3$ removes one, and the product of the other two character factors removes another. Everything cancels. At higher conductors, the number of characters grows rapidly. Directly carrying out such calculations is still possible in principle, but we need a better way to organize them. Before doing that, it is useful to examine what happens at the first level for a general prime $p$ .

Kummer’s criterion and Bernoulli numbers

The ordinary Bernoulli numbers are rational numbers defined by $B_0=1$ and the recurrence $\sum_{j=0}^{m}\binom{m+1}{j}B_j=0$ for $m\ge1$ . The first few are $B_1=-1/2$ , $B_2=1/6$ , $B_4=-1/30$ , and $B_6=1/42$ . The odd Bernoulli numbers beyond $B_1$ vanish. Why should these rational numbers have anything to do with class numbers of cyclotomic fields? The connection comes from a congruence between the generalized Bernoulli sums $B_{1,\chi}$ and ordinary Bernoulli numbers.

Let $\omega$ be the character modulo $p$ whose values are the $(p-1)$ -st roots of unity that reduce to the corresponding residue classes modulo a chosen prime above $p$ . Thus $\omega(a)\equiv a\pmod p$ . Every character modulo $p$ is a power of $\omega$ , and the odd characters are those with odd exponents. For an even integer $k$ satisfying $2\le k\le p-3$ , consider the odd character $\chi=\omega^{k-1}$ . The key congruence is

$\displaystyle B_{1,\omega^{k-1}}\equiv\frac{B_k}{k}\pmod p.$

We can understand the source of this congruence through a finite calculation involving multiplication of residues. Choose an integer $c$ whose residue generates the nonzero residues modulo $p$ . For each $a=1,\ldots,p-1$ , write the ordinary division algorithm as $ca=r_a+pq_a$ , where $1\le r_a\le p-1$ and $q_a=\lfloor ca/p\rfloor$ . Multiplication by $c$ permutes the nonzero residues, so the numbers $r_a$ run through $1,\ldots,p-1$ . Apply this permutation to the sum defining $B_{1,\chi}$ . Since $\chi(r_a)=\chi(c)\chi(a)$ , we have $pB_{1,\chi}=\chi(c)\sum_a(ca-pq_a)\chi(a)$ . Rearranging gives the exact identity

$\displaystyle (c\chi(c)-1)B_{1,\chi}=\chi(c)\sum_{a=1}^{p-1}\left\lfloor\frac{ca}{p}\right\rfloor\chi(a).$

This identity is useful because the denominator $p$ has disappeared. We have replaced the generalized Bernoulli number by a sum involving ordinary integers and floor functions. Now reduce modulo $p$ , with $\chi=\omega^{k-1}$ . Since $\chi(a)\equiv a^{k-1}$ , the right side becomes an ordinary power sum with floor weights. Also, $c\chi(c)-1\equiv c^k-1\pmod p$ . This is nonzero in our range of $k$ , because $c$ generates the nonzero residues and $k$ is not a multiple of $p-1$ . We can calculate the same floor sum from ordinary power sums. Write $S_k=\sum_{a=1}^{p-1}a^k$ . The finite power-sum formula, which follows from the defining recurrence of the Bernoulli numbers, gives $S_k\equiv pB_k\pmod{p^2}$ for $2\le k\le p-3$ . On the other hand, using $r_a=ca-pq_a$ and expanding the $k$ -th power modulo $p^2$ , we obtain $r_a^k\equiv c^ka^k-kpc^{k-1}a^{k-1}q_a\pmod{p^2}$ . Sum over $a$ . Since the $r_a$ permute the residues, their power sum is also $S_k$ . Therefore $(c^k-1)S_k\equiv kpc^{k-1}\sum_a a^{k-1}q_a\pmod{p^2}$ . Substituting $S_k\equiv pB_k$ and cancelling $p$ gives exactly the same expression for $B_k/k$ that we obtained for $B_{1,\omega^{k-1}}$ . This proves the congruence.

There is one exceptional odd character, namely $\omega^{-1}=\omega^{p-2}$ . Its Bernoulli sum has valuation $-1$ , because $pB_{1,\omega^{-1}}=\sum_{a=1}^{p-1}a\omega(a)^{-1}\equiv p-1\equiv-1\pmod p$ . This negative valuation cancels the explicit factor of $p$ in the class number formula. For the remaining odd characters, the congruence with ordinary Bernoulli numbers determines whether their factors are divisible by $p$ . We obtain Kummer’s criterion:

$\displaystyle p\mid h_1^-\quad\Longleftrightarrow\quad p\text{ divides the numerator of some }B_k,\quad k=2,4,\ldots,p-3.$

This is a remarkable simplification. We started with the class number of a field of degree $p-1$ , but we can test its divisibility by $p$ using a finite list of rational Bernoulli numbers. For example, $37$ is an irregular prime. The numerator of $B_{32}=-7709321041217/510$ is divisible by $37$ . Therefore the character $\omega^{31}$ contributes a factor divisible by $37$ in the class number formula, and $37\mid h_1^-$ . The important point for us is not the numerical example itself, but the method. We have related class numbers to finite character sums, and then related those character sums to ordinary power sums. We should now ask whether similar arithmetic relationships persist when the conductor increases.

At conductor $p^r$ , the unit group $(\mathbb Z/p^r\mathbb Z)^\times$ has order $(p-1)p^{r-1}$ . For odd $p$ , it is cyclic, and we can think of it as having two components: one of order $p-1$ , already present at conductor $p$ , and another of order $p^{r-1}$ , generated by $1+p$ . This allows us to organize the characters into families. Every odd character can be written as $\chi=\omega^{k-1}\psi$ , where $k=2,4,\ldots,p-1$ and $\psi$ is a character of the subgroup generated by $1+p$ . Such a character $\psi$ is determined by a $p^{r-1}$ -st root of unity $z$ , through $\psi(1+p)=z$ .

Thus there are only $(p-1)/2$ choices for the first part of an odd character, while the second part varies through characters of increasing $p$ -power order. This is the first indication that the higher-level calculations should be organized into a fixed number of families rather than treated as unrelated sums. For each character, we still have the explicit formula $B_{1,\chi}=p^{-r}\sum_{a\bmod p^r}a\chi(a)$ , with the sum over units. We could evaluate every sum separately, but the number of characters increases by a factor of $p$ each time. More importantly, doing so would hide the relations between consecutive conductors.

We need a way to express the character sums through integer tables that behave predictably when the modulus changes. The floor-sum identity from the previous section gives precisely such a construction. Choose a positive odd integer $c$ , coprime to $p$ , whose residue generates the nonzero residues modulo $p$ . For each unit $a$ modulo $p^r$ , define

$\displaystyle d_r(a)=\left\lfloor\frac{ca}{p^r}\right\rfloor-\frac{c-1}{2}.$

These are ordinary integers. The constant $(c-1)/2$ has been subtracted because it makes the tables at consecutive levels satisfy a simple summation identity. It does not change the nontrivial character sums, since the sum of any nontrivial character over the units is zero. The exact permutation argument from the preceding section works equally well modulo $p^r$ . It gives

$\displaystyle (c\chi(c)-1)B_{1,\chi}=\chi(c)\sum_{a\bmod p^r}d_r(a)\chi(a)$ ,

where the sum is over units. Thus the generalized Bernoulli number is obtained by taking a finite character sum of the integer table $d_r$ , followed by multiplication by the explicit factor $\chi(c)/(c\chi(c)-1)$ . Now consider how the table changes when we pass from modulus $p^r$ to modulus $p^{r+1}$ . Every unit $b$ modulo $p^r$ has exactly $p$ lifts, namely $b,b+p^r,\ldots,b+(p-1)p^r$ . A direct floor calculation shows

$\displaystyle \sum_{j=0}^{p-1}d_{r+1}(b+jp^r)=d_r(b).$

It is worth seeing why this identity holds. Put $q=p^r$ and write $cb=qD+s$ , where $0<s<q$ . The sum of the unadjusted floor terms at the higher level is $\sum_{j=0}^{p-1}\lfloor(D+s/q+cj)/p\rfloor$ . Since $c$ is coprime to $p$ , the residues of $D+cj$ run through all residue classes modulo $p$ . Summing the integer quotients gives $D+(p-1)(c-1)/2$ . Subtracting $p(c-1)/2$ leaves $D-(c-1)/2=d_r(b)$ , exactly as claimed. This is a very useful finite identity. It says that the old table can be recovered by adding the entries in the new table lying above each old residue class. Thus the tables at different conductors are not independent.

Notice that we have not introduced any infinite object. At each level we have a finite list of integers, and the relationship between two consecutive lists is an exact identity. We can check it simply by integer division. For the character families $\chi=\omega^{k-1}\psi$ with $2\le k\le p-3$ , the factor $c\chi(c)-1$ is a unit at $p$ . Indeed, modulo a prime above $p$ , every $p$ -power root of unity reduces to one, while $\omega(c)^{k-1}$ reduces to $c^{k-1}$ . Thus $c\chi(c)-1$ reduces to $c^k-1$ , which is nonzero. Consequently, in these families, the valuation of $B_{1,\chi}$ is exactly the valuation of the corresponding integer-table character sum. The exceptional family $k=p-1$ requires separate treatment because its denominator is divisible by $p$ , just as the exceptional character did at the first level. This gives a much better way to organize the class-number calculation. Rather than carrying the denominator $p^r$ through every Bernoulli sum, we work with compatible tables of integers and evaluate their finite Fourier transforms.

p-adic L-function

We can now give a concrete interpretation of the terminology without introducing the full theory. Fix one character family, say $\omega^{k-1}$ . Every unit $a$ modulo $p^r$ can be written uniquely as its component of order dividing $p-1$ times a power $(1+p)^{t_a}$ , where $0\le t_a<p^{r-1}$ . The character $\psi$ determined by $\psi(1+p)=z$ then satisfies $\psi(a)=z^{t_a}$ . Form the finite polynomial

$\displaystyle P_{r,k}(Z)=\sum_{\substack{a\bmod p^r\\p\nmid a}}d_r(a)\omega(a)^{k-1}Z^{t_a}.$

If we evaluate this polynomial at a $p^{r-1}$ -st root of unity $z$ , we obtain exactly the character sum $\sum_a d_r(a)\omega(a)^{k-1}\psi(a)$ . Up to the explicit factor $\chi(c)/(c\chi(c)-1)$ , this is the generalized Bernoulli value appearing in the class number formula. The compatibility of the tables $d_r$ gives a corresponding compatibility of the polynomials. Reducing the exponents modulo $p^{r-1}$ and adding the coefficients lying above each residue recovers the polynomial from the preceding level. Thus we can calculate character sums at many conductors using one family of compatible finite polynomials. It is convenient to replace $Z$ by $1+T$ . A character of order $p^j$ is then evaluated at $T=\zeta_{p^j}-1$ . This is the source of the expressions in powers of $T$ that appear in Iwasawa theory.

A cyclotomic $p$ -adic $L$ -function is a way of packaging these compatible finite character-sum calculations into one arithmetic object. At any prescribed modulus $p^m$ , we can work with finite coefficient calculations and congruences. The compatibility ensures that the calculations at different conductors agree when reduced to a common level. For our purpose, the important question is what this organization tells us about valuations. If the relevant character sums can be controlled by a fixed polynomial with a particular divisibility pattern, the growth of their products becomes easy to calculate. We can prove this last fact completely by elementary arithmetic, without needing the general theory of $p$ -adic $L$ -functions.

Growth Rate

Consider a polynomial of the form

$\displaystyle P(T)=T^\lambda+p b_{\lambda-1}T^{\lambda-1}+\cdots+p b_0,$

where the coefficients $b_i$ are integers. Its leading coefficient is one, and every lower coefficient is divisible by $p$ . We want to evaluate it at $T=\zeta-1$ , where $\zeta$ is a primitive root of unity of order $p^j$ . First we need to know the valuation of $\zeta-1$ . The minimal polynomial of $\zeta$ is $\Phi_{p^j}(X)=1+X^{p^{j-1}}+\cdots+X^{(p-1)p^{j-1}}$ . Substituting $X=1+T$ gives an Eisenstein polynomial at $p$ : its constant term is $p$ , its leading coefficient is one, and every other coefficient is divisible by $p$ . There is an especially concrete way to get the valuation. We have the exact identity $\prod_{p\nmid a}(1-\zeta^a)=p$ , where $a$ runs through the nonzero residues modulo $p^j$ . The factors $1-\zeta^a$ all have the same valuation, because $(1-\zeta^a)/(1-\zeta)$ is a unit. There are $\varphi(p^j)=p^{j-1}(p-1)$ factors, and their product is $p$ . Therefore

$\displaystyle v_p(\zeta-1)=\frac1{\varphi(p^j)}.$

Here the valuation is normalized so that $v_p(p)=1$ . It may take fractional values on algebraic numbers, but the valuation of a rational integer is the usual exponent of $p$ . Now evaluate $P$ at $T=\zeta-1$ . The leading term $T^\lambda$ has valuation $\lambda/\varphi(p^j)$ . Every other term is divisible by $p$ , so its valuation is at least one. Once $j$ is large enough that $\varphi(p^j)>\lambda$ , the leading term has valuation less than one. It is therefore strictly smaller in valuation than every other term, and no cancellation is possible. Hence, for every sufficiently large $j$ ,

$\displaystyle v_p(P(\zeta-1))=\frac{\lambda}{\varphi(p^j)}.$

Now multiply these values over all primitive roots $\zeta$ of order $p^j$ . There are $\varphi(p^j)$ such roots, and each contributes the same valuation. Therefore the valuation of their product is exactly $\lambda$ . This is the source of the linear term in Iwasawa’s formula. Each sufficiently high conductor level contributes exactly $\lambda$ to the total valuation. If we include all character levels up to $p^n$ , we obtain a total contribution of $\lambda n$ , apart from the finitely many initial levels. Those initial contributions are absorbed into a constant.

Now suppose that the polynomial expression also contains an overall factor $p^\mu$ . Every character evaluation then contributes an additional $\mu$ to the valuation. At conductor $p^j$ , there are $\varphi(p^j)$ primitive characters of the relevant cyclic component, so the additional contribution is $\mu\varphi(p^j)$ . Summing from $j=1$ to $n$ gives $\mu\sum_{j=1}^{n}\varphi(p^j)=\mu(p^n-1)$ . Thus a common factor $p^\mu$ produces growth proportional to $p^n$ , while a polynomial of degree $\lambda$ of the form above produces growth proportional to $n$ . The remaining initial contributions are constant. We obtain the numerical pattern

$\displaystyle \mu p^n+\lambda n+\nu.$

This calculation is the main point. We have not proved that the character sums from the class number formula are always controlled by such a fixed polynomial. That requires a deeper theorem. But we have proved exactly why this kind of polynomial control produces the two characteristic growth terms in Iwasawa’s formula.

Take $p=3$ and consider the polynomial $P(T)=T+3$ . For each $n\ge1$ , multiply $P(\zeta-1)$ over all nontrivial roots of unity whose orders divide $3^n$ . Call the resulting integer $R_n$ . Since $P(\zeta-1)=\zeta+2$ , the factorization of $X^{3^n}-1$ gives the exact formula

$\displaystyle R_n=\prod_{\substack{\zeta^{3^n}=1\\zeta\ne1}}(\zeta+2)=\frac{1-(-2)^{3^n}}3.$

For $n=1$ , we get $R_1=(1+8)/3=3$ , so $v_3(R_1)=1$ . For $n=2$ , we obtain $R_2=(1+512)/3=171=3^2\cdot19$ , so $v_3(R_2)=2$ . In fact, $v_3(R_n)=n$ for every $n\ge1$ . We can prove this directly from the binomial theorem. Suppose $u\equiv1\pmod3$ and $v_3(u-1)=a\ge1$ . Writing $u=1+3^av$ with $3\nmid v$ , the expansion of $u^3-1$ has first term $3^{a+1}v$ , while all remaining terms have strictly larger valuations. Hence $v_3(u^3-1)=a+1$ . Apply this repeatedly to $u=-2$ , for which $v_3(u-1)=1$ . We obtain $v_3((-2)^{3^n}-1)=n+1$ . Dividing by $3$ gives the claimed formula $v_3(R_n)=n$ . This is a particularly simple model of the $\lambda$ -term. Every time we increase $n$ by one, the valuation increases by exactly one. If instead we multiplied every factor $P(\zeta-1)$ by $3^\mu$ , the product would acquire the additional valuation $\mu(3^n-1)$ , producing the exponential term as well.

Nothing in this example requires an infinite construction. We are calculating the valuations of explicitly defined integers. We can now put the pieces together. The relative class number formula gives $v_p(h_r^-)=r+\sum_{\chi\text{ odd}}v_p(B_{1,\chi})$ . We then organize the odd characters into families $\omega^{k-1}\psi$ , where $\psi$ runs through characters of $p$ -power order. The floor-sum calculation replaces the generalized Bernoulli values, apart from explicit factors, by Fourier transforms of compatible integer tables. Evaluating those transforms at characters of order $p^j$ amounts to evaluating finite polynomials at $\zeta_{p^j}-1$ .

If these polynomial evaluations can be controlled by a fixed factor $p^\mu$ and a fixed polynomial of degree $\lambda$ whose lower coefficients are divisible by $p$ , this gives exactly the type of valuation growth appearing in Iwasawa’s theorem. The factor $p^\mu$ contributes once for each character, producing the exponential term. The degree $\lambda$ contributes the same amount at every sufficiently high conductor, producing the linear term. The finite initial calculations contribute the constant.

There are two important qualifications. First, the finite compatibility identities do not by themselves prove that the relevant arithmetic is controlled by a fixed polynomial in the required way. Establishing the needed control is a substantial part of Iwasawa theory. Second, the odd-character calculation directly concerns the relative class numbers $h_r^-$ , not the real class numbers $h_r^+$ . The full theorem must account for both parts.

For cyclotomic fields, there is an additional deep result. Ferrero and Washington proved that the invariant $\mu$ vanishes for cyclotomic extensions of abelian number fields. Consequently, in our setting, the exponent of $p$ in the full class number is eventually linear in the level:

$\displaystyle v_p(h_r)=\lambda(r-1)+\nu\qquad\text{for all sufficiently large }r.$

The vanishing of $\mu$ does not follow from the elementary Bernoulli congruences we proved. Those congruences explain the first-level divisibility of relative class numbers; the Ferrero–Washington theorem supplies a much stronger statement about all conductor levels. The main purpose of the calculations here has been to understand the numerical shape of Iwasawa’s formula. We began with the class number formula, which comes from counting ideals and understanding the geometry of units. For cyclotomic fields, the relative class number became a product of special $L$ -values. Those special values became finite character sums. Multiplication of residues related the character sums to ordinary Bernoulli numbers, and the compatibility between consecutive conductors let us organize the higher-level sums into polynomial evaluations at roots of unity. The final valuation calculation is elementary: a common factor $p^\mu$ contributes once per character, while a distinguished polynomial of degree $\lambda$ contributes $\lambda$ at each sufficiently high conductor. This explains the exponential and linear terms in the growth formula.

We have reached the point where the relative class number formula gives explicit finite character sums, and our calculation with roots of unity explains why terms of the form $\mu p^n+\lambda n+\nu$ appear. In fact, for the relative class numbers, we can go further than merely suggesting this pattern. The compatible integer tables constructed earlier give finite polynomials whose values at $T=\zeta_{p^j}-1$ are, up to explicit factors, the generalized Bernoulli numbers in the class number formula. Once we establish that their coefficients can be organized consistently at different conductors, we can remove the largest common power $p^\mu$ and look for the first coefficient not divisible by $p$ . Suppose it occurs at degree $\lambda$ . For sufficiently large $j$ , the corresponding term has strictly smaller valuation than all the others when evaluated at $\zeta_{p^j}-1$ . Its valuation is exactly $\lambda/\varphi(p^j)$ , in addition to the common contribution $\mu$ . Summing over the $\varphi(p^j)$ characters at that conductor gives $\mu\varphi(p^j)+\lambda$ . After treating the exceptional character family and the normalizing factors separately, this gives the eventual growth formula for $v_p(h_n^-)$ . Thus, with a further algebraic argument establishing the required compatibility of the coefficients, the finite character-sum method really can prove the relative class-number growth formula. We do not need to assume that the ideal class groups themselves have some previously established polynomial structure.

The remaining difficulty is the real class number $h_n^+$ . Since the ordinary class number is $h_n=h_n^+h_n^-$ , controlling only the relative part is not enough. Can we do something similar with the real subfields, perhaps using units instead of the odd-character Bernoulli sums? There is a natural reason to try. If $q=p^n$ and $K_n^+=\mathbb Q(\zeta_q+\zeta_q^{-1})$ , we can write down many explicit units using the expressions $(1-\zeta_q^a)/(1-\zeta_q)$ . After multiplying by a suitable power of $\zeta_q$ , these become real units. Let $C_n^+$ be the group generated by these cyclotomic units and $-1$ , and let $E_n^+$ be the group of all units of $K_n^+$ . The classical cyclotomic-unit index formula gives, for odd prime-power conductor,

$\displaystyle h_n^+=[E_n^+: C_n^+].$

This formula can be understood through the same class number formula we used earlier. The regulator of a group of units is the determinant of the logarithms of its independent generators under the real embeddings. For cyclotomic units, that determinant can be calculated explicitly: the entries involve $\log|1-\zeta_q^a|$ , and taking finite Fourier transforms separates the determinant into factors involving the even Dirichlet characters. Comparing the result with the analytic class number formula for $K_n^+$ gives the index formula. Thus the idea of using units is not merely speculative. It really does turn the real class number into a concrete arithmetic question: how many additional units are needed, beyond the explicitly known cyclotomic units, to generate the full unit group? We can also find exact relations between the cyclotomic units at consecutive conductors. For example, if $p\nmid a$ , the elementary factorization of $X^p-1$ gives

$\displaystyle N_{\mathbb Q(\zeta_{p^{n+1}})/\mathbb Q(\zeta_{p^n})}\left(\frac{1-\zeta_{p^{n+1}}^a}{1-\zeta_{p^{n+1}}}\right)=\frac{1-\zeta_{p^n}^a}{1-\zeta_{p^n}}.$

The corresponding real units satisfy the same compatibility. So this initially looks very promising: we have an explicit subgroup of units, an exact formula relating its index to the class number, and norm identities relating its generators at different levels. Could we now repeat the polynomial argument and prove Iwasawa’s formula for $h_n^+$ as well? Here we encounter the real difficulty. Knowing the cyclotomic units and their norm relations does not tell us how they sit inside the full unit group. The index $[E_n^+: C_n^+]$ depends on all the units, not only on the ones we can write down explicitly. Although the cyclotomic units generate a subgroup of finite index, we still have to determine how divisible their exponent vectors are inside the lattice of all units. This is already visible in a small example. For $p=5$ , the real field is $\mathbb Q(\sqrt5)$ , and the cyclotomic unit $1+\zeta_5+\zeta_5^{-1}=(1+\sqrt5)/2$ is a fundamental unit, so the index is one. But at conductor $25$ , the real field has degree ten and unit rank nine. We can explicitly construct nine independent cyclotomic units, but showing that they generate the full unit lattice up to a particular power of five is a different problem. Their explicit formulas and norm identities alone do not answer it. We could try to compare the quotients $E_{n+1}^+/C_{n+1}^+$ and $E_n^+/C_n^+$ using norms, just as we compared the character sums. But to obtain a useful relation between these quotients, we would have to know which units in the lower field are norms of units in the higher field, and whether any new divisibility relations appear among the cyclotomic units when we enlarge the field. A norm identity for the known generators does not establish either fact. In particular, it does not show that the index of the cyclotomic units is governed by a fixed number of relations independent of the conductor. That is precisely the additional structural statement we would need before applying our polynomial valuation argument. We can therefore locate the remaining gap quite precisely. For the relative class numbers, the class number formula has eliminated the unknown regulator and replaced the problem by finite character sums. Their compatibility allows us to study their valuations directly. For the real class numbers, the class number formula replaces the problem by an index of unit lattices, but only one of those lattices—the cyclotomic-unit lattice—is explicitly under control. To finish the proof by this route, we would need a further argument showing that the missing units, or equivalently the indices $[E_n^+, C_n^+]$ , satisfy comparably uniform relations at successive conductors. The formulas we have proved do not yet supply that argument. Thus the elementary character-sum calculations give a genuine proof strategy for the relative class-number growth, while the full Iwasawa theorem requires one more substantial idea to control the real unit indices. It is this precise difficulty, rather than any mystery about the appearance of $\mu p^n+\lambda n+\nu$ , that remains unresolved by our calculations.

Proof of full class number theorem

We have seen that the relative class number can be studied using finite character sums, while the real class number can be expressed as the index of the cyclotomic units in the full unit group. The difficulty with the second approach is that, although the cyclotomic units and their norm relations are explicit, we do not know how all the other units behave as the conductor increases. There is another way to proceed. Instead of trying to calculate the missing units, we can study the ideal class groups themselves, retaining their generators and relations rather than just their orders. This will also have the advantage of treating the full class number, without separating the real and relative parts. Let $p$ be an odd prime, put $K_n=\mathbb Q(\zeta_{p^{n+1}})$ , and let $A_n$ be the subgroup of its ideal class group consisting of elements of $p$ -power order. Thus $|A_n|=p^{e_n}$ , where $e_n=v_p(h(K_n))$ . Our aim is to prove that, for sufficiently large $n$ , there are fixed integers $\mu,\lambda,\nu$ for which

$\displaystyle e_n=\mu p^n+\lambda n+\nu.$

We will see that the real issue is not finding a polynomial that happens to produce this formula. It is showing that the relations among the actual ideal classes, at all conductors, can be reduced to finitely many persistent relations. Once this is established, the valuation calculation is quite elementary.

The fields $K_n$ form a tower, and their Galois groups contain a distinguished family of automorphisms. Let $\gamma$ act at level $n$ by $\gamma(\zeta_{p^{n+1}})=\zeta_{p^{n+1}}^{1+p}$ . Since $1+p$ has order $p^n$ modulo $p^{n+1}$ , we have $\gamma^{p^n}=1$ on $K_n$ . The automorphism acts on ideal classes by sending the class represented by an ideal $\mathfrak a$ to the class represented by $\gamma(\mathfrak a)$ . We write $T=\gamma-1$ , so that $Tc=\gamma c-c$ in additive notation for the class group. If $m\ge n$ , the subgroup fixing $K_n$ is generated by $\gamma^{p^n}$ .

There is also an ideal norm $N_{m,n}\to A_n$ . The norm of an ideal does not change if we first replace that ideal by one of its conjugates over $K_n$ . Therefore $N_{m,n}((\gamma^{p^n}-1)c)=0$ for every class $c\in A_m$ . The norm consequently factors through $A_m/(\gamma^{p^n}-1)A_m$ . This much is immediate from the definition of the ideal norm. What we need is the converse: if a class has trivial norm, must it be a product of differences between conjugate classes? For our cyclotomic tower the answer is yes:

$\displaystyle A_m/(\gamma^{p^n}-1)A_m\cong A_n.$

This is the norm-control relation, and it is the decisive arithmetic input. It deserves care because it does not follow from ordinary ideal factorization alone.

Let $E/F$ be one of the extensions $K_m/K_n$ , and let $G={\text{Gal}}(E/F)=\langle\sigma\rangle$ . It is cyclic of order $p^{m-n}$ . The only ramified prime of $F$ is the prime above $p$ , and it is totally ramified. In particular, every nontrivial intermediate field of $E/F$ is ramified at this prime. To prove the norm-control relation, we need the correspondence between ideal classes and everywhere-unramified abelian extensions. Let $H_E$ be the maximal everywhere-unramified abelian extension of $E$ of $p$ -power degree. The Hilbert class field theorem identifies ${\text{Gal}}(H_E/E)$ with $A_E$ , through the rule sending prime ideal classes to their Frobenius automorphisms. Principal ideals give the identity, and every element of the Galois group arises from an ideal class. This correspondence is a substantial arithmetic theorem; we are using it here, not proving all of class field theory from scratch. The field $H_E$ is also Galois over $F$ . Put $\mathcal G={\text{Gal}}(H_E/F)$ . There is an exact sequence $1\to A_E\to\mathcal G\to G\to1$ . Let $I$ be the inertia group at a prime of $H_E$ above the unique ramified prime of $F$ . Because $H_E/E$ is unramified and $E/F$ is totally ramified, the restriction map $I\to G$ is an isomorphism. Thus $\mathcal G=A_E\rtimes I$ : every element is a product of an element of $A_E$ and an element of the cyclic inertia group. Now $A_E$ is abelian, and $I$ is cyclic. If $\widetilde\sigma$ generates $I$ , the commutators of $\mathcal G$ are generated by $\widetilde\sigma a\widetilde\sigma^{-1}a^{-1}$ . Under the identification with ideal classes, this is precisely $(\sigma-1)a$ . Therefore the commutator subgroup is $(\sigma-1)A_E$ . We can now make the whole group abelian and then remove its inertia subgroup. The resulting quotient is $A_E/(\sigma-1)A_E$ . Its fixed field is an abelian extension of $F$ unramified at the distinguished prime, since we have killed inertia there. It is unramified at every other prime as well, because no other prime ramifies in $E/F$ or in $H_E/E$ . Why is this the largest such extension of $F$ ? Let $H_F$ be the maximal unramified abelian $p$ -extension of $F$ . The compositum $EH_F/E$ is unramified, so it lies in $H_E$ . Also $E\cap H_F=F$ , because a nontrivial intersection would be simultaneously ramified and unramified at the distinguished prime. Thus $H_F$ is contained in the extension obtained by killing inertia. Since that extension is itself unramified and abelian over $F$ , it must equal $H_F$ .

We therefore obtain $A_E/(\sigma-1)A_E\cong {\text{Gal}}(H_F/F)\cong A_F$ . Compatibility of the Frobenius correspondence with ideal norms identifies this isomorphism with the norm map. This proves the required relation. Notice exactly what was used. The group calculation was elementary, but the existence of $H_E$ with Galois group equal to the ideal class group was the arithmetic input. Without that correspondence, the argument would not yet be a proof of norm control.

Finite generators, relations

We can now work almost entirely with finite groups and their relations. At level $n$ , the relation $\gamma^{p^n}=1$ becomes $(1+T)^{p^n}-1=0$ . Write $\omega_n(T)=(1+T)^{p^n}-1$ . The norm-control theorem says that $A_m/\omega_n(T)A_m\cong A_n$ for every $m\ge n$ . Suppose the initial $p$ -class group satisfies $A_0/pA_0\cong(\mathbb F_p)^d$ . Choose $d$ classes $c_1,\ldots,c_d$ whose images form a basis. Since the norm maps are surjective, lift them to classes $x_1,\ldots,x_d$ at any higher level $A_n$ . Let $B$ be the subgroup generated by these classes and all their conjugates under $\gamma$ . We claim that $B=A_n$ .

Put $C=A_n/B$ . By norm control, $A_n/TA_n\cong A_0$ , and the chosen classes generate this quotient. Hence $C=TC$ . Reducing modulo $p$ , the relation $(1+T)^{p^n}=1$ becomes $T^{p^n}=0$ , because all intermediate binomial coefficients are divisible by $p$ . But $C=TC$ implies $C/pC=T(C/pC)$ . Repeating this equality gives $C/pC=T^{p^n}(C/pC)=0$ . Therefore $C=pC$ . Since $C$ is a finite $p$ -group, repeated multiplication by $p$ eventually gives zero. Thus $C=0$ , proving the claim. This gives our first strong uniformity statement:

$\displaystyle \boxed{\text{Every }A_n\text{ is generated by at most }d\text{ classes and their Galois conjugates.}}$

The group $A_n$ might be enormous, but we never need more than $d$ generators if we are allowed to apply the Galois action. In particular, if $A_0=0$ , all the higher $A_n$ vanish. This is why a regular prime cannot suddenly begin dividing the class number at a higher conductor. Let us make the finite presentation explicit. Choose actual fractional ideals $\mathfrak a_1,\ldots,\mathfrak a_d$ representing the selected generators at level $n$ . Every element of $A_n$ is represented by a product of their conjugates, and therefore by an expression of the form $\prod_{i=1}^{d}\prod_{j=0}^{p^n-1}\gamma^j(\mathfrak a_i)^{c_{ij}}$ . Such an expression represents the trivial class exactly when it is a principal fractional ideal, say $(\alpha)$ . These principal-ideal identities are precisely the relations among the generators. For example, if $\mathfrak a_1^p\gamma(\mathfrak a_2)\gamma^2(\mathfrak a_1)^{-1}=(\alpha)$ , then, in additive notation for the class group, we obtain $(p-T^2-2T)x_1+(1+T)x_2=0$ . We have used $\gamma=1+T$ to write the relation as a polynomial expression.

Thus every relation can be recorded as a vector of polynomials in $T$ . At any fixed conductor there are only finitely many possibilities, since both the class group and the Galois group are finite. We could, in principle, enumerate every vector of exponents modulo a sufficiently large power of $p$ and test whether the corresponding ideal product is principal. This is not an efficient way to calculate a large class group, but it shows concretely what the relation matrix represents. The more useful question is whether all these finite relation matrices can be controlled by one fixed collection of relations as the conductor changes. Since the norm maps are surjective, we can choose the generators compatibly: a generator at level $n+1$ is chosen so that its norm is the generator at level $n$ . We then consider families of classes whose components agree under all the norm maps. Call the collection of these families $X$ . A coefficient acting on such a family is specified by its residue at every finite conductor. The resulting coefficient ring is $\Lambda=\mathbb Z_p[[T]]$ . This notation does not require us to perform an infinite numerical calculation: to work modulo $p^a$ and $T^b$ , we only retain finitely many coefficients. At level $n$ , the additional relation is $\omega_n(T)=0$ . The norm-control identities give

$\displaystyle \boxed{X/\omega_n(T)X\cong A_n,\qquad X/TX\cong A_0.}$

To justify the first identity, suppose a compatible family maps to zero in $A_n$ . At every higher finite level, norm control says that its component is divisible by $\omega_n(T)$ . We need to choose these divisors compatibly. For any fixed finite collection of levels, this is a finite system of equations. Its solution sets are finite and nonempty, and the images of solutions from higher levels eventually stabilize. Choosing solutions from these stabilized images produces a compatible family of divisors. Thus the original family lies in $\omega_n(T)X$ , proving the identity without appealing to an unexplained exactness assertion. The finite-generation argument from the preceding section now shows that the compatible families $x_1,\ldots,x_d$ generate $X$ over $\Lambda$ . At each finite level, every class is a polynomial combination of their components. The coefficient choices at different levels can again be made compatible by stabilization of finite solution sets.

Why are there only finitely many relations? This is an important step that is often compressed into the statement that $\Lambda$ is a Noetherian ring. We can see directly why it is true. Since $X$ has $d$ generators, there is a map $\Lambda^d\to X$ sending the standard basis vectors to the chosen families. Let $R$ be its kernel. An element $(a_1(T),\ldots,a_d(T))$ belongs to $R$ precisely when $\sum_i a_i(T)x_i=0$ . Thus $R$ is exactly the collection of all compatible relations. We want to prove that finitely many of these vectors generate every other relation. First consider an ideal $I$ of $\mathbb Z_p[[T]]$ . For each $j\ge0$ , let $I_j$ be the set of coefficients that can occur at $T^j$ in elements of $I$ whose lower coefficients are all zero. Each $I_j$ is an ideal of $\mathbb Z_p$ . Moreover, $I_j\subseteq I_{j+1}$ , since multiplying an expression by $T$ shifts its first nonzero coefficient up by one degree. But every ideal of $\mathbb Z_p$ is of the form $p^a\mathbb Z_p$ , and an ascending chain of such ideals must eventually stabilize. Therefore there is some $N$ for which $I_j=I_N$ for all $j\ge N$ . For each $j\le N$ , choose an element $f_j(T)\in I$ whose first nonzero coefficient occurs at degree $j$ and generates $I_j$ ; omit an index if $I_j=0$ . We claim that these finitely many elements generate $I$ . Take any $f(T)\in I$ . Its constant coefficient lies in $I_0$ , so subtract a suitable multiple of $f_0$ to eliminate it. The remainder has zero constant coefficient. Its coefficient of $T$ lies in $I_1$ , so eliminate it using $f_1$ . Continue through degree $N$ . Beyond that degree, the leading-coefficient ideal no longer changes, so every subsequent coefficient can be eliminated using a suitable multiple of $T^{j-N}f_N(T)$ . Collecting the successive multiples gives an identity expressing $f(T)$ as a combination of the finitely many $f_j(T)$ , with power-series coefficients. The identity is exact because the coefficient of every power of $T$ has been matched. Thus every ideal of $\Lambda$ is finitely generated.

The same reasoning applies to submodules of $\Lambda^d$ . Project a submodule onto its first coordinate. Its image is an ideal, so it has finitely many generators. Lift those generators to vectors in the submodule. The remaining vectors have first coordinate zero and form a submodule of $\Lambda^{d-1}$ . Proceed by induction on $d$ . Hence our relation module $R$ has finitely many generators. In concrete terms, there are finitely many relation vectors $r_1(T),\ldots,r_s(T)$ such that every principal-ideal relation among compatible class families is a combination of them. We can arrange these vectors as the columns of a finite matrix and write

$\displaystyle \boxed{\Lambda^s\xrightarrow{;B(T);}\Lambda^d\longrightarrow X\longrightarrow0.}$

This is a finite presentation of the entire system. The entries of $B(T)$ are the coefficients of actual relations among the chosen class families. We have now justified both the finite number of generators and the finite number of relations, rather than assuming either. A finite presentation alone does not imply Iwasawa’s growth formula. For example, a system with no relations would have a free copy of $\Lambda$ , and its quotient by $\omega_n(T)$ would have infinitely many elements. The actual class groups are finite, so something must prevent this. We can construct a nonzero relation directly from the first class group. Since $A_0$ is finite, choose $b$ such that $p^bA_0=0$ . Because $X/TX\cong A_0$ , we have $p^bX\subseteq TX$ . Thus for each generator $x_i$ there exists $y_i\in X$ satisfying $p^bx_i=Ty_i$ . Express each $y_i$ in terms of the chosen generators, say $y_i=\sum_j a_{ij}(T)x_j$ . The relations become $p^bx_i-T\sum_j a_{ij}(T)x_j=0$ . Put $A(T)=(a_{ij}(T))$ and form the square matrix $M(T)=p^bI-TA(T)$ . Since this matrix annihilates the column of generators, multiplying by its adjugate shows that its determinant annihilates every generator. The determinant is not zero, because its constant term is $p^{bd}$ . We therefore have

$\displaystyle \boxed{D(T)X=0,\qquad D(T)=\det(p^bI-TA(T)),\qquad D(0)=p^{bd}\ne0.}$

This is the first nontrivial global relation we wanted. We have obtained it from the finiteness of $A_0$ , not from an assumed characteristic polynomial. However, a common annihilating expression is not the same thing as a complete description of the group. If $D(T)=p$ , for instance, both $\Lambda/(p)$ and $\Lambda/(p)\oplus\Lambda/(p)$ are annihilated by $D$ , but their finite-level quotients have orders $p^{p^n}$ and $p^{2p^n}$ , respectively. To calculate the class numbers, we must retain the multiplicities of the independent relations. This is why the finite relation matrix, rather than just its common annihilator, matters.

Let us first understand an individual nonzero expression $D(T)=a_0+a_1T+a_2T^2+\cdots$ . Remove the largest common power $p^\mu$ dividing all its coefficients, writing $D(T)=p^\mu G(T)$ . Let $s$ be the first degree at which the coefficient of $G(T)$ is not divisible by $p$ . Modulo $p$ , we can then write $G(T)=T^sV(T)$ , where $V(0)$ is nonzero. We would like to remove the invertible factor $V(T)$ , leaving a monic polynomial of degree $s$ . This can be done coefficient by coefficient. Modulo $p$ , divide $T^s$ by $G(T)$ , obtaining a quotient and zero remainder. Suppose we have performed the division correctly modulo $p^a$ , with remainder of degree less than $s$ . The remaining error is divisible by $p^a$ . Divide that error by $p^a$ , solve the corresponding division problem modulo $p$ , and use the result to correct the quotient and remainder modulo $p^{a+1}$ . Repeating gives compatible solutions at every finite precision. The result is an identity $T^s=Q(T)G(T)+R(T)$ , where $R(T)$ has degree less than $s$ and all its coefficients are divisible by $p$ . Moreover, the constant coefficient of $Q(T)$ is a unit, so $Q(T)$ is invertible. Rearranging gives

$\displaystyle {D(T)=p^\mu U(T)P(T),\qquad P(T)=T^s+pc_{s-1}T^{s-1}+\cdots+pc_0,}$

where $U(T)$ is invertible. This is the polynomial-preparation argument. It is important that the degree $s$ is fixed: it is determined by the first coefficient of $G$ that survives modulo $p$ . The infinitely many higher coefficients have been absorbed into a unit and do not affect the relevant valuations. We now know that every nonzero relation can be factored into powers of $p$ and distinguished polynomials, apart from invertible factors. But our system has several coupled relations, so we still need to understand how their multiplicities are determined.

Take the finite presentation matrix $B(T)$ . We know from the common annihilator that it has full rank when its entries are considered in the fraction field of $\Lambda$ . Therefore some square minor has nonzero determinant. Factor the nonzero minors into irreducible expressions using the polynomial preparation argument and ordinary polynomial factorization. Only finitely many irreducible factors can occur in the common annihilator. Fix one such factor, say $f(T)$ . Temporarily allow division by every expression not divisible by $f(T)$ . In this arithmetic, every nonzero element becomes a power of $f(T)$ times an invertible element. We have therefore reduced the problem to the same situation as ordinary integer matrices when only one prime is relevant. Choose an entry of the relation matrix with the smallest exponent of $f$ . It divides every other entry in this localized calculation. Use row and column operations to clear its row and column, leaving one diagonal entry $f^{b_1}$ . Repeat with the remaining matrix. Eventually we obtain diagonal relations $f^{b_1},\ldots,f^{b_t}$ . These exponents are the multiplicities of the relations at the factor $f$ . We repeat this process for every irreducible factor, including $p$ . This gives a list of elementary blocks of the forms $\Lambda/(p^a)$ and $\Lambda/(P(T)^b)$ , with $P$ an irreducible distinguished polynomial.

There is one point we should not conceal. The row and column operations at one irreducible factor may involve divisions that are not permitted at another. Thus the separate diagonalizations do not necessarily combine into one exact diagonalization over $\Lambda$ . They do, however, combine up to finite groups. Here is why the discrepancy is finite. Two distinct irreducible factors have no common divisor. Their quotient relations therefore leave only finitely many possibilities for both $p$ and $T$ . For instance, if one factor is $p$ and the other is a distinguished polynomial $P(T)$ of degree $s$ , then imposing both relations gives $\mathbb F_p[[T]]/(T^s)$ , a group with exactly $p^s$ elements. A similar conclusion for two distinct distinguished polynomials follows by polynomial division and their nonzero resultant. More explicitly, at each irreducible factor the localized diagonalization gives generators of cyclic modules. We can clear the denominators used in choosing those generators and lift them back to the original module. These lifts define a map from the direct sum of the cyclic blocks to the original module. At each irreducible factor, this map becomes an isomorphism. Thus any element in its kernel or cokernel can be annihilated by expressions not divisible by each of the finitely many relevant factors. Combining these annihilators with the original nonzero relation leaves powers of both $p$ and $T$ acting trivially. Since the module has finitely many generators, the kernel and cokernel are finite groups.

We have therefore obtained the structural description

$\displaystyle \boxed{X\ \sim\ \bigoplus_i\Lambda/(p^{a_i})\ \oplus\ \bigoplus_j\Lambda/(P_j(T)^{b_j}),}$

where the symbol $\sim$ means that the two modules are related by a homomorphism with finite kernel and cokernel. This is not an assertion that every relation matrix can be exactly diagonalized. The finite discrepancies are essential, but we will see that they affect only the constant term in the growth formula. We can now do the numerical calculation that explains Iwasawa’s formula. At level $n$ , impose $\omega_n(T)=(1+T)^{p^n}-1=0$ . First take the block $\Lambda/(p^a)$ . Since $\omega_n(T)$ is monic of degree $p^n$ , every element of the quotient can be represented uniquely by a polynomial of degree less than $p^n$ whose coefficients are taken modulo $p^a$ . Thus there are $p^n$ independent coefficients, each with $p^a$ possible values. The order is exactly $p^{ap^n}$ . This block therefore contributes $ap^n$ to the valuation. Now take a block $\Lambda/(P(T)^b)$ , where $P(T)=T^s+pc_{s-1}T^{s-1}+\cdots+pc_0$ . Because $P^b$ is monic of degree $bs$ , the quotient $\Lambda/(P^b)$ is a free $\mathbb Z_p$ -module with basis $1,T,\ldots,T^{bs-1}$ . Multiplication by $\omega_n(T)$ is therefore represented by an ordinary square matrix of size $bs$ . If its determinant is nonzero, the valuation of the order of the finite quotient is the valuation of that determinant.

Let $\alpha_1,\ldots,\alpha_s$ be the roots of $P$ , counted with multiplicity. The eigenvalues of multiplication by $T$ are these roots, each repeated $b$ times. Consequently, the determinant of multiplication by $\omega_n(T)$ is $\prod_{i=1}^{s}((1+\alpha_i)^{p^n}-1)^b$ . Every root $\alpha_i$ has positive valuation because $P(T)\equiv T^s\pmod p$ . We need the elementary fact that, if $v_p(u)>0$ and $1+u$ is not a root of unity of $p$ -power order, then for sufficiently large $n$ ,

$\displaystyle v_p((1+u)^{p^n}-1)=n+c(u).$

To prove this, expand $(1+u)^p-1=pu+\binom p2u^2+\cdots+u^p$ . The valuation strictly increases when we raise $1+u$ to the $p$ -th power. After sufficiently many repetitions, the new difference from one has valuation greater than $1/(p-1)$ . From that point onward, the linear term $pu$ has strictly smaller valuation than all the other terms, so each further raising to the $p$ -th power increases the valuation by exactly one. This proves the formula. A $p$ -power root of unity must be excluded because it eventually gives zero; such a root cannot occur in our elementary factors, since the actual class groups $X/\omega_nX$ are finite. Applying the lemma to all the roots gives $v_p(\det(\omega_n(T)))=bs,n+\text{constant}$ for sufficiently large $n$ . Thus a polynomial relation of degree $s$ repeated $b$ times contributes $bs$ at every sufficiently high conductor. We can now see the two terms of Iwasawa’s formula directly. The $p^a$ -relations contribute something proportional to $p^n$ , while the polynomial relations contribute something proportional to $n$ .

We must still justify why replacing the actual module by its elementary blocks does not alter the coefficients of $p^n$ and $n$ . Choose a map from the direct sum $Y$ of elementary blocks to $X$ , with finite kernel and cokernel. In fact, the kernel must be zero: the blocks $\Lambda/(p^a)$ have no nonzero finite subgroup stable under $T$ , because multiplication by $T$ is injective, while the blocks $\Lambda/(P^b)$ are torsion-free as $\mathbb Z_p$ -modules. Thus we may write an exact sequence $0\to Y\to X\to C\to0$ , where $C$ is finite. Because $C$ is a finite $p$ -group and $\gamma=1+T$ acts on it with $p$ -power order, there is some $N$ such that $\omega_n(T)C=0$ for all $n\ge N$ . Multiplication by $\omega_n(T)$ is injective on $Y$ , since none of its elementary factors is a factor of $\omega_n(T)$ . Comparing the exact sequence with itself after multiplication by $\omega_n(T)$ gives a relation between the finite quotients. The kernel of multiplication by $\omega_n$ on $X$ is a subgroup of $C$ . Moreover, these kernels increase with $n$ , because $\omega_n(T)$ divides $\omega_{n+1}(T)$ . As $C$ is finite, the kernels eventually stabilize. Counting the kernels and cokernels in the resulting exact sequence gives $|X/\omega_nX|=|Y/\omega_nY| |X[\omega_n]|$ . The last factor is eventually constant, by the stabilization just proved. Therefore the finite discrepancies can change only the constant exponent $\nu$ .

We have now justified every term in the eventual growth law. Suppose the elementary blocks of the compatible class-group system are $\Lambda/(p^{a_1}),\ldots,\Lambda/(p^{a_u})$ and $\Lambda/(P_1(T)^{b_1}),\ldots,\Lambda/(P_v(T)^{b_v})$ . The first group of blocks contributes $(a_1+\cdots+a_u)p^n$ to the valuation. The second contributes $(b_1\deg P_1+\cdots+b_v\deg P_v)n$ , together with a fixed constant. The finite discrepancies give another fixed constant. Thus, defining $\mu=\sum_i a_i$ and $\lambda=\sum_j b_j\deg P_j$ , we obtain

$\displaystyle v_p\bigl(h(\mathbb Q(\zeta_{p^{n+1}}))\bigr)=\mu p^n+\lambda n+\nu$

for all sufficiently large $n$ . The important point is what we have now proved about the source of this formula. The finite generation is not a vague consequence of some abstract stability principle: it follows by taking a quotient $C$ , showing that $C=TC$ , reducing modulo $p$ , and using the finite Galois relation to conclude that $C=0$ . The finite presentation follows from a coefficient-by-coefficient argument showing that every ideal of $\mathbb Z_p[[T]]$ is finitely generated. The common annihilating relation is constructed by taking the determinant of explicit relations among the generators. The elementary blocks are obtained by reducing the relation matrix one irreducible factor at a time. Finally, the growth formula comes from counting the quotients of these blocks, using ordinary polynomial division, determinants, and the binomial theorem.

We should nevertheless keep the arithmetic input separate from the algebra. The class number formula and our finite Bernoulli calculations do not establish the norm-control identity. In the proof above, that identity came from the Hilbert class field correspondence, which we explained but did not derive from first principles. Once norm control is established, however, the remaining argument gives a concrete route from the actual class groups to Iwasawa’s growth formula, without assuming the formula in advance and without needing to calculate the real unit indices. This also clarifies the relationship between the two approaches we have developed. The special-value method turns class numbers into products of explicit character sums and is especially effective for the relative class number. The structural method instead retains the relations among ideal classes and proves that those relations are controlled by finitely much data. The first is better suited to calculating particular factors and congruences; the second explains why the full class-number sequence must eventually obey one uniform numerical law. They are two ways of studying the same arithmetic, but they retain different information about it.

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