4 Structure of Iwasawa module
4.1 Iwasawa algebra
Let \(p\) be a prime, \(\Gamma \) be a topological abelian group, isomorphic to \({\mathbb {Z}}_p\) as a topological abelian group, and let \(\gamma \in \Gamma \) be a topological generator of \(\Gamma \) (i.e. the subgroup \(\{ \gamma ^n\mid n\in {\mathbb {Z}}\} \) is dense in \(\Gamma \)). For each \(n\geq 0\) let \(\Gamma _n:=\Gamma /\Gamma ^{p^n}\).
The Iwasawa algebra is defined as the completed group algebra
where the transition map \({\mathbb {Z}}_p[\Gamma _{n+1}]\to {\mathbb {Z}}_p[\Gamma _n]\) is induced by the natural projection \(\Gamma _{n+1}\to \Gamma _n\). Each \({\mathbb {Z}}_p[\Gamma _n]\) is a free \({\mathbb {Z}}_p\)-module of rank \(p^n\), we endow it with the \(p\)-adic topology, and endow \(\Lambda \) with the subspace topology of the product topology of \(\prod _{n=0}^\infty {\mathbb {Z}}_p[\Gamma _n]\).
Let \(A\) be a ring, \({\mathfrak {a}}\) be an ideal of \(A\). Let \(M,N\) be two \(A\)-modules, and \(\varphi :M\to N\) be an \(A\)-module homomorphism. Then \(\varphi \) is continuous if we endow \(M,N\) with \({\mathfrak {a}}\)-adic topology.
In particular, if \(\varphi \) is an \(A\)-module isomorphism, then it is a homeomorphism of topological spaces if we endow \(M,N\) with \({\mathfrak {a}}\)-adic topology.
Let \(x\in M\) and \(y:=\varphi (x)\in N\). Let \(U\) be any open neighborhood of \(y\) in \(N\). Then there exists some \(n\) such that \(y+{\mathfrak {a}}^nN\subset U\). Take \(V=x+{\mathfrak {a}}^nM\) then it is an open neighborhood of \(x\) in \(M\), and we have \(\varphi (V)=y+\varphi ({\mathfrak {a}}^nM)=y+{\mathfrak {a}}^n\varphi (M)\subset y+{\mathfrak {a}}^nN\subset U\). Therefore \(\varphi \) is continuous.
For each \(n\geq 0\), there is an isomorphism of \({\mathbb {Z}}_p\)-algebras
They are all free \({\mathbb {Z}}_p\)-modules of finite rank, we endow them with \(p\)-adic topology. By Proposition 4.2 we know that it is a homeomorphism of topological spaces.
We have \(\Gamma /\Gamma ^{p^n}\cong {\mathbb {Z}}/p^n{\mathbb {Z}}\) as an abelian group, and the image of \(\gamma \in \Gamma \) in it is a generator of it. By abuse of notation we still denote the image of \(\gamma \in \Gamma \) in it by \(\gamma \). Then \({\mathbb {Z}}[\Gamma /\Gamma ^{p^n}]\) as a \({\mathbb {Z}}\)-module is free of rank \(p^n\) and \(\{ \gamma ^k\} _{0\leq k\leq p^n-1}\) is a basis of it. Now it’s easy to see that as a \({\mathbb {Z}}_p\)-module homomorphism, \({\mathbb {Z}}_p[\Gamma /\Gamma ^{p^n}]\xrightarrow \sim {\mathbb {Z}}_p[T]/\big((1+T)^{p^n}-1\big)\), \(\gamma ^k\mapsto (1+T)^k\) is well-defined and is a \({\mathbb {Z}}_p\)-module isomorphism, and preserves multiplication. Therefore it is an isomorphism of \({\mathbb {Z}}_p\)-algebras.
There is an isomorphism of topological rings \({\mathbb {Z}}_p[[T]]\xrightarrow \sim \Lambda \) sending \(1+T\) to \(\gamma \), where \({\mathbb {Z}}_p[[T]]\) is endowed with \((p,T)\)-adic topology.
We prove that there is a natural isomorphism of \({\mathbb {Z}}_p\)-algebras
and which is a homeomorphism of topological spaces, where the topology of the left hand side is the subspace topology of the product topology of the topology defined in Proposition 4.3, and the topology of the right hand side it the \((p,T)\)-adic topology. From which we can obtain the isomorphism of topological rings \(\Lambda \xrightarrow \sim {\mathbb {Z}}_p[[T]]\) given by \(\gamma \mapsto 1+T\).
For simplicity of notation, denote \(\varphi _n(T):=(1+T)^{p^n}-1\). It’s easy to see that for each \(n\geq 1\) and for all \(1\leq i\leq p^n-1\), we have \(\binom {p^n}{i}\in p^n{\mathbb {Z}}\), hence \(\varphi _n(T)\in T^{p^n}+p^n{\mathbb {Z}}[T]_{\deg \leq p^n-1}\), in particular \(\varphi _n(T)\) is a distinguished polynomial of degree \(p^n\).
It’s clear that \({\mathbb {Z}}_p[T]_{\deg \leq p^n-1}\xrightarrow \sim {\mathbb {Z}}_p[T]/(\varphi _n(T))\) is an isomorphism of \({\mathbb {Z}}_p\)-modules, and the Weierstrass division (Proposition 4.7) implies that \({\mathbb {Z}}_p[T]_{\deg \leq p^n-1}\xrightarrow \sim {\mathbb {Z}}_p[[T]]/(\varphi _n(T))\) is also an isomorphism of \({\mathbb {Z}}_p\)-modules, therefore the natural ring homomorphism \({\mathbb {Z}}_p[T]\hookrightarrow {\mathbb {Z}}_p[[T]]\) induces an isomorphism of \({\mathbb {Z}}_p\)-algebras \({\mathbb {Z}}_p[T]/(\varphi _n(T))\xrightarrow \sim {\mathbb {Z}}_p[[T]]/(\varphi _n(T))\) (Corollary 4.8).
Since each \({\mathbb {Z}}_p[[T]]/(\varphi _n(T))\) is a free \({\mathbb {Z}}_p\)-module of finite rank endowed with the \(p\)-adic topology, we have \({\mathbb {Z}}_p[[T]]/(\varphi _n(T))\cong \varprojlim _m{\mathbb {Z}}_p[[T]]/(p^m,\varphi _n(T))\), where each \({\mathbb {Z}}_p[[T]]/(p^m,\varphi _n(T))\) is endowed with discrete topology. Therefore we have the following isomorphisms of rings as well as topological spaces:
here \((*)\) holds because \(\{ (p^m,\varphi _n(T))\} _{m\geq 1,n\geq 1}\) and \(\{ (p,T)^k\} _{k\geq 1}\) are cofinal. In fact, for each \(k\geq 1\), and for any \(m\geq k\) and \(n\geq k\), we have \((p^m,\varphi _n(T))\subset (p,T)^k\); conversely, for each \(m\geq 1\) and \(n\geq 1\), we have \((p,T)^{p^n}\subset (p,\varphi _n(T))\), and \((p,\varphi _n(T))^m\subset (p^m,\varphi _n(T))\), therefore for any \(k\geq p^nm\), we have \((p,T)^k\subset (p,T)^{p^nm}\subset (p,\varphi _n(T))^m\subset (p^m,\varphi _n(T))\).
4.2 Weierstrass preparation theorem
This is WIP in #21944.
If \((A,{\mathfrak {m}},k)\) is a local ring, then a polynomial \(f(X)=\sum _{i=0}^na_iX^i\in A[X]\) is called a distinguished polynomial if \(a_n=1\) and \(a_i\in {\mathfrak {m}}\) for all \(0\leq i\leq n-1\).
(Mathlib: Polynomial.IsDistinguishedAt)
Let \(A\) be a ring, and let \(f(X)=\sum _{n=0}^\infty a_nX^n\in A[[X]]\) be a formal power series. Then \(f(X)\in A[[X]]^\times \) if and only if \(a_0\in A^\times \).
(Mathlib: PowerSeries.isUnit_iff_constantCoeff)
If \(f(X)\in A[[X]]^\times \), then there exists \(g(X)=\sum _{n=0}^\infty b_nX^n\in A[[X]]\) such that \(f(X)g(X)=1\), therefore by considering constant term we obtain \(a_0b_0=1\), hence \(a_0\in A^\times \). Conversely, if \(a_0\in A^\times \), by multiplying \(a_0^{-1}\) to \(f(X)\) if necessary, we may assume that \(a_0=1\) and \(f(X)=1-Xf_1(X)\) for some \(f_1(X)\in A[[X]]\). Since \(A[[X]]\) is \((X)\)-adically complete and separated, it’s easy to see that \(1+\sum _{k=0}^\infty X^kf_1(X)^k\) converges \((X)\)-adically in \(A[[X]]\) and which is the inverse of \(f(X)\).
Let \((A,{\mathfrak {m}},k)\) be a complete local ring, \(g(X)=\sum _{i=0}^\infty a_iX^i\in A[[X]]\setminus {\mathfrak {m}}[[X]]\) be a formal power series such that not all of its coefficients are in \({\mathfrak {m}}\). Let \(n\geq 0\) be the integer such that \(a_n\in A\setminus {\mathfrak {m}}=A^\times \) and \(a_i\in {\mathfrak {m}}\) for all \(0\leq i\leq n-1\). Then for any \(f\in A[[X]]\), there exists a unique formal power series \(q(X)\in A[[X]]\) and a unique polynomial \(r(X)\in A[X]\) of degree \(\leq n-1\) such that \(f=gq+r\).
Write \(g(X)=\sum _{i=0}^{n-1}a_iX^i+X^ng_1(X)\) for some \(g_1(X)\in A[[X]]^\times \), and \(f(X)=\sum _{i=0}^{n-1}b_iX^i+X^nf_1(X)\) for some \(f_1(X)\in A[[X]]\). We construct a sequence \((q_k)_{k=1}^\infty \) inductively in \(A[[X]]\), such that \(f-gq_k\in A[X]_{\deg \leq n-1}+{\mathfrak {m}}^k[[X]]\), and such that \(q_{k+1}(X)-q_k(X)\in {\mathfrak {m}}^k[[X]]\). We construct \(q_1(X):=f_1(X)g_1(X)^{-1}\). Since \(a_i\in {\mathfrak {m}}\) for all \(i\leq n-1\), we have
Suppose \(q_k(X)\) is constructed, then we may write \(f(X)-g(X)q_k(X)=\sum _{i=0}^{n-1}b_i^{(k)}X^i+X^ns_k(X)\) for some \(s_k(X)\in {\mathfrak {m}}^k[[X]]\), and we construct \(q_{k+1}(X):=q_k(X)+s_k(X)g_1(X)^{-1}\). Then we have
Since \(A[[X]]\) is complete and separated according to the sequence \(\{ {\mathfrak {m}}^k[[X]]\} _{k\geq 1}\) of ideals, there exists a unique limit \(q(X)\in A[[X]]\) of the sequence \((q_k)_{k=1}^\infty \), which satisfies \(r:=f-gq\in A[X]_{\deg \leq n-1}\).
To prove the uniqueness, suppose \(q(X)\in A[[X]]\) and \(r(X)\in A[X]_{\deg \leq n-1}\) such that \(gq=r\), then we prove by induction that for any \(k\geq 0\) we have \(q,r\in {\mathfrak {m}}^k[[X]]\), which implies that \(q=r=0\). When \(k=0\) there is nothing to prove. Suppose \(q,r\in {\mathfrak {m}}^k[[X]]\) for some \(k\geq 0\). Then we have \((\sum _{i=0}^{n-1}a_iX^i)q(X)+X^ng_1(X)q(X)=r(X)\), since \(a_i\in {\mathfrak {m}}\) for all \(i\leq n-1\), we obtain \(r(X)\in {\mathfrak {m}}^{k+1}[X]_{\deg \leq n-1}\). Multiply \(g_1(X)^{-1}\) to both side, we obtain \((\sum _{i=0}^{n-1}a_iX^i)q(X)g_1(X)^{-1}+X^nq(X)=r(X)g_1(X)^{-1}\), therefore \(q(X)\in {\mathfrak {m}}^{k+1}[[X]]\).
Let \((A,{\mathfrak {m}},k)\) be a complete local ring, \(g(X)=\sum _{i=0}^n a_iX^i\in A[X]\) be a polynomial such that \(a_n\in A\setminus {\mathfrak {m}}\) and \(a_i\in {\mathfrak {m}}\) for all \(i{\lt}n\). Then the natural map \(A[X]/(g)\to A[[X]]/(g)\) is an isomorphism.
Let \(f\in A[[X]]\). Then by Proposition 4.7, we may find a unique formal power series \(q(X)\in A[[X]]\) and a unique polynomial \(r(X)\in A[X]\) of degree \(\leq n-1\) such that \(f=gq+r\). Then \(r\) is the unique inverse of \(f\) under the natural map \(A[X]/(g)\to A[[X]]/(g)\).
Let \((A,{\mathfrak {m}},k)\) be a complete local ring. Let \(g(X)\in A[[X]]\setminus {\mathfrak {m}}[[X]]\) be a formal power series such that not all of its coefficients are in \({\mathfrak {m}}\). Then there is a unique distinguished polynomial \(f(X)\in A[X]\) and a unique invertible formal power series \(h(X)\in A[[X]]^\times \) such that \(g=fh\).
Take \(f(X)=X^n\) in Proposition 4.7, we obtain \(q(X)\in A[[X]]\) and \(r(X)\in A[X]_{\deg \leq n-1}\) such that \(X^n=g(X)q(X)+r(X)\). Since \(g(X)=\sum _{i=1}^{n-1}a_iX^i+X^ng_1(X)\) with \(a_i\in {\mathfrak {m}}\) for all \(i\leq n-1\) and \(g_1(X)\in A[[X]]^\times \), we have \(r(X)\in {\mathfrak {m}}[X]_{\deg \leq n-1}\), and by the construction in (ii) we have \(q(X)\in g_1(X)^{-1}+{\mathfrak {m}}[[X]]\subset A[[X]]^\times \). Therefore take \(h(X):=q(X)^{-1}\in A[[X]]^\times \) and \(f(X):=X^n-r(X)\), then \(f(X)\) is a distinguished polynomial of degree \(n\), and \(g(X)=f(X)h(X)\) holds.
To prove the uniqueness, suppose \(f(X)\) and \(f'(X)\) are two distinguished polynomials and \(u(X)\in A[[X]]^\times \) such that \(f'(X)=f(X)u(X)\). Then \(\overline u(X)\in k[[X]]^\times \) and we have \(\overline f{}'(X)=X^{\deg (f')}=\overline f(X)\overline u(X) =X^{\deg (f)}\overline u(X)\in k[[X]]^\times \), which forces that \(\deg (f)=\deg (f')\) and \(\overline u(X)=1\). Therefore \(f'(X)-f(X)\in A[X]_{\deg \leq \deg (f)-1}\) and \(f'(X)=f(X)+(f'(X)-f(X))\) is a Weierstrass division of \(f'\) by \(f\), on the other hand, \(f'(X)=f(X)u(X)\) is also a Weierstrass division of \(f'\) by \(f\), hence by the uniqueness of Weierstrass division we have \(u(X)=1\) and \(f'(X)=f(X)\).
(Another proof. It is possible to prove Weierstrass preparation theorem using a form of Hensel’s lemma presented in https://ncatlab.org/nlab/show/Hensel's+lemma.)
4.3 Characteristic ideal
\(\Lambda \cong {\mathbb {Z}}_p[[T]]\) is a Noetherian regular local ring of Krull dimension \(2\).
Hence by Proposition 2.4, \(\Lambda \) is a UFD (or maybe one can check directly that the \(I\)-adic completion of a UFD is a UFD).
Hence by Proposition 2.3, any height \(1\) prime \({\mathfrak {p}}\) of \(\Lambda \) is principal.
(i) If the generator of \({\mathfrak {p}}\) is in \(p{\mathbb {Z}}_p[[T]]\), then we must have \({\mathfrak {p}}=(p)\).
(ii) If the generator of \({\mathfrak {p}}\) is not in \(p{\mathbb {Z}}_p[[T]]\), then by Proposition 4.9 such \({\mathfrak {p}}\) has a unique generator which is a distinguished polynomial.
Therefore, we have
If \(X\) is a finitely generated torsion \(\Lambda \)-module, then there exists a pseudo-isomorphism
where \(f_1,\cdots ,f_m\) are distinguished polynomials. The characteristic ideal \(\operatorname{char}_\Lambda (X)\) is generated by \(p^{\sum _{j=1}^sn_j}\prod _{i=1}^mf_i^{b_i}\) which is contained in \(p^{\sum _{j=1}^sn_j}{\mathbb {Z}}_p[[T]]\) but not in \(p^{1+\sum _{j=1}^sn_j}{\mathbb {Z}}_p[[T]]\).
(i) The \(\mu \)-invariant of \(X\) is defined to be \(\mu (X):=\sum _{j=1}^sn_j\), and the \(\lambda \)-invariant of \(X\) is defined to be \(\lambda (X):=\sum _{i=1}^mb_i\deg f_i\).
(ii) If \(f\in {\mathbb {Z}}_p[[T]]\) is not zero, then define \(\mu (f)\) be the integer such that \(f\in p^{\mu (f)}{\mathbb {Z}}_p[[T]]\) but \(f\notin p^{1+\mu (f)}{\mathbb {Z}}_p[[T]]\), define \(\lambda (f)\) be the leading degree of \((p^{-\mu (f)}f\bmod p)\in {\mathbb {F}}_p[[T]]\).
Clearly, \(\mu (X)\) and \(\lambda (X)\) are equal to \(\mu (f)\) and \(\lambda (f)\) if \(\operatorname{char}_\Lambda (X)=(f)\).
We have \(\mu (X)=\sum _{i=0}^\infty \operatorname{rank}_{{\mathbb {F}}_p[[T]]}X[p^{i+1}]/X[p^i]\), and \(\lambda (X)=\operatorname{rank}_{{\mathbb {Z}}_p}X/X[p^\infty ]\).
......
4.4 Growth of coinvariant part
If \(X\) is a finitely generated torsion \(\Lambda \)-module such that \(X/(\gamma ^{p^n}-1)X\) is finite for any \(n\geq 0\), then there exists some constant \(\nu =\nu (X)\) such that for all sufficiently large \(n\), \(\operatorname{ord}_p(\# (X/(\gamma ^{p^n}-1)X))=\mu (X)p^n +\lambda (X)n+\nu (X)\).