On the Iwasawa Algebra

Question

I am reading Joaquin Rodrigues Jacinto's and Chris Williams' notes on $p$-adic $L$-functions http://www.ltcc.ac.uk/media/london-taught-course-centre/documents/Number-Theory---Full-Lecture-Notes-2017-18.pdf and I have questions regarding the topological structure of the Iwasawa Algebra. I assume that:

• $L/\mathbb{Q_p}$ (finite) is a non-archimedean local field with ring of integers $\mathcal{O}_L$ and max.ideal $\mathfrak{p} = (\pi)$.
• $G$ is a pro-finite abelian group.

Then the Iwasawa Algebra of $G$ with coefficients in $\mathcal{O}_L$ is defined as (Def.2.4., p.13 from above lecture notes) $$ \Lambda(G) := \lim_{\substack{\longleftarrow \\ U \subset G \text{ open}}} \mathcal{O}_L[G/U]. $$

If $G = \mathbb{Z}_p$, then $\Lambda(\mathbb{Z}_p) \cong \mathcal{O}_L[[T]]$ via the Amice-Transform. Thus $\Lambda(\mathbb{Z}_p)$, should be a local ring with maximal ideal generated by $(\mathfrak{p}, T)$ (assuming question 1 is true).

Question(s):

1. Is the Amice Transform an isomorphism of topological rings?
2. What is the topology on $\Lambda(G)$? Since $U \subset G$ open implies of finite index, is the topology on each $\mathcal{O}_L[G/U] \cong \mathcal{O}^{[G \colon U]}_L$ just the (finite) product topology and thus $\Lambda(G)$ carries the (proj.) limit-topology?
3. To understand the topology on $\Lambda(G)$ better: I imagine the elements as in Remark 7.8. on p.50; as formal power series $$ \sum_{g \in G} a_g [g], $$ but with the restriction that every projection $\pi_U \colon \mathcal{O}_L[[G]] \to \mathcal{O}_L[G/U]$ being well-defined and probably open continuous maps. So.. is the topology on $\mathcal{O}_L[[G]]$ the $\mathfrak{m}$-adic topology, where $\mathfrak{m} = (\mathfrak{p}, I(G))$ and where $I(G)$ is the augmentation ideal of $\Lambda(G)$? In other words, is it correct to state that $\pi$ and every element in $I(G)$, so specially every $[g] - [0]$ (being $1 = [0]$ in $\Lambda(G)$) are topologically nilpotent?
4. Is the Iwasawa Algebra in general an integer domain? I would say no, since Lemma 3.8. the authors state a condition on $\mu \in \Lambda(\mathbb{Z}^{\times}_p)$ being a non-divior. It therefore surprises me that $\Lambda(\mathbb{Z}_p)$ is a domain.
5. Is it in general true, that $\Lambda(G)$ is a local ring as it is for the case of $G = \mathbb{Z}_p$?

Answer 1

There are a lot of questions in your post - here are answers to some of them!

Q1: Yes, it is. There are two natural topologies one can put on $\Lambda(\mathbb{Z}_p)$, most easily seen by considering this as the space of measures on $\mathbb{Z}_p$. One has: the strong topology (that of uniform convergence); and the weak topology (that of pointwise convergence). Under the Amice transform these correspond to the $\mathfrak{p}$-adic and $(\mathfrak{p},T$)-adic topologies on $\mathcal{O}[[T]]$.

Q2&3: As highlighted by Q1 there will not generally be a unique natural topology on $\Lambda(G)$, so probably the topology you prefer will depend on any applications you have in mind. The topology you write down looks like the general analogue of the weak topology from Q1. I'm reluctant to assert anything more precise without more careful thought. (In Q3, I should point out that Remark 7.8 of our notes was sloppy, and should be viewed more as convenient intuitive tool than a precise description of the elements of $\Lambda(G)$.)

Q4: It will not in general be a domain. Indeed, as you intuit from that Lemma, $\Lambda(\mathbb{Z}_p^\times)$ is not a domain. Rather, it can be viewed as the product of $p-1$ copies of $\Lambda(\mathbb{Z}_p)$, corresponding to the $p-1$ characters of $(\mathbb{Z}_p/p\mathbb{Z}_p)^\times$. (In the theory of Hida families, this corresponds to the weight space being a disjoint union of $p-1$ discs).

I agree that at first glance it seems strange that $\Lambda(\mathbb{Z}_p)$ is a domain whilst $\Lambda(\mathbb{Z}_p^\times)$ is not. This is because $\Lambda(\mathbb{Z}_p^\times) \subset \Lambda(\mathbb{Z}_p)$ is not a subalgebra; the multiplication structures on these two groups are different (on the left, multiplication on $\Lambda(\mathbb{Z}_p^\times)$ is induced by multiplication on $\mathbb{Z}_p^\times$; on $\Lambda(\mathbb{Z}_p)$, it's induced by addition on $\mathbb{Z}_p$, and corresponds to convolution of measures).

Q5: It will not in general be local. $\Lambda(\mathbb{Z}_p^\times)$ is a counterexample, since it has $p-1$ maximal ideals (corresponding to the $p-1$ copies of $\Lambda(\mathbb{Z}_p)$).