I am a student learning Iwasawa theory. I am so sorry if this post is too trivial for this site. I posted it on math.stackexchange yesterday but obtained no responce.

A quite basic object is the Iwasawa algebra. A basic version is as follows:

Let $L/\mathbb{Q}_p$ be a finite extension (i.e. a local number field) where $p$ is a prime (if necessary, we may assume $p \geq 3$ or $p \geq 5$.) Let $\mathcal{O}_L$ be its ring of integer. Then we define $\Lambda := \mathcal{O}_L[[T]]$ as the ring of formal power series of one indeterminate $T$ with $\mathcal{O}_L$-coefficient, called an Iwasawa algebra.

I have see in many papers that people regard $\Lambda$ as a disk, or regard the "rational Iwasawa algebra" $\Lambda_{\mathbb{Q}}:= \Lambda \otimes_{\mathbb{Z}_p} \mathbb{Q}_p$ (I am still wondering if this is correct, as a tensor product over $\mathbb{Z}_p$?) as a disk. Moreover, people often say that some kind of a $p$-adic L-function $\mathcal{L}_p$ is in the disk, and "a small neighborhood" of the $\mathcal{L}_p$ satisfies certain properties.

MY QUESTION: How to make such statements precise? In other words, how can people view the Iwasawa algebra as a disk, talk about its elements as points and the neighboorhood of the points.

Moreover, quite recently, there is a notion called "shrinked Iwasawa algebra" $\Lambda_{m}:= \mathcal{O}_L[[p^{-m}T]] \subset \Lambda$. The author stated that this can be regarded as shrinking the disk into smaller radius, just as the name shows.

My further question: How do people regard this "shrinked Iwasawa algebra" as a disk of smaller radius?

Put a step further, in more applications, we may consider $\mathbb{Z}_p$-extensions $\mathcal{K}_{\infty}/\mathcal{K}$ of Galois group $\mathbb{Z}_p^d$ for $d > 1$. Then the corresponding Iwasawa algebra should be $\Lambda_{(d)} := \mathcal{O}_L[[T_1, \ldots, T_d]]$. Then

Can we also regard $\Lambda_{(d)}$ as some kind of disk and further generalize the "shrinked Iwasawa algebra"?

Thank you all for commenting and answering! Any references on explaining these are also welcome!


2 Answers 2


The correct viewpoint is not "$\Lambda$ is like a disc", but "$\Lambda$ is like the functions on a disc".

To see this, ask yourself: given an element $f \in \mathbb{Z}_p[[T]]$, what values can we plug in for $T$ such that the series will converge? It's pretty easy to see that if we put $T = x$ for some $x \in \mathbb{Q}_p$, then the series will converge if $|x| < 1$ but (usually) diverge otherwise.

With a little bit more effort, you can convince yourself that if we put $T = x$ for some $x$ lying in a finite field extension of $\mathbb{Q}_p$, then again we will have convergence for $|x| < 1$ and divergence otherwise.

So we can interpret an element of $\mathbb{Z}_p[[T]]$ as a function on the set of $x \in \overline{\mathbb{Q}}_p$ (or its completion $\mathbb{C}_p$) satisfying $|x| < 1$.

As for "shrinked" (shrunken?) Iwasawa algebras: $\mathbb{Z}_p[[p^{-m}T]]$ is not a subring of $\mathbb{Z}_p[[T]]$; it is a bigger ring, because it corresponds to functions satisfying a weaker condition -- that they converge on a smaller region. The exercise you should do is to work out for which $x \in \overline{\mathbb{Q}}_p$ the "set $T = x$" map makes sense on $\mathbb{Z}_p[[p^{-m}T]]$.

  • $\begingroup$ Thank you so much! Your answers have always been of great help! Still one more doubt that often comes to me: regarding elements in $\Lambda$ as functions from $\mathbb{C}_p$-disk to $\mathbb{C}_p$, then what is the corrrsponding interpretation after tensoring $\Lambda$ with $\mathbb{Q}_p$? Besides, I have always seen people tensoring $\Lambda$ with other things, for example $\mathcal{O}_L^{ur}$, or rational $\mathbb{Q}_p$, but sadly, I cannot feel the real difference after additional tensorings. $\endgroup$
    – Hetong Xu
    May 27, 2023 at 1:32
  • $\begingroup$ $\Lambda = $ analytic functions on the unit disc whose norm is bounded by 1. On the other hand $\Lambda \otimes \mathbb{Q}_p$ = bounded analytic functions on the unit disc (but the bound can be anything). $\endgroup$ May 27, 2023 at 1:36
  • $\begingroup$ (Important to note that $\mathbb{Z}_p[[T]] \otimes_{\mathbb{Z}_p} \mathbb{Q}_p$ is not the same as $\mathbb{Q}_p[[T]]$ -- the former is much smaller. If this isn't obvious to you, then you should reflect on the definitions a bit until it becomes so.) $\endgroup$ May 27, 2023 at 1:38
  • $\begingroup$ Ah, I see! So here by saying norm, I guess you mean the norm of a function $f$ defined in Alexey Do's answer? Thank you! $\endgroup$
    – Hetong Xu
    May 27, 2023 at 1:52
  • $\begingroup$ Yes, I meant the norm $\|f\|$ as defined in Alexey Do's answer (this is generally called the "Gauss norm"). $\endgroup$ May 27, 2023 at 15:14

I give this answer just to make David's answer a little more general. Precisely, your $\Lambda$ should be thought as functions with norm less than $1$ in some Banach $\mathbb{Q}_p$-algebras. We fix a complete non-archimedean field $K$, its valuation ring $R$, and the residue field $k$. The theory of analytic manifolds is somehow not a satisfying theory because the theory arises from the valuation of $K$ is totally disconnected. Rigid geometry is roughly a way to fix this, though it is not original motivation of the theory. As in scheme theory, one should start with "functions on spaces". If the basic building block of scheme theory is affine space $\mathbb{A}^n$, then the basic building block of rigid geometry is a unit polydisc $$\mathbb{B}^n = \left \{(x_1,...,x_n) \in K^n \mid \left |x_i \right| \leq 1 \right \}.$$ However, as in scheme theory, if $X$ is a variety over a field $F$, then closed subvariety of $X$ maybe empty if $F$ is not algebraically closed, so one should base change to $\overline{F}$ and we do not lose information by doing this since $X_{\overline{F}}/\mathrm{Aut}(\overline{F}/F) = X$. Therefore the correct notion of polydisc shouldbe $$\mathbb{B}^n(\overline{K}) = \left \{(x_1,...,x_n) \in \overline{K}^n \mid \left |x_i \right| \leq 1\right \}.$$ where here we have extended the valuation on $K$ to $\overline{K}$, though this valuation is not complete (unless we restrict to a finite extension of $K$ inside $\overline{K}$). Whatever kind of functions on the polydisc, they "should" lie inside $$\left \{f= \sum_{I = (i_1,...,i_n) \in \mathbb{N}^n} a_{i_1,...,i_n}X_1^{i_1}\cdots X_n^{i_n} \in K[[X_1,...,X_n]] \right \}.$$ Let's consider two different situtations about convergence. I note here the technical reasons why there are differences between Isawasa algebras and Tate algebras, it lies in the fact that, for non-archimedean field $K$, a series $\sum a_n$ is convergent if and only if its terms converge to zero, i.e. $\lim \left|a_n \right| = 0$.

A formal power series $\sum a_I X^I$ is convergent iff $\lim \left|a_I\right| \left |X^I \right|=0$.

  • If we restrict ourself to the open polydisc $\left \{\left|x_i \right| < 1 \forall i = \overline{1,n} \right \}$, then the terms $X^I$ are already convergent to $0$ so no need to impose conditions on coefficients $a_I$. This situation leads to Iwasawa algebras.
  • If we stick to the closed polydisc $\left \{\left|x_i \right| \leq 1 \forall i = \overline{1,n} \right \}$, then the series is convergent if and only if it is convergent at $(1,,..,1)$, that is, the series $\sum a_I$ itself must be convergent, this one leads to Tate algebras.

Both types of algebras have norms defined by supremums of coefficients. Restricting to functions with coefficients in the valuation ring $R$ gives us functions with norms less than $1$.

  • 4
    $\begingroup$ I am not sure the last paragraph here is quite correct. Iwasawa algebras are not the same as (bounded subrings of) Tate algebras: they don't have the $\lim_{|I| \to \infty} |a_I| = 0$ condition (which is why the Iwasawa algebra corresponds to functions on the set $\{ x : |x_i| < 1\, \forall i\}$, whereas the Tate algebra corresponds to $\{ x : |x_i| \le 1\, \forall i\}$) $\endgroup$ May 26, 2023 at 14:52
  • $\begingroup$ @DavidLoeffler Hi, I was mistaken, thanks for pointing out, edited! $\endgroup$
    – Alexey Do
    May 27, 2023 at 16:29

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