Let $p$ be a prime number and $\zeta_{p^n}$ be a primitive $p^n$-th root of unity. We know that there is a unique subfield $\mathbb{Q}_1$ of $\mathbb{Q}(\zeta_{p^2})$ such that $[\mathbb{Q}_1:\mathbb{Q}]=p$ (the first layer of the cyclotomic $\mathbb{Z}_p$-extension of $\mathbb{Q}$).

Here are some basic things I know about $\mathbb{Q}_1$:

  1. Since $[\mathbb{Q}(\zeta_{p^2}):\mathbb{Q}]=p(p-1)$ and Gal$(\mathbb{Q}(\zeta_{p^2})/\mathbb{Q})$ is cyclic we know that $\mathbb{Q}_1$ is contained in the maximal real subfield $\mathbb{Q}(\zeta_{p^2})^+$ of $\mathbb{Q}(\zeta_{p^2})$.

  2. Since $p$ is prime, we have that $\mathbb{Q}_1$ contains no other subfields.

  3. We know that $p$ is totally ramified in $\mathbb{Q}_1$.

If $k$ is an imaginary quadratic field such that the discriminant $m$ of $k$ is co-prime to $p$, then the first layer $k_1$ of the cyclotomic $\mathbb{Z}_p$-extension of $k$ is the compositum $k\mathbb{Q}_1$ (this is also true for $k_n$ and $k\mathbb{Q}_n$).

Let $\lambda = \lambda_p$ be Iwasawas lambda invariant for the cyclotomic $\mathbb{Z}_p$-extension $k \subseteq k_1 \subseteq k_2 \dots k_{\infty}$, and $A(k_n)$ be the $p$-part of the class group of $k_n$.

In this paper, Sands has shown that Iwasawa's Theorem usually kicks in at an early sage for Imaginary quadratic fields. In particular, if $\lambda < p-1$, then $|A(k_1)| = |A(k)|p^{\lambda}$. So, it seems to me if we know enough about $k_1$, we may have a shot at knowing about $\lambda$ (provided we know about $A(k)$ and $A(k_1)$, which is another question altogether). But from the above, I feel that knowing about $\mathbb{Q}_1$ in general might be worthwhile since again $k_1 = k\mathbb{Q}_1$.

After trying to work out a few examples, it seems pretty difficult in general to figure out the first what the first layer $\mathbb{Q}_1$ is.

Some questions I have:

  1. Have people thought about this before?
  2. Is there anything in the literature that may help with this?
  3. Are there any other obvious properties about $\mathbb{Q}_1$ that I've overlooked?

Any help is appreciated.

  • 4
    $\begingroup$ One way to generate this field explicitly is to use the polynomial whose roots are $\sum_{n \in c} \zeta_{p^2}^n$ where $c$ ranges over the $p$ cosets of ${(\mathbb{Z}/p^2\mathbb{Z})^\times}^p$ in $(\mathbb{Z}/p^2\mathbb{Z})^\times$. For $p=3,5,7,11$ this gives $x^3-x+1$ (for $\zeta_9^n + \zeta_9^{-n}$), $x^5 - 10x^3 - 5x^2 + 10x - 1$, $x^7 - 7x^6 + 49x^4 - 98x^2 - 49x + 7$, and $x^{11}-11x^{10}+363x^8-1089x^7-1089x^6+6413x^5+242x^4-11616x^3-2178x^2+6534x+2673$. $\endgroup$ Mar 13, 2021 at 4:40
  • $\begingroup$ @NoamD.Elkies Does this come from Kummer theory? Or am I totally off? $\endgroup$ Mar 13, 2021 at 4:56
  • $\begingroup$ typo: $x^3-3x+1$ $\endgroup$ Mar 13, 2021 at 10:32
  • 1
    $\begingroup$ By the way, one can also compute anticyclotomic ones : numdam.org/item/CM_1976__32_2_157_0/?source=CM_1975__30_3_259_0 by Carroll and Kisilevsky and higher layers for $p=3$ in arxiv.org/abs/1806.10473. Though that was not asked. $\endgroup$ Mar 13, 2021 at 11:07
  • $\begingroup$ @ChrisWuthrich Thanks for pointing me towards the paper. $\endgroup$ Mar 13, 2021 at 15:04

1 Answer 1


Let $p$ be a prime number and let $\mathbb{Q}_{n}$ be the $n$th layer of the cyclotomic $\mathbb{Z}_{p}$-extension of $\mathbb{Q}$. Then $A(\mathbb{Q}_{n})$, the $p$-part of the class group of $\mathbb{Q}_{n}$, is trivial for all $n$. (In particular, the $\lambda$ and $\mu$ invariants of the cyclotomic $\mathbb{Z}_p$-extension of $\mathbb{Q}$ are both zero.)

This follows from Theorem 10.4(b) in Washington's "Introduction to cyclotomic fields" (second edition). Alternatively, see Proposition 1.1.4 in Ralph Greenberg's online book available here: https://sites.math.washington.edu/~greenber/book.pdf

So my feeling is that explicit knowledge of $\mathbb{Q}_{1}$ doesn't really help with the problem you're interested in. Of course, I'd be interested to know what's going on if this intuition turns out to be incorrect.

  • $\begingroup$ My real aim was to find the initial layer of an imaginary quadratic field $\mathbb{Q}(\sqrt{-m})$, so I wanted to find $\mathbb{Q}_1$ and then just adjoin $\sqrt{-m}$. $\endgroup$ Mar 13, 2021 at 15:07
  • $\begingroup$ Okay, but I'm not clear in what sense you want to "find" $\mathbb{Q}_{1}$. Something along the lines of Noam's comment, or in some other sense? $\endgroup$ Mar 13, 2021 at 17:04
  • $\begingroup$ I wanted to get a better picture of what $\mathbb{Q}_1$ is like, so Noam's comment is more or less what I was looking for. $\endgroup$ Mar 13, 2021 at 17:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.