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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.

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    $\begingroup$ 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
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    $\begingroup$ 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

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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.

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  • $\begingroup$ 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

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