How hard is it to find the first layer of this basic $\mathbb{Z}_p$-extension?

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:

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.

• 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 $x^{11}-11x^{10}+363x^8-1089x^7-1089x^6+6413x^5+242x^4-11616x^3-2178x^2+6534x+2673$. Mar 13, 2021 at 4:40
• @NoamD.Elkies Does this come from Kummer theory? Or am I totally off? Mar 13, 2021 at 4:56
• typo: $x^3-3x+1$ Mar 13, 2021 at 10:32
• 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. Mar 13, 2021 at 11:07
• @ChrisWuthrich Thanks for pointing me towards the paper. Mar 13, 2021 at 15:04

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.)
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.
• 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 $\sqrt{-m}$. Mar 13, 2021 at 15:07
• 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? Mar 13, 2021 at 17:04
• 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. Mar 13, 2021 at 17:28