There exists a unique abelian extension such that $[K:\mathbb{Q}]=p$ and $disc(K)=p^n$

Question

I am reading this PDF: http://www.math.uchicago.edu/~may/VIGRE/VIGRE2007/REUPapers/FINALFULL/Culler.pdf

On page $7$ it states (and proves) the following assertion:

If $p$ is an odd prime, then there is a unique abelian extension $K/\mathbb{Q}$ of degree $p$ with discriminant a power of $p$; in particular, it is the unique subfield of $\mathbb{Q}\left (\zeta\right )$ of degree $p$ over $\mathbb{Q}$, where $\zeta$ is a $p^2$th root of unity.

I am trying to read the proof but I really find it incomprehensible. The proof is the following, I will be stopping in order to explain which things I do not understand.

Proof. Let $K$ be the unique subfield of the $p^2$th cyclotomic field of order $p$. Then $K$ is ramified only at the prime $p$, which shows existence of an extension with the desired properties.

Now suppose that $K'$ is another such extension. We want to show that $K = K'$. To do this, first take the composite $K'L$ with the $p$th cyclotomic field $L = \mathbb{Q}(\zeta)$. Since $L$ contains the $p$th roots of unity, the standard results of Kummer theory apply, so $K'L = L( \sqrt[p]{\alpha})$ for some $\alpha \in L$. For example, if $K = K'$, then $\alpha$ could be $\zeta$, or a number of the form $\zeta^k\beta^p$ for some $k\in \mathbb{Z}$ not divisibleby $p$ and some $\beta \in L$.

Well, here I think I understand. I really do not know too much about Kummer Theory, but I know that $K'L = L( \sqrt[p]{\alpha})$ because of Theorem 6.2 on Lang's Algebra, "Cyclic extensions" section. Also, because of what is going next, we need $\alpha$ integral, and we clearly can achieve this: there exists $m\in \mathbb{N}$ such that $m\alpha$ is integral, and $L( \sqrt[p]{\alpha})=L( m^p\sqrt[p]{\alpha})=L( \sqrt[p]{m\alpha})$.

Let $\lambda=1-\zeta$. Then $N(\lambda)=p$, so $\lambda$ generates the unique prime ideal of $L$ lying over $p$.

I did not understand this, but I know that if $\omega$ is a $p^n$-rooth of unity then $p\mathbb{Z}[\omega]=(1-\omega)^{\varphi\left (p^n\right )}$ so I am OK with that.

We will show that $\alpha$ can be chosen to be an algebraic integer satisfying $\alpha\equiv 1\pmod {\lambda^p}$. First we show can choose $\alpha$ to be prime to $p$. To see this, we use the fact that $K'L$ is abelian. Consider a generator $\tau$ for $\text{Gal}(L/\mathbb{Q})$ and extend it to an automorphism $\tau\in \text{Gal}(K'L/L)$. Since $\sigma$ and $\tau$ commute, we have:

$\sigma \tau(\sqrt[p]{\alpha})=\tau\sigma(\sqrt[p]{\alpha})=\tau(\zeta \sqrt[p]{\alpha})=\zeta^l\tau(\sqrt[p]{\alpha})$

for some primitive root modulo $p$. This shows that $\sqrt[p]{\alpha}$ is an eigenvector of $\sigma$ with eigenvalue $\zeta^l$.

I think this last sentence is wrong. What is true is that $\sqrt[p]{\alpha}$ is an eigenvector of $\sigma$ with eigenvalue $\zeta$ and that $\tau (\sqrt[p]{\alpha})$ is an eigenvector of $\sigma$ with eigenvalue $\zeta^l$. So let's assume that and continue:

Hence $\tau(\alpha)=\tau(\sqrt[p]{\alpha})^p=\left (c\sqrt[p]{\alpha^l}\right )^p=c^p\alpha^l$.

Who is $c$? This is my attempt to explain what he tried to do: since the $L$-linear transformation $\sigma :K'L\to K'L$ between $p$-dimensional vector spaces admits $\sqrt[p]{\alpha}$ as an eigenvector with eigenvalue $\zeta$, then we easily obtain that $\sqrt[p]{\alpha^i}$ is an eigenvector with eigenvalue $\zeta^i$ for $i=0,1,\cdots ,p-1$. Since the $\zeta^i$ are all pairwise distinct because $\zeta$ is a primitive $p$th root of unity, we obtained $p$ different eigenvectors, and therefore each space of eigenvectors has dimension $1$. Since the eigenvectors with eigenvalue $\zeta^l$ are generated by $\sqrt[p]{\alpha^l}$ and $\tau (\sqrt[p]{\alpha})$ is an eigenvector of $\sigma$ with eigenvalue $\zeta^l$, then there exists $c\in L$ such that $\tau \left (\sqrt[p]{\alpha}\right )=c\sqrt[p]{\alpha^l}$.

But this argument has a little problem: we need (because of what is going next) $c$ integral, and it is not clear to me that $c$ is an algebraic integer. Anyone?

Now it is clear that $\alpha$ can be chosen to be prime to $p$. Simply replace $\alpha$ by $\frac{\tau(\alpha)}{\alpha}$. Since the ideal generated by $\lambda$ is invariant under $\tau$, any factor of $\lambda$ dividing $\alpha$ cancels out, leaving something prime to $p$. Note that once $\alpha$ is prime to $p$, we can also force $\alpha$ to be congruent to $1$ mod $\lambda$ by raising $\alpha$ to a suitable power, since the multiplicative group of a finite field is cyclic.

Why did he say that the multiplicative group of a finite field is cyclic? I mean... Yes, it is true, but didn't it suffice to say just that it is a finite group and just use that $g^{|G|}=1$ for every $g$ in a finite group $G$?

Also, using the fact that $\zeta^a\equiv 1-a\lambda\pmod{\lambda^2}$ we can force $\alpha$ to be congruent to $1$ mod $\lambda^2$ by multiplying by a suitable power of $\zeta$.

Why? I mean, we know that $\alpha = 1+\lambda s$ with $s\in \mathbb{Z}[\zeta]$ and $\zeta^a\equiv 1-a\lambda\pmod{\lambda^2}$, therefore $\zeta^a\alpha \equiv 1+\lambda (s-a)\pmod {\lambda^2}$, hence we are saying that for every $s\in \mathbb{Z}[\zeta]$ there exists $a\in \mathbb{N}_0$ such that $\lambda \mid s-a$. Why it is true?

Finally, we use induction to obtain the desired congruence. Say we have already shown that $\alpha \equiv 1+ a\lambda^e\pmod{\lambda^{e+1}}$. Now we use again the fact that $K'L$ is abelian. We have the congruence $\sigma(\alpha)\equiv c^p\alpha^l\pmod{\lambda^{e+1}}$ which, given our assumption, gives that $c\equiv c^p\equiv 1\pmod{\lambda}$. And therefore $c^p\equiv 1\pmod p$.

As a consequence we have $1+a(l\lambda)^e\equiv \sigma(\alpha)\equiv \alpha^l\equiv 1+al(\lambda^e)\pmod{\lambda^e}$ and hence $l^e\equiv l\pmod{\lambda}$. But $l$ was supposed to be a primitive root modulo $\lambda$, and $e$ was greater than $1$. The inductive step works as long as $e$ is less than $p$, so we have shown $\alpha\equiv 1+a\lambda^p\pmod{\lambda^{p+1}}$ or in other words, $\alpha \equiv 1\pmod {\lambda^p}$, as desired.

I could not understand even a simple word. Why do we have the congruence $\sigma(\alpha)\equiv c^p\alpha^l\pmod{\lambda^{e+1}}$? Why does it imply that $c\equiv c^p\equiv 1\pmod{\lambda}$? And what is the meaning of $c^p\equiv 1\pmod p$? Wasn't $c$ an element in $L$? Who is $a$? As you may suspect, I have several questions about the rest of the argument. I really do not understand a simple step, so what I really need here is a detailed explanation.

That $K=K'$ follows immediately from this. To see why, consider the number $\xi=\frac{1-\sqrt[p]{\alpha}}{\lambda}$. It is a root of the polynomial $f(x)=\left (x-\frac{1}{\lambda}\right )^p-\frac{\alpha}{\lambda^p}$.

Well, I would take $f(x)=\left (x-\frac{1}{\lambda}\right )^p+\frac{\alpha}{\lambda^p}$ instead.

It is clear that this polynomial is monic, and that all but the constant term are algebraic integers. But by the preceding argument, $1−\alpha$ is divisible by $\lambda^p$. Hence the constant term is also an algebraic integer. Hence $\xi$ is an algebraic integer.

Since $\xi\in \mathcal{O}_{K'KL}$, the discriminant of $KK'L$ over $KL$ must contain the ideal generated by $\pm N(f'(\xi))=\pm N\left (p\left (\xi-\frac{1}{\lambda}\right )^{p-1}\right )=\epsilon \alpha^{p-1}$ for some unit $\epsilon$.

I completely understood the first paragraph, but what about the second? It is clear that if $f(x)\in KL[x]$ then the discriminant of $KK'L$ over $KL$ must contain the ideal generated by $\pm N(f'(\xi))$, but why is it true that $f(x)\in KL[x]$? Moreover, why $\pm N\left (p\left (\xi-\frac{1}{\lambda}\right )^{p-1}\right )=\epsilon \alpha^{p-1}$ for some unit $\epsilon$?

In particular, this discriminant is prime to $p$, so $p$ is unramified in the extension $KK'L/KL$. Hence $p$ is unramified in the inertial field $T/\mathbb{Q}$, and this extension is nontrivial. But $p$ was the only ramified prime in $K$, $K'$, and $L$, and therefore no prime other than $p$ can be ramified in $T$. Hence $T/\mathbb{Q}$ is unramified. But this is a contradiction, since there are no nontrivial unramified extensions of $\mathbb{Q}$.

I understood the argument, but we obtained a contradiction from what? What were we supposing that we obtained a contradiction?

Since my questions are just too much (I really cannot understand anything of the proof, as you may have seen) I will summarize and enumerate them:

1) Why is $c$ an algebraic integer?

2) Why did he say that the multiplicative group of a finite field is cyclic? Wasn't it enough to say that it was a finite group?

3) Why using the fact that $\zeta^a\equiv 1-a\lambda\pmod{\lambda^2}$ we can force $\alpha$ to be congruent to $1$ mod $\lambda^2$ by multiplying by a suitable power of $\zeta$? If for every $s\in \mathbb{Z}[\zeta]$ there exist $b\in \mathbb{N}_0$ such that $s\equiv b\pmod{\lambda}$ then we are done, but why it is true?

4) How did the author prove that $\alpha \equiv 1\pmod {\lambda^p}$?

5) Why $f(x)\in KL[x]$?

6) Why $\pm N\left (p\left (\xi-\frac{1}{\lambda}\right )^{p-1}\right )=\epsilon \alpha^{p-1}$ for some unit $\epsilon$?

7) The author says at the end that we obtained a contradiction. But I do not know which contradiction we achieved because I do not know which assumption we made in order to reach that contradiction. I suspect it must be that $K\neq K'$, but the author neither says something like "let's assume that $K=K'$" (or some other assumption) nor explicits where he is making use of that assumption. So... What assumption are we making and where are we making use of it?

Answer 1

The main reason why you find the proof incomprehensible is probably your lack of background. Honestly, if you don't know that $1-\zeta$ generates the prime ideal above $p$ in the field of $p$-th roots of unity or if you don't know the basics of Kummer theory then you should not study proofs of the Kronecker-Weber theorem. The books by Washington and Marcus should help.

One result that you have repeatedly missed is the fact that ${\mathcal O}_L/(\lambda) \simeq {\mathbb Z}/p{\mathbb Z}$. This closes quite a few of your gaps. In particular, $c \equiv c^p \bmod \lambda$ is just Fermat's Little Theorem.

Ad 7: As you have quoted, the author writes "Now suppose that $K'$ is another such extension". This is the assumption that contradicts the fact derived at the end.

The eigenvalue calculation apparently serves to prove something like Lemma 2.1 in here.

Similarly, the congruence $\alpha \equiv 1 \bmod \lambda^p$ is related to the decomposition law in Kummer extensions (see Hecke's book on algebraic number theory). And the induction hypothesis is part of any proof by induction.

Answer 2

If you interpret the question as: "What is explicitly an abelian extension $k$ of $\mathbf Q$ of degree $p$ (hence cyclic) unramified outside $(p)$", class field theory gives an easy answer.

Let us first state a general result. For a number field $K$ and a prime number $p$ (assume $p$ odd to avoid petty trouble), let $S$ be the set of primes of $K$ above $p$, and denote by $G_S (K)$ the Galois group of the maximal abelian pro-$p$-extension $K_S$ of $K$ unramified outside $S$. Then CFT asserts that $G_S (K)$ has $\mathbf Z_p$-rank $1+r_2+\delta$, where $r_2$ is the number of pairs of complex embeddings of $K$, and $\delta \ge 0$ is the defect of Leopoldt's conjecture (see e.g. Washington's book, §13.1, thm. 13.4). Assuming this conjecture (which is proved for abelian fields), the $\mathbf Z_p$-torsion $T_S (K)$ of $G_S (K)$ can be described in terms of Iwasawa theory. Anyway, if $K=\mathbf Q$, coroll. 13.6 of Washhington suffices to show that this torsion vanishes and $G_S (\mathbf Q)\cong \mathbf Z_p$, more precisely $\mathbf Q_S$ coincides with $\mathbf Q^{cyc}$, the so called cyclotomic $\mathbf Z_p$-extension of $\mathbf Q$, characterized by the fact that its compositum with $\mathbf Q(\mu_p)$ is just $\mathbf Q(\mu_{p^{\infty}})$, where $\mu_p$ (resp. $\mu_{p^{\infty}}$) is the group of $p$-th roots (resp. all $p^n$-th roots) of unity. Then your extension $k$ is the unique cyclic extension of $\mathbf Q$ of degree $p$ contained in $\mathbf Q^{cyc}$, and by Galois theory, it is straightforward that $k$ is the unique cyclic extension of degree $p$ of $\mathbf Q$ contained in $\mathbf Q(\mu_{p^2})$ .

If you don't want to use CFT, you could do "reverse engineering" on the above proof, but I did not try.