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

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?

• The statement is only correct if you add the assumption that $K$ be abelian. Otherwise take $K = {\mathbb Q}(\sqrt[p]{p})$ or suitable subfields of the Hilbert class field of an irregular cyclotomic field. Dec 8, 2017 at 13:04
• Fixed. Also, when the author says "discriminant a power of $p$" I think he means "the only prime dividing the discriminant is $p$" (which is not the same, because the discriminant could be, for example, $-4$ which is not a prime power). Indeed, the only think he uses is that the only ramified prime is $p$. But those are just small details. Dec 8, 2017 at 15:40
• 3) Since $s \equiv b \bmod \lambda$ for an integer $b$, we have $\alpha \equiv 1 + b\lambda \bmod \lambda^2$. Now multiply by $\zeta^b$..This works because the residue class ring modulo $\lambda$ is isomorphic to ${\mathbb Z}/p{\mathbb Z}$. Dec 8, 2017 at 16:54
• 7) The assumption is that $K$ is a cyclic extension of prime degree $p$ ramified only at $p$ that is not cyclotomic. Dec 8, 2017 at 16:55
• @franzlemmermeyer For (3): That is exactly what I am asking. I said that if I knew that for every $s\in \mathbb{Z}[\zeta]$ there exists an integer $b$ such that $s\equiv b\pmod {\lambda}$ then I would understand the argument, but I do not know why it is true. Dec 8, 2017 at 19:16

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.

• I will be answering your first paragraph: Firstly, I am precisely reading Marcus's book. I read the whole first four chapters and I am working through the exercises on chapter $4$. The final exercises on chapter $4$ are precisely intended to prove Kronecker-Weber (and I have already done the previous exercises), so according to Marcus I have no lack of background. Secondly, I would like to know why you think that I do not know that $1-\zeta$ generates the prime ideal above $p$. I know it perfectly well, as I explicitly said on my post. And finally... (next comment) Dec 10, 2017 at 14:47
• And finally, I have been asking this question to some of my professors on my university. One of them is a PhD working specifically on number theory, and he also wasn't able to understand the proof (I asked him to sit down with me and read the proof at my side so that I could ask him the questions concerning the things I did not understand). So, I think that the real problem is that the proof is written in a very ununderstandable way (the professor agreed with me), and with big errors such as that of the eigenvector. Dec 10, 2017 at 14:52
• (The other professors also were not able to understand the proof, by the way, I am looking for help with other professors but also I am looking for help in this forum, and I really appreciate your contribution). Dec 10, 2017 at 14:53
• Specifically on your answer: I understand why $c\equiv c^p\pmod {\lambda}$, but why $c\equiv c^p\equiv 1\pmod{\lambda}$? Also, remember that all of this is possible providing that $c$ is integral, and I don't know why it is true. Lemma 2.1 on the article you cited explicitly says "$\xi \in F^{\times}$" an we need $\xi \in \mathcal{O}_F\setminus \left \{0\right \}$. Dec 10, 2017 at 15:09
• When the author said "Let's assume that $K'$ is another such extension" I thought he mean "another, possibly the same", as in almost all proofs for uniqueness. If the assumption was that $K\neq K'$, as I suspected, then where is he making use of that assumption? Dec 10, 2017 at 15:10

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.

• Thank you. I must confess that I do not know anything about CFT and I don't know anything about a large number of things you mentioned in your proof. I really appreciate your contribution (in fact I will come back here to read it when I know CFT), but I am not looking for another proof, I am trying to understand the one I posted (which is written in a "language" I can understand). That's the reason why I asked very concrete questions which more or less explain what I am missing, but of course the real problem is that I do not understand the proof the way it is written. Dec 9, 2017 at 23:18
• OK, without CFT you can (must) use Kummer theory as follows . I stick to your notations. To show that $K$ is unique, take $K'$ with the desired properties and show that $K'':= K'L$ (where $L=Q(\zeta_p)$), which is a cyclic extension of degree $p$ of $L$, coincides with $Q(\zeta_{p^2})$. By Kummer theory, $K''$ is of the form $L(\sqrt [p] \alpha)$, with $\alpha\in L^*/{L^*}^p$. We'll abuse notation by writing also $\alpha\in L^*$ according to the context. There are 2 things to prove: 1) An algebraic property, $K''=Q(\zeta_{p^2})$, which is equivalent, by Kummer... Dec 10, 2017 at 14:02
• ... to the existence of an integer $j$ prime to $p$ s.t. $\alpha\zeta^j \in {L^*}^p$. This is the easy part, which you can find in any textbook on Galois theory. 2) An arihmetic property, which is a necessary and sufficient condition on $\alpha$ for $K''$ to be unramified outside the set of $p$-primes of $L$. You can find this conveniently in G. Gras' book "CFT-From theory to practice" (Springer), chapter I, §6, thm.6.3. Sufficient conditions are also provided by Cassels-Fröhlich's "Algebraic Number Theory", exercise 2.12. Check whether these are the same as the ones given by VIGRE2007. Dec 10, 2017 at 14:05
• @nguyenquangdo Could you explain why we have the uniqueness of $K$ if we have proven $K''$ coincides with $\mathbb{Q}(\zeta_{p^2})$? Thanks.
– user685167
May 4, 2020 at 7:45
• What a commendable perseverance ! You said "I'll be back", and here you are, more than two years later ! Sticking to my notations$L,K'$ etc., if I understand well, you ask why two extensions $K , K'$ with the desired properties, must coincide when the "shifted" extensions $K(\zeta_p), K'(\zeta_p)$ coincide ? This is simply a Galois/Kummer problem, which I'll try to explain in the most general possible setting. Let us ask when a Kummer extension $L(\sqrt [p] A)$, where $A$ is a subgroup of $L^*/{L^*}^p$ , is galois over $\mathbf Q$. May 4, 2020 at 12:40