Algebraic Number Theory

Let $k\in \mathbb{Z}$ with $\mathit{𝑗𝑘}\equiv 1modn$. Then

is an algebraic integer since ${\zeta }_n$ is. The same being true after reversing $i$ and $j$, we have the result. □

Note that $[\mathbb{Q}({\mu }_{p^r}):\mathbb{Q}]\le \mathrm{deg}{\mathrm{\Phi }}_{p^r}=p^{r-1}(p-1)$. We have

which forces $1-{\zeta }_{p^r}^i$ to be divisible by a prime $𝔭$ over $p$ for some $i$ with $p\nmid i$. By Lemma 3.1.12, this implies that each $1-{\zeta }_{p^r}^j$ with $p\nmid j$ is divisible by $𝔭$. Therefore, ${𝔭}^{p^{r-1}(p-1)}$ divides $(p)$, which forces $[\mathbb{Q}({\mu }_{p^r}):\mathbb{Q}]=p^{r-1}(p-1)$ and

Finally, note that

which is a product of terms of the form ${\zeta }_{p^r}^i-{\zeta }_{p^r}^j$ for $i\ne j$ both not divisible by $p$, which from what we have seen is only divisible by the prime $𝔭=(1-{\zeta }_{p^r})$. In other words, no prime other than $p$ can ramify. □

If we write $n=p_1^{r_1}\cdots p_k^{r_k}$ for distinct primes $p_i$ and $r_i\ge 1$, then

Hence, the second statement implies the first. In $n$ is even and not divisible by $4$, then $\mathbb{Q}({\mu }_n)=\mathbb{Q}({\mu }_{n∕2})$ and $\phi (n)=\phi (n∕2)$, so we may assume that either $n$ is odd or divisible by $4$.

The result holds for $k=1$ by Lemma 3.1.13. Write $n=m{p}^r$ for some $m,r\ge 1$ and prime $p$ not dividing $m$. As $p$ does not not ramify in $\mathbb{Q}({\mu }_m)$ by induction on $k$ but is totally ramified in $\mathbb{Q}({\mu }_{p^r})$ by Lemma 3.1.13, the fields $\mathbb{Q}({\mu }_m)$ and $\mathbb{Q}({\mu }_{p^r})$ are linearly disjoint. So by induction, the primes which divide $n$ are exactly those which ramify in $\mathbb{Q}({\mu }_n)=\mathbb{Q}({\mu }_m)\mathbb{Q}({\mu }_{p^r})$, and we have

□

We first consider $n=p^r$ for a prime $p$ and $r\ge 1$. In this case, we know that the absolute value of $\mathrm{disc}(\mathbb{Z}[{\mu }_{p^r}])$ is a power of $p$, say $p^m$. In particular, ${𝔣}_{\mathbb{Z}[{\mu }_{p^r}]}$ divides $(p^m)$. Let ${\lambda }_r=1-{\zeta }_{p^r}$, which generates the unique prime over $(p)$ in $\mathbb{Q}({\mu }_{p^r})$. Since $(p)$ is totally ramified in $\mathbb{Q}({\mu }_{p^r})∕\mathbb{Q}$, we have that

In particular,

Replacing ${𝒪}_{\mathbb{Q}({\mu }_{p^r})}$ on the right-hand side of (3.1.1) using the formula for ${𝒪}_{\mathbb{Q}({\mu }_{p^r})}$ given by (3.1.1) itself, we have

Repeatedly replacing ${𝒪}_{\mathbb{Q}({\mu }_{p^r})}$ on the right, we eventually obtain

the latter step by definition of the conductor.

For the general case, we write $n=p_1^{r_1}\cdots p_k^{r_k}$ for distinct primes $p_i$ and $r_i\ge 1$. Then $\mathbb{Q}({\mu }_n)$ is the compositum of the $\mathbb{Q}({\mu }_{p_i^{r_i}})$ and the discriminants of the fields $\mathbb{Q}({\mu }_{p_i^{r_i}})$ are relatively prime, we have by Proposition 1.4.28 that the elements

with $0\le i_t\le p_t^{r_t-1}(p_t-1)-1$ for each $1\le t\le k$ form an integral basis of ${𝒪}_{\mathbb{Q}({\mu }_n)}$. Since these elements are all contained in the order $\mathbb{Z}[{\mu }_n]$ in $\mathbb{Q}({\mu }_n)$, we have the result. □

First, we note that $p$ is totally ramified in $\mathbb{Q}({\mu }_{p^r})$ of degree $\phi (p^r)$ and unramified in $\mathbb{Q}({\mu }_m)$. It follows from Remark 2.5.7 that the ramification index of $p$ in $\mathbb{Q}({\mu }_n)$ is then $\phi (p^r)$, and the residue degree of $p$ in $\mathbb{Q}({\mu }_n)$ is the residue degree of $p$ in $\mathbb{Q}({\mu }_m)$.

So, let $𝔮$ be a prime ideal over $p$ in $\mathbb{Z}[{\mu }_m]$. Its residue field is

As $F^{\times }$ is cyclic of order $|F|-1$, we have $F={𝔽}_{p^f}$ for the smallest $f\ge 1$ such that $m\mid (p^f-1)$, which is to say that the order of $p$ in ${(\mathbb{Z}∕\mathit{𝑚\mathbb{Z}})}^{\times }$. Finally, $g$ is constrained to be $f^{-1}\phi (m)$ by the degree formula. □

By Proposition 3.1.21, to say that a prime $p$ splits completely is exactly to say that $\phi (p^r)=1$ and the order of $p$ in ${(\mathbb{Z}∕(n∕p^r)\mathbb{Z})}^{\times }$ is $1$. If $p$ is odd, then this means $r=0$ and $p\equiv 1modn$. If $p=2$, this forces $n=1$ or $n=2$, which is to say that $\mathbb{Q}({\mu }_n)=\mathbb{Q}$. □

Recall that $\mathrm{Gal}(\mathbb{Q}({\mu }_p)∕\mathbb{Q})$ is cyclic of order $p-1$, so it has a unique quotient of order $2$. As $p$ is totally ramified in $\mathbb{Q}({\mu }_p)∕\mathbb{Q}$, it is ramified in $K$, so $K=\mathbb{Q}(\sqrt{p^{\prime }})$, where $p^{\prime }=\pm p$. Moreover, no other prime is ramified in $\mathbb{Q}({\mu }_p)∕\mathbb{Q}$. If $p^{\prime }\equiv 3mod4$, then $K$ has ring of integers $\mathbb{Z}[\sqrt{p^{\prime }}]$, and we have

which means that $2$ ramifies in $K$. Since $p^{\ast }\equiv 1mod4$, we must have $p^{\prime }=p^{\ast }$. □

The prime $q$ splits into an even number of primes in $\mathbb{Q}({\mu }_p)$ if and only if the decomposition group $G_q$ of any prime over $q$ has even index in $\mathrm{Gal}(\mathbb{Q}({\mu }_p)∕\mathbb{Q})$. Since the latter Galois group is cyclic with unique subgroup of index $2$ having fixed field $\mathbb{Q}(\sqrt{p^{\ast }})$ by Lemma 3.2.2, this occurs if and only if $G_q$ fixes $\mathbb{Q}(\sqrt{p^{\ast }})$, which is to say, if and only if $q$ splits in $\mathbb{Q}(\sqrt{p^{\ast }})$. □

Note that

if and only if

which is to say if and only if $(\frac{p^{\ast }}{q})=1$. This says exactly that there exists $a\in \mathbb{Z}$ such that

But then $x^2-p^{\ast }$ factors modulo $q$, so $(q)$ splits in $\mathbb{Z}[\sqrt{p^{\ast }}]$. Now, Proposition 3.2.3 tells us that this happens if and only if $(q)$ splits in $\mathbb{Z}[{\mu }_p]$ into an even number of primes. By Proposition 3.1.21, the prime $(q)$ splits into $(p-1)∕f$ primes in $\mathbb{Q}({\mu }_p)$, where $f$ is the order of $q$ modulo $p$. So, it splits into an even number if and only if $f$ divides $(p-1)∕2$, which is to say that

In other words, $(\frac{p}{q})={(-1)}^{\frac{p-1}{2}\frac{q-1}{2}}$ if and only if $(\frac{q}{p})=1$, as desired. □

If $𝔮$ is a prime ideal of $\mathbb{Z}[{\mu }_p]$ dividing both $x+{\zeta }_p^{i}y$ and $x+{\zeta }_p^{j}y$ for $0\le i<j<p$, then $𝔮$ divides both $({\zeta }_p^j-{\zeta }_p^i)y$ and $({\zeta }_p^j-{\zeta }_p^i)x$. If $𝔮$ does not lie over $p$, then it must divide both $x$ and $y$, which is impossible. Thus, $𝔮$ lies over $p$, so it equals $(1-{\zeta }_p)$. On the other hand, $x+{\zeta }_p^{i}y\equiv x+ymod(1-{\zeta }_p)$, which implies that $x+y\equiv 0modp$, since $x+y$ is an integer. Thus, no such $𝔮$ can exist. □

Note that $\sigma (\overline{𝜖}{𝜖}^{-1})$ has absolute value $1$ for every $\sigma \in \mathrm{Gal}(\mathbb{Q}({\mu }_p)∕\mathbb{Q})$, so it is a root of unity by Corollary 4.4.2 below, hence a $2p$th root of unity. If the lemma did not hold, then each

would have to be nontrivial for every $j$, which means that $\overline{𝜖}{𝜖}^{-1}$ would not be a $p$th root of unity. Thus, we would have $\overline{𝜖}{𝜖}^{-1}=-{\zeta }_p^i$ for some $i\in \mathbb{Z}$. We may write $𝜖\equiv cmod(1-{\zeta }_p)$ for some $c\in \mathbb{Z}$, and so $\overline{𝜖}\equiv cmod(1-{\zeta }_p)$ as well. On the other hand,

so $(1-{\zeta }_p)$ in fact divides $𝜖$, which is a contradiction as $𝜖$ is a unit. □

We have

and the images of any $p-1$ among $1,x,\dots ,x^{p-1}$ are ${𝔽}_p$-linearly independent in the last term. □

Suppose that $x^p+y^p=z^p$ for some integers $x,y,z$ with $p\nmid \mathit{𝑥𝑦𝑧}$ and $(x,y)=(1)$. For $p=3$, we note that the set of nonzero cubes modulo $9$ is $\{\pm 1\}$, and no sum of two elements of this set equals a third element modulo $9$. Thus, we may assume that $p\ge 5$ from now on.

Note that if $x\equiv y\equiv -zmodp$, then $2x^p\equiv -x^{p}modp$, so $p$ divides $3x^p$. As $p\nmid \mathit{𝑥𝑦𝑧}$, this cannot happen. Therefore, if $p\mid x-y$, then $p\nmid x+z$. Switching the roles of $y$ and $-z$ by writing $x^p+{(-z)}^p={(-y)}^p$ if needed, we may therefore assume that $p\nmid x-y$.

In $\mathbb{Z}[{\mu }_p]$, the quantity $x^p+y^p$ factors, and we have

Note that $z\equiv x+ymodp$, so $p$ does not divide $x+y$. By Lemma 3.3.1, we have that the ideals $(x+{\zeta }_p^{i}y)$ are coprime, so in order that their product be a $p$th power of an ideal, each must itself be a $p$th power. Write

for some ideal $𝔞$ of $\mathbb{Z}[{\mu }_p]$. Since we have assumed that ${\mathrm{Cl}}_{\mathbb{Q}({\mu }_p)}$ has no elements of order $p$, the ideal $𝔞$ is principal, so let $\alpha \in \mathbb{Z}[{\mu }_p]$ be a generator of $𝔞$. We then have

for some $𝜖\in \mathbb{Z}{[{\mu }_p]}^{\times }$.

Now, note that we may write $\alpha =c+d(1-\zeta )$ for some $c\in \mathbb{Z}$ and $d\in \mathbb{Z}[{\mu }_p]$, and

By Lemma 3.3.3, we that there exists $j\in \mathbb{Z}$ such that ${𝜖}^{\prime }={\zeta }_p^j𝜖$ is fixed by complex conjugation. We thus have

so

We therefore have

If ${\zeta }_p^{2j}$, ${\zeta }_p^{2j+1}$, $1$ and ${\zeta }_p^{-1}$ are distinct $p$th roots of unity, then since $p-1\ge 4$, they are linearly independent modulo $p$. This would force both $x$ and $y$ to be divisible by $p$, a contradiction.

It follows that $1={\zeta }_p^{2j}$, ${\zeta }_p^{2j+1}$ or ${\zeta }_p^{2j+2}$. In the first of these cases, we have

so $p$ divides $y$, a contradiction. In the second case, we have

so $p$ divides $x-y$, again a contradiction. In the final case, we have

so $p$ divides $x$, a contradiction, finishing the proof. □