$\DeclareMathOperator{\disc}{disc}$
- Suppose that $α$ is an algebraic integer of degree $n$, with monic minimal polynomial $m_α ∈ 𝐙[X]$. Let $K=𝐐(α)$. Show that \begin{equation}\label1 \disc_{K/𝐐}(1, α, …, α^{n-1})=(-1)^{n(n-1) / 2} N_{K/𝐐}(m_α'(α)) \end{equation} where $m_α'$ denotes the derivative. Using this, compute $\disc_{K/𝐐}(1,α,α^2)$, where $K=𝐐(α)$ with $α=2^{1 / 3}$.
Q2-Q6 discuss the cyclotomic field $K=𝐐(ζ)$, where $ζ≔e^{2 π i/p}$ and $p$ is an odd prime.
- Show that $[K: 𝐐]=p-1$.
In B3.1 we proved minimal polynomial of ζ is $Φ_p(x)={x^p-1\over x-1}$, so $[K: 𝐐]=\degΦ_p=p-1$. - Evaluate $N_{K/𝐐}(1-ζ)$.
The conjugates of $1-ζ$ are roots of the polynomial $Φ_p(1-x)$ which has degree $p-1$. Since $2|p-1$, product of roots equals its constant term $Φ_p(1)=p$, so $N_{K/𝐐}(1-ζ)=p$.
Generally, when $n=p^k$, $N_{K/𝐐}(1-ζ_{p^k})=\left.\frac{x^{p^k}-1}{x^{p^{k-1}}-1}\right|_{x=1}=p$, and is 1 when $n$ is not a prime power. See here - Show that $\frac1p (ζ-1)^{p-1}∈𝒪_K$.
Expanding ${(1+(ζ-1))^p-1\overζ-1}=0$ we get \[\sum_{k=1}^p\binom pk(ζ-1)^{k-1}=0\] so \[(ζ-1)^{p-1}=-\sum_{k=1}^{p-1}\binom pk(ζ-1)^{k-1}\] dividing by $p$ \[\frac1p(ζ-1)^{p-1}=-\sum_{k=1}^{p-1}\frac1p\binom pk(ζ-1)^{k-1}\] For each $k$, $\frac1p\binom pk∈𝐙$ and $(ζ-1)^{k-1}∈𝒪_K$. Since $𝒪_K$ is a ring, $\frac1p(ζ-1)^{p-1}∈𝒪_K$. - Evaluate $\disc_{K/𝐐}(1, ζ, …, ζ^{p-2})$.
Computing derivative$$Φ_p'(ζ)={pζ^{p-1}(ζ-1)-(ζ^p-1)\over(ζ-1)^2}=-{pζ^{-1}\over1-ζ}$$ Since $N_{K/𝐐}(ζ)=1$, $N_{K/𝐐}(1-ζ)=p$ by Q3 and norm is multiplicative, $$N_{K/𝐐}(Φ_p'(ζ))=(-1)^{p-1}{p^{p-1}N_{K/𝐐}(ζ)^{-1}\over N_{K/𝐐}(1-ζ)}=p^{p-2}$$ Substituting into \eqref{1}$$\disc_{K/𝐐}(1, ζ, …, ζ^{p-2})=(-1)^{(p-1)(p-2)/2} p^{p-2}$$ Since $(-1)^{p-2}=-1$ $$\disc_{K/𝐐}(1, ζ, …, ζ^{p-2})=(-1)^{(p-1)/2} p^{p-2}$$ - Suppose that $c_{0}, c_{1}, …, c_{p-2}$ are integers and that
$$\frac{1}{p}\left(c_{0}+c_{1}(ζ-1)+⋯+c_{p-2}(ζ-1)^{p-2}\right) ∈ 𝒪_{K}$$
Show that all the $c_{i}$ are divisible by $p$.
Suppose not, and let $r$ be minimal such that $p ∤ c_r$. Since $p$ divides $c_0,…,c_{r-1}$ $$\frac{1}{p}\left(c_r(ζ-1)^r+⋯+c_{p-2}(ζ-1)^{p-2}\right) ∈ 𝒪_{K}$$ Multiply by $(ζ-1)^{p-2-r}$ $$\frac{1}{p}c_r(ζ-1)^{p-2}+\frac{1}{p}(ζ-1)^{p-1}\left(c_{r+1}+⋯+c_{p-2}(ζ-1)^{p-3-r}\right)∈ 𝒪_{K}$$ By Q4, $\frac1p(ζ-1)^{p-1}∈ 𝒪_{K}$, so the second term$∈𝒪_{K}$, so $$\frac{1}{p}c_r(ζ-1)^{p-2}∈𝒪_{K}$$ By Q3, $N_{K/𝐐}(1-ζ)=p$, computing the norm $$\frac{1}{p^{p-1}}c_r^{p-1}p^{p-2}∈𝐙⇒\frac{c_r^{p-1}}p∈𝐙$$ but $c_r∈𝐙$, so $c_r$ is divisible by $p$, contradiction. - Show that $1, ζ, …, ζ^{p-2}$ is an integral basis for $𝒪_{K}$.
If $1, ζ, …, ζ^{p-2}$ is not an integral basis, by 2.21 ∃ prime $q$, $q^2|\disc(1, ζ, …, ζ^{p-2})$ and $m_1,…,m_{p-2}∈𝐙_q$ not all zero, such that$$\frac{1}{q}(m_{0}+m_{1}ζ+⋯+m_{p-2}ζ^{p-2})∈𝒪_{K}$$ $\disc(1, ζ, …, ζ^{p-2})=(-1)^{p-1\over2}p^{p-2}$ by Q5.
Since $q^2|(-1)^{p-1\over2}p^{p-2}$ and both $p,q$ are prime we have $p=q$. So $$\frac{1}{p}(m_{0}+m_{1}ζ+⋯+m_{p-2}ζ^{p-2})∈𝒪_{K}$$ replacing $ζ$ by $(ζ-1)+1$ we get a polynomial $f∈𝐙_p[x]$ of degree≤$p-2$ such that$$\frac1pf(ζ-1)∈𝒪_K$$ contradicting part (i).
- Suppose that $c_{0}, c_{1}, …, c_{p-2}$ are integers and that
$$\frac{1}{p}\left(c_{0}+c_{1}(ζ-1)+⋯+c_{p-2}(ζ-1)^{p-2}\right) ∈ 𝒪_{K}$$
Show that all the $c_{i}$ are divisible by $p$.
- Let $K$ be a number field. We say that $K$ is norm-Euclidean if $𝒪_{K}$ is a Euclidean domain with respect to the norm function: that is, given $a, b ∈ 𝒪_{K} ∖\{0\}$ we may find $q, r ∈ 𝒪_{K}$ such that $a=q b+r$ with ${|N_{K/𝐐}(r)|}<{|N_{K/𝐐}(b)|}$.
- Show that a Euclidean domain is a principal ideal domain.
Suppose $R$ is a Euclidean domain and $I$ is an ideal of $R$ and $I≠\{0\}$, we can take $a≠0$ in $I$ with $N(a)$ minimal. For any $b ∈ I$, write $b = qa + r$ with $r = 0$ or $N(r) < N(a)$; but $r=b-qa ∈ I$ and so by minimality of $N(a)$, $r = 0$; thus $a|b$ and $I = ⟨a⟩$. - Let $K=𝐐(\sqrt{-7})$. Show that $K$ is norm-Euclidean.
Since $-7≡1\pmod4,𝒪_{𝐐(\sqrt{-7})}=𝐙\left[\frac{1+\sqrt{-7}}{2}\right]$. For any $x + iy∈𝐂$ we must find $r,s∈𝐙$ $$z = r + s\frac{1+\sqrt{-7}}{2}=\left(r+\frac s2\right)+ i\frac{s\sqrt{7}}2\text{ such that }N(x + yi - z)<1$$ To minimize $N(x + yi - z)$, first take $s$ as the integer closest to $2y/\sqrt{7}$ \[ \left| \frac{2y}{\sqrt{7}} - s \right|⩽\frac12 ⇒ \left| y - \frac{s\sqrt7}2 \right|⩽\frac{\sqrt7}4 \] Then take $r$ as the integer closest to $(x - s/2)$ \begin{align*} \left|x - r - \frac s2\right| ⩽ \frac12. \end{align*} Putting both inequalities together \begin{align*} N(x + yi - z)&=\left(x - r -\frac s2\right)^2 +\left(y - s\frac{\sqrt{7}}2\right)^2 \\&⩽ 1/4 + 7/16 \\&< 1\end{align*}as desired. Hence, $𝒪_K$ is Euclidean domain.
- Show that a Euclidean domain is a principal ideal domain.
- Let $K=𝐐(\sqrt{-p})$, where $p$ is a prime, $p≡1\pmod 4$. By considering factorisations of 2, show that $𝒪_{K}$ is not a principal ideal domain.
Since $-p≡3\pmod4$, we know $𝒪_{K}=𝐙[\sqrt{-p}]$.
We compute$$(2,1+\sqrt{-p})(2,1+\sqrt{-p})=(4,2(1+\sqrt{-p}),1-p+2\sqrt{-p})$$ Since $2(1+\sqrt{-p})-(1-p+2\sqrt{-p})-4⋅\frac{p-1}4=2$ we get $(2)⊂(2,1+\sqrt{-p})^2$
but $(2,1+\sqrt{-p})^2⊂(2,1+\sqrt{-p})⊂(2)$
so $(2)=(2,1+\sqrt{-p})^2$.
But $N((2))=4$, so $N(2,1+\sqrt{-p})=2$ is prime, so $(2,1+\sqrt{-p})$ is prime.
Since the norm of any element $a+b\sqrt{-p}$ is $a^2+pb^2$, which does not take the value 2.
so $(2,1+\sqrt{-p})$ is not principal.
so $𝒪_{K}$ is not a PID.