In this note, we will mention some examples about the ring of integers and its integral basis.
1. Cyclotomic fields. We will fix some notions: $p$ is an odd prime number, $\zeta=e^{2\pi i/p}$ is the primitive $p^{\text{th}}$ root of unity, $K=\mathbb{Q}(\zeta)$. Recall that $\mathscr{O}_K=\mathbb{Z}[\zeta]$, and $\Delta_K=(-1)^{(p-1)/2}p^{p-2}$.
One can see that $\zeta+\zeta^{-1}=e^{2\pi i/p}+e^{-2\pi i/p}=2\cos2\pi/p\in \mathbb{R}$. Hence, $L:=\mathbb{Q}(\zeta+\zeta^{-1})$ is a real subfield of $K$. We will prove that $[K:L]=2$. We first see that $G:=Gal(K/\mathbb{Q})\cong \mathbb{F}_P^\times$, a cyclic group of order $p-1$, an even number. And hence, $G$ has only one subgroup of order 2. Furthermore, the automorphism $K\to K$ sending $\zeta\mapsto \zeta^{-1}$ is an element of order 2 in $G$, and $L$ is fixed by this automorphism.
On the other hand, $\cos 2\pi/p\ne \cos 2\pi k/p$, for any integer $k$ such that $1<k<p-1$, i.e. $\zeta+\zeta^{-1}\ne \zeta^k+\zeta^{-k}$ for all $1<k<p-1$. Say another words, $L$ is not fixed by any automorphism of $K$ sending $\zeta$ to $\zeta^k$. Hence, by the Galois' correspondence, $Gal(K/L)$ corresponds to the field extension $K/L$. This yields $[K:L]=2$.
Also, one can see that for any real subfield $F$ of $K$, $F$ is fixed by the automorphism $\zeta\mapsto \zeta^{p-1}$. By the Galois' correspondence again, $F\subset L$. We have the following
Proposition 1. Let $K,L$ be fields as above, then $L$ is the maximal real subfield of $K$. Also, $\mathscr{O}_L=\mathbb{Z}[\zeta+\zeta^{-1}]$.
Proof. Let $\psi:=\zeta+\zeta^{-1}. $It is easy to see that $\mathbb{Z}[\psi]\subset \mathscr{O}_L$, because $\psi$ is integral over $\mathbb{Z}$. Assume that we have the strict inclusion, and there exists $\alpha:=a_0+a_1\psi+...+a_n\psi^{n}\in \mathscr{O}_L, n\le (p-1)/2$, such that $\forall j, 1\le j\le n, a_j\in\mathbb{Q}$, and there exists $a_i\notin \mathbb{Z}$. If $a_n\in \mathbb{Z}$, we can subtract $\alpha-a_n\psi^n$ to obtain another element in $\mathscr{O}_L$, and we continue this process to eliminate all leading coefficients in $\mathbb{Z}$. Hence, without loss of generality, we can assume $a_n\notin \mathbb{Z}$. Now,
$$\zeta^n\alpha=\zeta^n(a_0+a_1(\zeta+\zeta^{-1})+...+a_n(\zeta+\zeta^-1)^n)=...+a_n\zeta^{2n}$$
That means, the leading coefficient of $\zeta^n\alpha$ by representing it as a polynomial of variable $\zeta$ is $a_n\zeta^{2n}$. However, $\mathscr{O}_L\subset\mathscr{O}_K=\mathbb{Z}[\zeta]$, and hence, both $\zeta,\alpha\in \mathbb{Z}[\zeta]$. This yields $a_n\in \mathbb{Z}$, a contradiction to our assumption. Therefore, $\mathscr{O}_L=\mathbb{Z}[\zeta+\zeta^{-1}]$.
(Q.E.D)
2. Non-monogenic ring of integers. One can see that both $\mathscr{O}_K, \mathscr{O}_L$ are monogenic, i.e. they have the form $\mathbb{Z}[\alpha]$ for some $\alpha$ integral in $K$. But this does not hold in general. We will consider a counter example by Dedekind.
Proposition 2. Let $\alpha$ be a root of an equation $f(x):=x^3 - x^2-2x-8=0$, and $K:=\mathbb{Q}(\alpha)$, then $\mathscr{O}_K$ is not monogenic.
Proof. Denote $d(\mathbb{Z}[\alpha])=D(1,\alpha,\alpha^2)$, then it can be seen that $d(\mathbb{Z}[\alpha])=-4.503$, and because $503$ is a prime, one can see by our Part 1 that the possible primes that are ramified in $\mathscr{O}_K$ are $2$ or $503$. By using Kummer-Dedekind's theorem, we can detect that $2$ is singular on $\mathbb{Z}[\alpha]$. By our previous note, we also know that $d(\mathbb{Z}[\alpha])=m^2\Delta_K$, where $m$ is the index $[\mathscr{O}_K:\mathbb{Z}[\alpha]]$ as $\mathbb{Z}$-module. Hence, $\Delta_K=-503$. From this, one can see $2$ is unramified in $\mathscr{O}_K$. Because $N(2)=8$, we have three cases:
(1) $2$ is inert in $\mathscr{O}_K$.
(2) $(2) = \mathfrak{p}_2\mathfrak{p}_4$ with $\mathfrak{p}_i$ denotes the prime ideal of norm $i$
(3) $(2) =\mathfrak{p}_2\mathfrak{q}_2\mathfrak{r}_2$, i.e. product of some prime ideals of norm 2.
Note that the minimum polynomial of $\alpha+1$ is $(x+1)^3-(x+1)^2-2(x+1)-8=0$, hence, $N(\alpha+1)=-10$, i.e. it can be factorized as $\mathfrak{p_2}{q}_5$, because $\mathscr{O}_K/(\alpha+1)=N(\alpha+1)=10$, and no finite field has 10 element. That means, there exists $\mathfrak{p}_2$ lying over $2$, and (1) cannot happen. Assume that (2) happens, i.e. $(2)=\mathfrak{p}_2\mathfrak{p}_4$, and they are all prime ideals lying over 2 in $\mathscr{O}_K$. We can also see that $N(\alpha)=8$, and $(\alpha)$ cannot be a prime ideal, since otherwise, $\mathfrak{p}=(\alpha)$ lies over 2, with $N(\mathfrak{p})=N(2)$, i.e. (2) is a prime ideal, that cannot happen. Hence, we must have $(\alpha)=\mathfrak{p}_2^3$, or $\alpha=\mathfrak{p}_2\mathfrak{p}_4$. In any case, we have $\gcd((\alpha),(\alpha+1))=\mathfrak{p}_2$, a contradiction, because the two ideal are co-prime. This yields (1) and (2) cannot happen.
And hence, $(2)$ is a product of three distinct prime ideals of norm 2 in $\mathscr{O}_K$. If there exists $\beta\in \mathscr{O}_K$, such that $\mathscr{O}_K=\mathbb{Z}[\beta]$, we can see, by Kummer-Dedekind's theorem that if $f_\beta(x)$ is the minimum polynomial of $\beta$, then $f_\beta$ will have three distinct root in $\mathbb{F}_2$, which is absurd, because $\mathbb{F}_2$ just has 2 elements. This shows $\mathscr{O}_K$ is not monogenic.
(Q.E.D)
Actually, we can point out the integral basis for $\mathscr{O}_K$, by noting that $\beta:=(\alpha+\alpha^2)/2$ is in $\mathbb{O}_K$, and $D(1,\alpha,\beta)=-503=\Delta_K$. Hence, $\mathbb{Z}[1,\alpha,\beta]=\mathscr{O}_K$.
3. Short note on integral basis of cubic fields. We begin this section by an interesting
Lemma 3.1. Let $f(x)=x^n+a_{n-1}x^{n-1}+...+a_1x+a_0\in\mathbb{Z}[x]$ be an Eisentein polynomial, i.e., there exists a prime number $p$ such that $p|a_i$ for all $0\le i\le n-1$, and $p^2\nmid a_0$, then $p$ is not singular in $\mathbb{Z}[\alpha]$, where $\alpha$ is a root of $f(x)=0$, or equivalently, $p\nmid [\mathscr{O}_K:\mathbb{Z}[\alpha]]$.
Proof. It follows directly from the Kummer-Dedekind's theorem (Q.E.D)
Now, we turn to the integral basis of a cubic field $\mathbb{Q}(\alpha)$, where $\alpha$ is a root of $f(x):=x^3 -r$, for some $r=ab^2$ is an integer, where $ab$ is square free (which does imply $a,b$ are square free and $\gcd(a,b)=1$), and if $3|r$, then $3|a$.
It can be seen first that the discriminant of $f$, i.e. $D(1,\alpha,\alpha^2)=N(f'(\alpha))=-27r^2=-3^3r^2$. And hence, $-27r^2=m^2\Delta_K$, where $m$ is the index $[\mathscr{O}_K:\mathbb{Z}[\alpha]$. First, if $p|a$, then $f(x)$ is an Eisentein polynomial respect to $p$, i.e. $p\nmid m$, by the previous lemma, and hence, $p^2|\Delta_K$, which yields, $a^2|\Delta_K$. If $3|r$, then $3|a$, and hence, $27a^2|\Delta_K$.
Because $\alpha^3 = ab^2$, we have $\alpha^6 = a^2b^4$, i.e. $\frac{\alpha^2}{b}=a^2b$. Let $\beta:=\frac{\alpha^2}{b}$, $\beta$ has the minimum polynomial $g(x):=x^3 - a^2b$. Hence, $\beta\in \mathscr{O}_K$. Let $m'$ be the index $[\mathscr{O}_K:\mathbb{Z}[\beta]]$, we have $d(\mathbb{Z}[\beta])=-27a^4b^2=m'^2\Delta_K$. For any $p|b$, $g(x)$ is an Esisentein polynomial with respect to $p$, by Lemma 3.1, we have $p\nmid m'$, and $p^2|\Delta_K$, which yields $b^2|\Delta_K$.
In short, if $3|r$, we have $\Delta_K = -27r^2=d(\mathbb{Z}[\alpha])$, or equivalently, $\mathscr{O}_K=\mathbb{Z}[\alpha]$, when $r\equiv 0\mod 3$. Otherwise, $\Delta_K=-3r^2$, or $\Delta_K=-27r^2$. That means, in any case, $3$ is ramified in $\mathscr{O}_K$, and $(3)=\prod_{i=1}^n\mathfrak{p}_i^{e_i}$, with some $e_i>1$. Furthermore, in the first and the third case, we can see $m=1$, and in the second case $m=3$, i.e. 3 is singular on $\mathbb{Z}[\alpha]$. And one can realize that to compute explicitly $\Delta_K$ by given $r$, we must look at 3 in $\mathbb{Z}[\alpha]$. The polynomial $f(x) = x^3 - r\in \mathbb{F}_3[x]$ can be factorized as $x^3, (x-1)^2(x+1)$ or $(x+1)^2(x-1)$, with respect to three cases $r\equiv 0,1,2\mod 3$. The case $3|r$ is treated by our above argument. We consider $r\equiv 1,-1\mod 3$. If $r\equiv 1\mod 3$, then by Kummer-Dedekind's theorem, the remainder when we divide $x^3 - r$ by $x-1$ is $r-1$, and $3$ is singular on $\mathbb{Z}[\alpha]$ iff $r\equiv1\mod 9$. Similarly, when $r\equiv -1\mod 3$, then $3$ is sunglar on $\mathbb{Z}[\alpha]$ iff $r\equiv -1\mod 9$.
In short, what we have done so far can be summarized as follows. If $r\not\equiv\pm1\mod 9$, then $\mathscr{O}_K=\mathbb{Z}[\alpha]$, and the integral basis of $\mathscr{O}_K$ in this case is just $1,\alpha,\alpha^2$. Otherwise, if $r\equiv\pm 1$, then $\Delta_K=-3r^2$, and $(3)$ is singular on $\mathbb{Z}[\alpha]$.
If $r\equiv 1\mod 9$, then it can be seen that
$$\beta:=\frac{1+\alpha+\alpha^2}{3}=\frac{\alpha^3-1}{3(\alpha-1)}=\frac{r-1}{3(\alpha-1)}$$
is in $\mathscr{O}_K$. And it is easy to see that $D(1,\alpha,\beta)=\Delta_K$, and in the integral basis in this case is $(1,\alpha,\frac{1+\alpha+\alpha^2}{3})$.
If $r\equiv-1\mod 9$, then it can be seen that $\frac{1-\alpha+\alpha^2}{3}$ is in $\mathscr{O}_K$, and the integral basis in this case is $(1,\alpha,\frac{1-\alpha+\alpha^2}{3})$.
4. The integral basis of multi-quadratic fields. If $m$ is a square free integers, and $K:=\mathbb{Q}(\sqrt{m})$ be a quadratic field, it can be easily proved that $\mathscr{O}_K=\mathbb{Z}[\sqrt{m}]$ if $m\equiv 2,3\mod 4$, and $\mathscr{O}_K=\mathbb{Z}[\frac{1+\sqrt{m}}{2}]$ if $m\equiv 1\mod 4$. Let $n$ be a square free integer, what is the integral basis for $\mathscr{O}_L$, for $L=\mathbb{Q}(\sqrt{m},\sqrt{n})$? We will partly answer this question in this section. We begin with the following important
Proposition 4.1. Let $K, L$ be two number fields, and $KL$ the smallest field containing both $K, L$. Let $d=\gcd(\Delta_K,\Delta_L)$, then $\mathscr{O}_{KL}\subset \frac{1}{d}\mathscr{O}_K\mathscr{O}_L$.
Proof.
Via this proposition, we obtain
Corollary 4.2. Let $K,L, KL$ be the same in Proposition, and $(\alpha_1,...,\alpha_n)$ the integral basis for $\mathscr{O}_K$, and $(\beta_1,...,\beta_m)$ the integral basis for $\mathscr{O}_L$, then $(\alpha_i\beta_j)_{ij}$ is the integral basis for $\mathscr{O}_{KL}$ if $\gcd(\Delta_K,\Delta_L)=1$.
Proof. Easily deduce from the proof of Proposition 4.1.
And we easily obtain the following
Corollary 4.3. Let $p,q$ be two distinct prime numbers, with $p,q\equiv 1\mod 4$, then $\mathscr{O}_{\mathbb{Q}(\sqrt{p},\sqrt{q})}=\mathbb{Z}[\alpha,\beta,\alpha\beta]$, where $\alpha=\frac{1+\sqrt{p}}{2}, \beta=\frac{1+\sqrt{q}}{2}$.
Proof. It can be easily seen that $\mathscr{O}_{\mathbb{Q}(\sqrt{p})}=\mathbb{Z}[\alpha]$, and $\mathscr{O}_{\mathbb{Q}(\sqrt{q})}=\mathbb{Z}[\beta]$. Furthermore, $\Delta_{\mathbb{Q}(\sqrt{p})}=p$, and $\Delta_{\mathbb{Q}(\sqrt{q})}=q$, and hence, $\gcd(p,q)=1$. It follows by Corollary 4.2 that $\mathscr{O}_{\mathbb{Q}(\sqrt{p},\sqrt{q})}=\mathbb{Z}[\alpha,\beta,\alpha\beta]$.
(Q.E.D)
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