Let us consider the free $\mathbb{Z}$-module of rank $n$, $\mathfrak{D}^{-1}:=\mathbb{Z}\omega_1^*\oplus...\oplus\mathbb{Z}\omega_n^*$, we can see that for any $x\in \mathfrak{D}^{-1}$, we can represent $x=\sum_{i=1}^n a_i\omega_i^*$, where $a_i\in \mathbb{Z}$. For all $y\in \mathscr{O}_K$, we can represent $y=\sum_{i=1}^n b_i\omega_i$, where $b_i\in \mathbb{Z}$. This yields by the property of dual basis that $tr(xy)=\sum_{i,j}a_ib_j\sigma_{ij}=\sum_{i=1}^na_ib_i\in \mathbb{Z}$. That means, for any $x\in \mathfrak{D}^{-1}$, and $y\in \mathscr{O}_K$, we always have $tr(xy)\in\mathbb{Z}$.
Conversely, for any $x\in K$, such that $tr(xy)\in\mathbb{Z}$, for all $y\in \mathscr{O}_K$, we will prove that $x\in \mathfrak{D}^{-1}$. By our previous argument, there exists $a_i\in \mathbb{Q}$, such that $x=\sum_{i=1}^na_i\omega_i^*$. Because $tr(xy)\in\mathbb{Z}$, for all $y\in \mathscr{O}_K$, in particular, $tr(x\omega_i)=a_i\in\mathbb{Z}$. That means, $x\in \mathfrak{D}^{-1}$. In short what we have proved is
Proposition 1.1. Let the notations be as above, then $\mathfrak{D}^{-1}=\{x\in K|tr(xy)\in\mathbb{Z}, \forall y \in \mathscr{O}_K\}$ is a fractional $\mathscr{O}_K$ module.
Proof. We just need to prove that $\mathfrak{O}_K$ is a fractional $\mathscr{O}_K$ module. We can see that each $\omega_i^*$ is an algebraic number, hence, there exists $m\in \mathbb{Z}\setminus\{0\}$ "clearing the denominator" of $\mathbb{Z}\omega_i$, for all $i$. That means, $m\mathfrak{D}^{-1}\in\mathscr{O}_K$. Hence, $\mathfrak{D}^{-1}$ is the fractional $\mathscr{O}_K$-module. (Q.E.D)
We are now ready to define the different ideal $\mathfrak{D}$ as an inverse element of $\mathfrak{D}^{-1}$ in the Picard group, it is called the different ideal of $\mathscr{O}_K$.
Proposition 1.2. $\mathfrak{D}$ is an ideal of $\mathscr{O}_K$, and $N(\mathfrak{D})=|\Delta_K|$.
Proof. By our definition $\mathfrak{D}=\{x\in K|xy\in \mathscr{O}_K, \forall y\in \mathfrak{D}^{-1}\}$. Take any $x\in \mathfrak{D}$, we can see $x\omega_j^*\in \mathscr{O}_K$. One can represent $x=\sum_{i=1}^na_i\omega_i$, where $a_i\in\mathbb{Q}$, and $tr(x\omega_i^*)=a_i\in\mathbb{Z}$, i.e. $x\in \mathscr{O}_K$. That means, $\mathfrak{D}$ is an ideal of $\mathscr{O}_K$.
For the second part, one can represent $\omega_i^*=\sum_{j=1}^na_{ij}\omega_j$, for some $a_{ij}\in\mathbb{Q}$, i.e. Let $A=(a_{ij})_{ij}$ be a matrix we have just obtained, and $B=(b_{ij})_{ij}$ is its inverse, we have $\sum_{j=1}^nb_{ij}\omega_j^*=\omega_i$. That means,
$$tr(\omega_i\omega_k)=tr((\sum_{j=1}^nb_{ij}\omega_j^*)\omega_k)=\sum_{j=1}^nb_{ij}tr(\omega_j^*\omega_k)=b_{ik}$$
This yields $\det B=\Delta_K$. And hence, $\det A = 1/\Delta_K$. On the other hand, we have
$$tr(\omega_i^*\omega_k^*)=tr((\sum_{j=1}^na_{ij}\omega_j)\omega_k^*)=\sum_{j=1}^na_{ij}tr(\omega_j\omega_k^*)=a_{ik}$$
This yields, $D(\omega_1^*,...,\omega_n^*)=\det (a_{ij})_{ij}=1/\Delta_K$. And our results from Part 1, we have $D(\omega_1^*,...,\omega_n^*)=N(\mathfrak{D}^{-1})^2\Delta_K$. This yields $N(\mathfrak{D}^{-1})=1/\Delta_K^2$, and hence, $N(\mathfrak{D})=|\Delta_K|$. (Q.E.D)
The propositions above give us the following
Proposition 1.3. Let $p\in\mathbb{Z}$ be a prime, and $\mathfrak{p}$ the ideal of $\mathscr{O}_K$ lying above $p$, and $e$ the ramification index of $\mathfrak{p}$. Then $\mathfrak{p}^{e-1}|\mathfrak{D}$.
Proof. It can be seen that $(p)=\mathfrak{p}I$, for some ideal $I\in \mathscr{O}_K$, with $\gcd(I,\mathfrak{p})=1$. We will prove that for all $x\in\mathfrak{p}I$, $p^{-1}x\in \mathfrak{D}^{-1}$, so that $p^{-1}\mathfrak{p}I\subset \mathfrak{D}^{-1}$. From this, $\mathfrak{D}p^{-1}\mathfrak{p}I\subset \mathscr{O}_K$, so that $D\subset p\mathfrak{p}^{-1}I^{-1}=\mathfrak{p}^{e-1}$. That means $\mathfrak{p}^{e-1}|\mathfrak{D}$.
Now, we are left to prove for all $x\in\mathfrak{p}I$, $p^{-1}x\in\mathfrak{D}^{-1}$, i.e. for all $y\in \mathscr{O}_K$, we will show that $tr(p^{-1}xy)\in \mathbb{Z}$. Because $x,y\in \mathscr{O}_K$, it suffices to prove that $tr(x)\in p\mathbb{Z}$. Now, we can represent $x=\sum_{j=1}^np_ji_j$, where $i_j\in I, p_j\in\mathfrak{p}$. And
$$x^p\equiv $\sum_{j=1}^np_j^pi_j^p\mod p$$
Similarly
$$x^{p^m}\equiv\sum_{j=1}^np_j^{p^m}i_j^{p^m}\mod p$$
For $m$ is large enough, i.e. $p^m>e$ we have $x^{p^m}\equiv 0\mod p$. This follows $tr(x^{p^m})\equiv tr(x)^{p^m}\mod p$, hence, $tr(x)\in p\mathbb{Z}$. (Q.E.D)
Corollary 1.4. If $p$ is ramified in $\mathscr{O}_K$, then $p|\Delta_K$. In particular, there exists only finitely prime $p$, such that $p$ is ramified in $\mathscr{O}_K$.
Proof. If $p$ is ramified in $\mathscr{O}_K$, then there exists $\mathfrak{p}$ lying over $p$ with ramification index $e\ge 2$. This yields, $\mathfrak{p}^{e-1}|\mathfrak{D}$, i.e. $\mathfrak{D}\subset\mathfrak{p}^{e-1}$, and $N(\mathfrak{p}^{e-1})|N(\mathfrak{D})=\Delta_K$, i.e. $p|\Delta_K$. (Q.E.D)
Note the the converse of Corollary 1.4 also holds, but we do not prove it here. In the next section, we will prove it in the case $\mathscr{O}_K$ is monogenic. The finiteness of ramification points is also important in Algebraic Geometry, when we want to define the pull-back map on the divisor class group.
2. Different ideal in the monogenic case. We now determine the different ideal in the important case when $\mathscr{O}_K$ is monogenic, i.e. $\mathscr{O}_K=\mathbb{Z}[\alpha]$, for some $\alpha\in\mathscr{O}_K$. To do this, we first need to compute explicitly $\mathfrak{D}^{-1}$. Similar to the first section, in this section, we fix $K$ the algebraic number field of degree $n$, and its ring of integers is monogenic.
Proposition 2.1. Let $\mathscr{O}_K=\mathbb{Z}[\alpha]$, and $f(x)$ the minimal polynomial of $\alpha$. If we write $f(x) = (x-\alpha)(b_{n-1}x^{n-1}+...+b_0)$, for $b_i\in \mathscr{O}_K$, then $(\frac{b_i}{f'(\alpha)})_{i=0}^{n-1}$ is the dual basis of $(\alpha^j)_{j=0}^{n-1}$ via the trace form.
Proof. It is equivalent to prove that $tr(\frac{b_i}{f'(\alpha)}\alpha^r)=\delta_{ir}$. Now, the celever idea comes, i.e. we prove that $\sum_{i=0}^{n-1}tr(\frac{b_i}{f'(\alpha)}\alpha^r)x^i=x^r$ (*). If we define the extension of the trace map, $tr: K[x]\to \mathbb{Q}[x]$, such that $tr(a_0+...+a_mx^m)=tr(a_0)+...+tr(a_m)x^m$. It can be checked that $tr$ is a $\mathbb{Q}$-linear map, and it is an extension of our usual trace map.
And (*) is equivalent to $tr(\sum_{i=0}^{n-1}\frac{b_i}{f'(\alpha)}{\alpha^r}x^i)=x^r$. Also,
$$\frac{b_i}{f'(\alpha)}\alpha^r x^i=\frac{\alpha^r}{f'(\alpha)}(\sum_{i=0}^{n-1}b_ix^i)=\frac{\alpha^r}{f'(\alpha)}\frac{f(x)}{x-\alpha}$$
If we denote $\alpha_1,...,\alpha_n$ conjugates of $\alpha$, then using the extension properties of trace map, we get
$$tr(\sum_{i=0}^{n-1}\frac{b_i}{f'(\alpha)}{\alpha^r}x^i)=tr(\frac{\alpha^r}{f'(\alpha)}\frac{f(x)}{x-\alpha})=tr(\sum_{i=1}^n(\frac{\alpha_i^r}{f'(\alpha_i)}\frac{f(x)}{x-\alpha_i}))$$
And it suffices to prove that
$$\sum_{i=1}^n(\frac{\alpha_i^r}{f'(\alpha_i)}\frac{f(x)}{x-\alpha_i})=x^r$$
Let $g(x):=\sum_{i=1}^n(\frac{\alpha_i^r}{f'(\alpha_i)}\frac{f(x)}{x-\alpha_i})-x^r$, which is a polynomial of degree $r\le n-1$. Because $f(x)$ is separable over $\mathbb{Q}$, all $\alpha_i$ are distinct, if we prove $g(\alpha_i)=0$ for all $i$, then $g(x)\equiv 0$.
Note that $f'(\alpha_1)=\prod_{j\ne 1}(\alpha_1-\alpha_j)=(\frac{f(x)}{x-\alpha_1})_{x=\alpha_1}$. And for all $j\ne 1, (\frac{f(x)}{x-\alpha_1})_{x=\alpha_j}=0$ Hence,
$$g(\alpha_1)=\sum_{i=1}^n(\frac{\alpha_i^r}{f'(\alpha_i)}\frac{f(x)}{x-\alpha_i})-\alpha_1^r=0$$
The similar conclusion holds for all $\alpha_i$, i.e. $g(x)\equiv0$, and our proposition follows. (Q.E.D)
Corollary 2.2. Let $\mathfrak{D}$ be the different ideal of $\mathscr{O}_K$, where $\mathscr{O}_K=\mathbb{Z}[\alpha]$, then $\mathfrak{D}=(f'(\alpha))$.
Proof. We can see from the previous proposition that $\mathfrak{D}^{-1}\subset \frac{1}{f'(\alpha)}\mathscr{O}_K$. It yields $\mathfrak{D}^{-1}f'(\alpha)\subset \mathscr{O}_K$, i.e. $f'(\alpha)\subset \mathfrak{D}$. By Proposition 1.2, $N(\mathfrak{D})=|\Delta_K|$, and by our results of Part 1, $|\Delta_K|=|N(f'(\alpha))|$, which yields $\mathfrak{D}=(f'(\alpha))$. (Q.E.D)
We now turn to the ramification of monogenic ring of integers.
Proposition 2.3. Let $\mathscr{O}_K=\mathbb{Z}[\alpha]$, then $p$ is ramified in $\mathscr{O}_K$ iff $p|\Delta_K$.
Proof. The direction "$\Rightarrow$" have been proved in Section 1. For "$\Leftarrow$", we just use Kummer-Dedekind's to prove that $f$ has double root in $\mathbb{F}_p[x]$. (Q.E.D)
3. Examples. We will prove that if $a$ is a square free integer, $p$ is a prime, that is coprime to $a$, $\alpha:=a^{1/p}$, and $K:=\mathbb{Q}(\alpha)$. Then $\mathscr{O}_K=\mathbb{Z}[\alpha]$ iff $a^{p-1}\not\equiv1\mod p^2$.
One can see that if $\mathscr{O}_K=\mathbb{Z}[\alpha]$, then due to the Kummer-Dedekind's theorem, because $x^p-a=(x-a)^p\in\mathbb{F}_p[x]$, we have $(p)=\mathfrak{p}^p$, where $\mathfrak{p}=(p,\alpha-a)$ in $\mathscr{O}_K$. It can be seen from this that $\alpha-a\in \mathfrak{p}$, and it is not in $\mathfrak{p}^2$. Hence, $(\alpha-a)=\mathfrak{p}I$, where $\gcd(\mathfrak{p}, I)=1$, which yields $\gcd(N(\mathfrak{p}), N(I))=1$. Taking the norm, we have $N(\alpha-a)=N(\mathfrak{p})N(I)$, and hence, $N(\alpha-a)\equiv0\mod p$, and $N(\alpha-a)\not\equiv0\mod p^2$. Also, the minimal polynomial of $\alpha-a$ is $(x+a)^p-a$, this follows $N(\alpha-a)=a^p-a\not\equiv0\mod p^2$.
Conversely, we first see $d(\mathbb{Z}[\alpha])=(-1)^{p(p-1)/2}a^{p-1}p^{p}=m^2\Delta_K$, where $m$ is the index of $[\mathscr{O}_K:\mathbb{Z}[\alpha]]$. And the possible singular primes are primes dividing $a$ or $p$. Let $q$ be a prime divisor of $a$, then $x^p-a$ is Eisentein respect to $q$. By our results from Part 2, we can see $q\nmid m$, i.e. $q$ is not singular on $\mathbb{Z}[\alpha]$. Also, by Kummer-Dedekind's theorem again, the remainder when we divide $x^p-a$ to $x-a$ is $a^p-a$, which is not divisible by $p^2$. Hence, $p$ is also not singular on $\mathbb{Z}[\alpha]$. This yields $p\nmid m$. Hence, $m=1$, and $\mathbb{Z}[\alpha]=\mathscr{O}_K$.
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