Let $R$ be a number ring, we already know that $R$ is of dimension 1 and Noetherian. However, in general, $R$ is not a Dedekind domain. That means, ideals in $R$ fail to be expressed as products of prime ideals. Or equivalently, $R$ is not integrally closed. As a consequence, there are some prime ideals of $R$ that are not invertible, and hence, the localization at these prime ideals are not DVR. These prime ideals with this property are called singular prime of $R$.
For curves, we already know that a point on an algebraic curve is smooth iff the localization of the coordinate ring at the corresponding maximal ideal is a local P.I.D, or a DVR. Therefore, if a localization at a point (or equivalently, the ring of regular functions at this point), is not a DVR, then this point is not smooth, or say, it is singular point. This is the very similar properties between number rings and algebraic curves. It is natural to ask these questions
1. How to detect singular primes in a number ring $R$? For curves, we already know the classical method is just check the vanishing of partial derivatives at some points in our curve.
2. Does there exists finitely many singular primes for a number ring $R$? For curves, by using Berzout's theorem, one can deduce that an irreducible algebraic curve has only finitely many singular points.
3. How to resolve the singularities? Basically, one will find the normalization of both number rings and coordinate rings of curves. But I am not sure if they are the same in nature. We will not discuss this deep topic in this post.
Let $R$ be a number ring, then we already know that any non-zero prime ideal of $R$ lies over some non-zero prime ideal of $\mathbb{Z}$, which generated by a prime number $p$ in $\mathbb{Z}$. From $p$, one may look up, i.e. find prime ideals of $\mathbb{R}$ that lie above it, and give a method to measure how far $pR$ cannot be expressed by product of prime ideals lying over it. We will deal with this question by restricted ourselves in the case $R=\mathbb{Z}[\alpha]$ first, where $\alpha$ is integral over $\mathbb{Z}$.
Theorem 1 (Kummer-Dedekind). Let $\alpha$ be integral over $\mathbb{Z}$, and $f(x)\in\mathbb{Z}[x]$ is the minimum polynomial of $\alpha$. Let $\overline{f}(x)$ be the reduction of $f(x)$ modulo $p$, and $\overline{f}(x)=\prod_{i=1}^m \overline{g_i}^{e_i}$ is the factorizations of $\overline{f}(x)$ into product of irreducible factors in $\mathbb{F}_p[x]$, and $g_i(x)$ be polynomials in $\mathbb{Z}[x]$ with the reduction modulo $p$ are $\overline{g_i}(x)$. Then the following holds
1. All prime ideals of $R=\mathbb{Z}[\alpha]$ that lie over $p$ are of the form $\mathfrak{p}_i=pR+g_i(\alpha)R$, and $\prod_{i=1}^m\mathfrak{p}_i^{e_i}\subset pR$.
2. The equality $pR = \prod_{i=1}^m\mathfrak{p}_i^{e_i}$ holds iff all $\mathfrak{p}_i$ are invertible.
3. $\mathfrak{p}_i$ is singular iff $e_i>1$ or the remainder $f(x)\mod g(x)$ is divisible by $p^2$.
Proof.
1. It can be seen that $\mathbb{Z}[\alpha]\cong \mathbb{Z}[x]/(f(x))$ by sending $\alpha$ to $x$, and from this $R/pR\cong \mathbb{Z}[x]/(p,f(x))\cong \mathbb{F}_p[x]/(\overline{f}(x))$. Hence, any prime ideal of $R$ lying over $p$ corresponds to the prime ideal of $\mathbb{F}_p[x]$ that contain $\overline{f}(x)$. From this, one can see that all of such prime ideals are of the form $\overline{g_i}(x)$. From this, one can see $\mathfrak{p}_i$ is the kernel of the map from $R$ to $\mathbb{F}_p[x]/(\overline{g_i}(x))$, which is identical to the kernel of the composition of two canonical homomorphism from
$$\mathbb{Z}[x]/(f(x))\rightarrow \mathbb{F}_p[x]/(\overline{f}(x))\rightarrow \mathbb{F}_p[x]/(\overline{g_i}(x))$$
The first map is the reduction modulo $p$, and the second is the reduction modulo $\overline{g_i}(x)$. Hence, the kernel will be $(p,g_i(x))$, and $\mathfrak{p}_i$ is $(p,g_i(\alpha))=pR+g_i(\alpha)R$. Moreover, we can see $\prod_{i=1}^m (pR+g_i(\alpha)R)^{e_i}\subset pR + \prod_{i=1}^m g_i(\alpha)^{e_i}R$. Via the homomorphism between $R$ and $\mathbb{F}_p[x]/(\overline{f}(x))$, $\prod_{i=1}^m g_i(\alpha)^{e_i}$ will be sent to $\overline{0}$, because $\prod_{i=1}^m \overline{g_i}(x)^{e_i}=\overline{f}(x)$. Hence $\prod_{i=1}^m g_i(\alpha)^{e_i}\in pR$. This implies $\prod_{i=1}^m (pR+g_i(\alpha)R)^{e_i}\subset pR$. And we are done the first part.
2. We have several ways to prove this fact, and all of them are beautiful at their own. First of all, one can see if $\prod_{i=1}^m\mathfrak{p}_i^{e_i}=pR$, then $p_i(\frac{1}{p}\prod_{j\ne i}\mathfrak{p}_j^{e_j})=R$, and both $p_i$ and $\frac{1}{p}\prod_{j\ne i}\mathfrak{p}_j^{e_j}$ are fractional ideals. Hence, $p_i$ is invertible. Conversely, if all $\mathfrak{p}_i$ are invertible, we have the following lemma
Lemma 2. Let $R$ be a local noetherian domain with maximal ideal $M$, then $R$ is a local P.I.D iff $M/M^2\cong R/M$.
Proof. If $R$ is a local P.I.D, or equivalently, a D.V.R, then we know that there is no ideal between $M$, and $M^2$, hence, the vector space $M/M^2$ over $R/M$ is of dimension 1, and it is isomorphic to the base field. Conversely, if $M/M^2\cong R/M$, then we know that $M\ne M^2$. Hence, there exists $t\in M\setminus M^2$, and $t+M^2$ is the generator for the vector space $M/M^2$ over $R/M$. The scalar multiplication is actually induced from the multiplication in $R$. Hence, the space generated by $t+M^2$ is actually $R(t+M^2)$, and $R(t+M^2)/M^2\cong R/M\cong M/M^2$. This yields $(t)+M^2=M$. Also, $M(M/(t))=(M^2+(t)/(t))$ (recall that as $R$-module, $m_1$ acts on $m_2+(t)$ as $m_1m_2+(t)$) $=M/(t)$. Because $R$ is Noetherian, all ideals we are considering are finitely generated. Hence $M(M/(t))=M/(t)$ implies that $M/(t)=0$, by Nakayama's lemma. This shows $M$ is actually a principal. And hence, $R$ is a local P.I.D. (Q.E.D)
Now, back to the problem, we can see if $\mathfrak{p}_i$ is invertible the localization $R_{\mathfrak{p}_i}$ is a local P.I.D. Applying the previous lemma, we have $\mathfrak{p}_i^n/\mathfrak{p}_i^{n+1}\cong R_{\mathfrak{p}_i}/\mathfrak{p}_iR_{\mathfrak{p}_i}$ (by just replacing $2$ by $n$ everywhere). And hence, $\#R_{\mathfrak{p}_i}/\mathfrak{p}_i^nR_{\mathfrak{p}_i}=(\#R_{\mathfrak{p}_i}/\mathfrak{p}_iR_{\mathfrak{p}_i})^n$. Furthermore, via the extension and contraction map, $R/\mathfrak{p}_i^{n}\cong R_{\mathfrak{p}_i}/\mathfrak{p}_i^nR_{\mathfrak{p}_i}$. From this, $\#R/\mathfrak{p}_i^nR=(\#R/\mathfrak{p}_iR)^n$.
By Chinese Remainder Theorem, and the recent remark, we have
$$\#R/\prod_{i=1}^m\mathfrak{p}_i^{e_i}=\prod_{i=1}^m\#R/\mathfrak{p}_i^{e_i}=\prod_{i=1}^m(\#R/\mathfrak{p}_i)^{e_i}$$
By the isomorphism $R/\mathfrak{p}_i\cong\mathbb{F}_p[x]/(\overline{g_i}(x))$, we have $\#R/\mathfrak{p}_i=p^{\deg g_i}$. Hence, $\#R/\prod_{i=1}^m\mathfrak{p}_i^{e_i}=p^{\sum_{i=1}^m e_i\deg g_i}=p^{\deg f}$. Similarly, via the isomorphism, $R/pR\cong \mathbb{F}_p[x]/(f(x))$, we have $\#R/pR=p^{\deg f}$. From this, we can see both $\prod_{i=1}^m\mathfrak{p}_i^{e_i}\subset pR$ are ideal of $R$ with finite index, they must be equal (recall $I\subset J$ be two ideals of $R$ then $(R/I)/(J/I)\cong R/J$). This finishes our second part.
3. We first see that the choice of $g_i(x)$ is arbitrary modulo $p$. Hence, without loss of generality, we can choose $g_i(x)$ as $\overline{g}_i(x)$. Then one can see if $\overline{f}(x)$ is irreducible in $\mathbb{F}_p[x]$, then the ring $R/pR\cong \mathbb{F}_p[x]$ is a field, or $pR$ is maximal ideal of $R$, and the factorization is trivial in this case, and this is also a principal ideal, and hence, invertible. We now treat the case $\overline{f}(x)$ is not irreducible in $\mathbb{F}_p[x]$. Because of that, we can assume $g_i(x)$ are monic, and $\deg g_i<\deg f$.
Because $f(x)-\prod_{i}g_i(x)^{e_i}\equiv \overline{f}(x)-\prod_{i}g_i(x)^{e_i}=0\mod p$, there exists $q_i(x),s_i(x)$, such that $f(x)=q_i(x)g_i(x)+ps_i(x)$ (actually, $q_i(x)=\prod_{j\ne i}g_j(x)^{e_j}$), with $\deg s_i<\deg g_i$, and $q_i(x)$ is also monic. Let $x=\alpha$, we have $-ps_i(x) = q_i(x)g_i(x)$. From the monic of $q_i,g_i$, we can see that both $q_i(\alpha),g_i(\alpha)$ is not in $pR$. Otherwise, via the isomorphism $R/pR\cong \mathbb{F}_p[x]/(\overline{f}(x))$, $q_(x)$ or $g_i(x)$ is divisible by $\overline{f}(x)$, which is impossible.
Now, let us assume that $\mathfrak{p}_i$ is singular, and $e_i=1$, then it can be seen that $q_i(x)$ is not divisible by $g_i(x)$, hence $q_i(\alpha)$ is not in $g_i(\alpha)R$ (via the isomorphism $R\cong \mathbb{Z}[x]/(f(x))$), and by previous argument, it is also not in $pR$. And hence, $g_i(\alpha)$ is invertible in the local ring $R_{\mathfrak{p}_i}$. And hence, $g_i(x)$ does not divides $s_i(x)$, or $s_i(\alpha)$ is not in $\mathfrak{p}_iR$. This yields $g_i(\alpha)\in pR_{\mathfrak{p}_i}$. Hence, the unique maximal ideal $\mathfrak{p}_iR_{\mathfrak{p}_i}$ in the local ring $R_{\mathfrak{p}_i}$ is generated by $p$. And it is a local P.I.D, or $\mathfrak{p}_i$ is invertible, a contradiction. Hence, $e_i=1$ in this case.
Assume that $\mathfrak{p}_i$ is singular and $s_i(x)\ne 0\mod p$ (or equivalently $p^2|f(x)\mod g(x)$). We can see $g_i(x)$ is irreducible modulo $p$ implies that it is irreducible over $\mathbb{Z}[x]$. These imply $s_i(\alpha)$ is invertible in the local ring $R_{\mathfrak{p}_i}$. And hence, $p\in g_i(\alpha)$, which implies the local ring $R_{\mathfrak{p}_i}$ is generated by $g_i(\alpha)$, and it is a local P.I.D, and also, $\mathfrak{p}_i$ is invertible in this case.
Conversely, if both $e_i>1$, and $p^2|s_i(x)$, then we will prove $[\mathfrak{p}_i:\mathfrak{p}_i]=\{x\in \mathbb{Q}(\alpha)|x\mathfrak{p}_i\subset \mathfrak{p}_i\}\ne R$, and it suffices to conclude $\mathfrak{p}_i$ is not invertible. When $e_i>1$, then $q_i(\alpha)\in p_i(\alpha)R$, and $s_i(\alpha)\in pR$. Also, $\deg q_i<\deg f$ implies $q_i(\alpha)\ne 0$, and at the second paragraph of 3, we have proved that $q_i(\alpha)\notin pR$ . This yields, $q_i(\alpha)/p$ does not lie in $R$, but
$$\frac{q_i(\alpha)}{p}(pR+g_i(\alpha)R) = q_i(\alpha)R-s_i(\alpha)R\in g_i(\alpha)R+pR\subset \mathfrak{p}_i$$
And hence $q_i(\alpha)/p$ does lie in $[\mathfrak{p}_i:\mathfrak{p}_i]$ this finishes our proof.
(Q.E.D)
As an application of Kummer-Dedekind's theorem, let $R=\mathbb{Z}[\zeta]$, where $\zeta$ is the $p^n$ primitive root of unity. We now give the factorization of $pR$, and prove that it is actually a Dedekind domain. By basic group theory, we can see there are $\varphi(p^n) = p^{n-1}(p-1)$ such primitive roots. Let us denote $S$ the set all primitive $p^n$ roots of unity. Moreover, from basic group theory again, for any $\zeta \in S$, $\zeta^{p^{n-1}}\ne 1$, and for any $\zeta_i,\zeta_j\in S$, there exists $k$, such that $\gcd(k,n)=1$, and $\zeta_i^k = \zeta_j$. From this, we can see all $\zeta_i\in \mathbb{Z}[\zeta]$, and the ideals $(1-\zeta_i)$ and $(1-\zeta_j)$ are actually the same in the $\mathbb{Z}[\zeta]$ (*).
Furthermore, it is obvious to see that a complex number $\zeta$ is a primitive $p^n$ root of unity iff it is the root of the polynomial
$$F(X) = \frac{X^{p^n}-1}{X^{p^{n-1}}-1}=X^{(p^{n-1})(p-1)} +...+X^{(p^{n-1})2}+X^{p^{n-1}}+1$$
Because there are $p^{n-1}(p-1)$ primitive $p^n$ roots of unity, and all of them are distinct roots of $F(X)$, we have the identity
$$F(X) = X^{(p^{n-1})(p-1)} +...+X^{(p^{n-1})2}+X^{p^{n-1}}+1=\prod_{\zeta\in S}(X-\zeta)$$
Taking $X = 1$, we have the $p=\prod_{\zeta\in S}(1-\zeta)$. From (*), we obtain an identity of ideals $p\mathbb{Z}[\zeta] = (1-\zeta)^{\varphi(p^n)}$ (**). In $\mathbb{F}_p[x]$, we have $X^{p^n}-1=(X-1)^{p^n}$, and $X^{p^{n-1}}-1=(X-1)^{p^{n-1}}$, and this implies the reduction of $F(X)$ in $\mathbb{F}_p[x]$ can be expressed $\overline{F(X)} = (X-1)^{p^{n-1}(p-1)}$. By the Kummer-Dedekind theorem, all prime ideals lie above $p$ in $\mathbb{Z}[\zeta]$ is of the form $p\mathbb{Z}[\zeta] + (1-\zeta)\mathbb{Z}[\zeta]$. But it can be seen from (**) that $p\mathbb{Z}[\zeta] \subset (1-\zeta)$. Or the only prime ideal lying over $p$ in $\mathbb{Z}[\zeta]$ is principal with generator $(1-\zeta)$. The factorization of $p\mathbb{Z}[\zeta]$ follows that there is no singular prime lie above $p$ via the Kummer-Dedekind theorem.
Let $q$ be another prime number different from $p$, we can see the factorization of $X^{p^n}-1$ in $\mathbb{F}_q[x]$ has no power occurring larger than 2, otherwise, it has repeated root in the $\overline{\mathbb{F}_q}$, but it is impossible, since its derivative $(X^{p^n}-1)'=p^nX^{p^n-1}$ is co-prime with itself. By the Kummer-Dedekind's theorem, no prime of $\mathbb{Z}[\zeta]$ lies above $q$ are singular.
Therefore, $\mathbb{Z}[\zeta]$ is a Dedekind domain.
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