Lemma 1.1. Let $\alpha$ be integral over $\mathbb{Z}$, then there exists only finitely many primes $p\in\mathbb{Z}$ such that $\mathbb{Z}[\alpha]$ is singular above $p$ (i.e. there exists singular primes of $\mathbb{Z}[\alpha]$ that lies above $p$).
Proof. Let $f(x)=a_0+a_1x+...+x^n$ be the minimum polynomial of $\alpha$, where $a_i\in\mathbb{Z}$. By Kummer-Dedekind's theorem, $\mathbb{Z}[\alpha]$ is singular above $p$ only if $\gcd(\overline{f},\overline{f}')$ is not $1$ in $\mathbb{F}_p[x]$ (because the power of some prime factor in factorization is greater than 1). However, $f(x)$ is irreducible over $\mathbb{Q}[x]$ implies that $\gcd(f,f')=1$, and there exists $u,v\in\mathbb{Q}[x]$ such that $uf+vf'=1$. By clearing the denominator of $u,v$ with suitable multiplication with $k\in\mathbb{Z}$, we have $(ku)f+(kv)f'=k$, and $k$ is just not invertible in many finitely primes, whose prime divisor of $k$. It follows there exists only finitely many primes $p$ such that $\mathbb{Z}[\alpha]$ is singular above $p$. (Q.E.D)
From the lemma, one can see that there exists only finite prime number $p\in\mathbb{Z}$ such that $\mathbb{Z}[\alpha]$ is singular above $p$, and the fact that $\mathbb{Z}[\alpha]/p\mathbb{Z}[\alpha]$ is finite implies that there exists only finitely many primes of $\mathbb{Z}[\alpha]$ that lies over $p$. And hence, we obtain
Proposition 1.2. Let $\alpha$ be integral over $\mathbb{Z}$, then there exists only finitely many singular primes in $\mathbb{Z}[\alpha]$.
Let $R$ be a number ring, and $K$ the fraction field of $R$. Then $K/\mathbb{Q}$ is a finite extension, and it is well-known that such extension is also simple, and there exists a primitive element $\alpha\in K$ such that $K=\mathbb{Q}(\alpha)$. Let $f(x)=a_0+a_1x+...+a_nx^n$ be the minimum polynomial of $\alpha$, where $a_i\in\mathbb{Z}$, then $a_0a^{n-1} + a_1a_n^{n-2}(a_nx)+...+(a_nx)^n$ is the minimum polynomial of $a_n\alpha$. This implies $a_n\alpha$ is integral over $\mathbb{Z}$, and obviously $\mathbb{Z}[a_n\alpha]\subset R$, and both have the same fraction field. If we start with a non-zero prime ideal $\mathfrak{p}$ in $\mathbb{Z}[a_n\alpha]$, then it can be seen that there exists only finitely many prime ideal of $R$ lying over $\mathfrak{p}$ (to see this, just look at the finite ring $R/\mathfrak{p}R$). If we can prove that singular primes in $R$ actually lie above singular primes in $\mathbb{Z}[a_n\alpha]$, then by Proposition 2, we have proved there exists only finitely many singular primes in $R$.
Proposition 1.3. Let $R_1\subset R_2$ be two number rings with the same field of fraction. Let $\mathfrak{p}_2$ be a non-zero prime in $R_2$ and $\mathfrak{p}_1=\mathfrak{p}_2\cap R_1$ the regular prime in $R_1$. Then $\mathfrak{p}_2$ is regular in $R_2$.
Proof. We can see $(R_1)_{\mathfrak{p}_1}\subset (R_2)_{\mathfrak{p}_2}$, because $R_1\subset R_2$, and $R_1\setminus(R_1\cap P_2)=R_1\setminus P_2\subset R_2\setminus P_2$. If $(R_1)_{\mathfrak{p}_1}\subsetneq (R_2)_{\mathfrak{p}_2}$, then there exists $x\in (R_2)_{\mathfrak{p}_2}\setminus (R_1)_{\mathfrak{p}_1}$, and $(R_1)_{\mathfrak{p}_1}[x]\subset (R_2)_{\mathfrak{p}_2}$. However, because $\mathfrak{p}_1$ is regular, the local ring $(R_1)_{\mathfrak{p}}$ is a DVR, and for any $x\in K, x\notin (R_1)_{\mathfrak{p}_1}$, $(R_1)_{\mathfrak{p}_1}[x]=K$ as the following lemma points out.
Lemma 1.4. Let $R$ be a DVR with fraction field $K$. Let $x\in K\setminus R$, then $R[x]=K$.
Proof. Because $R$ is a DVR, for any $x\in K$, $x\in R$ or $1/x\in R$. By the hypothesis, $x\notin R$, $1/x\in R$, or we can represent $1/x=a/s$, where $s$ is invertible in $R$, and $a\in R$. If $a$ is also invertible, then both $a/s$ and $s/a$ lies in $R$, a contradiction. This implies there exists a positive integer $n$ such that $n=\pi^n$, where $\pi$ is the generator of the unique maximal ideal of $R$. Then $x=\frac{s}{\pi^n}$. This imlies $\frac{s}{\pi}=\frac{s}{\pi^n}\pi^{n-1}\in R$, or $\frac{s}{\pi}s^{-1}=\frac{1}{\pi}\in R$. Hence $K=R[1/\pi]\subset R[x]$, or $R[x]=K$. (Q.E.D)
Back to the problem, because $R_2$ is not $K$, and $\mathfrak{p}_2$ is non-zero, $(R_2)_{\mathfrak{p}_2}\subsetneq K$. This implies $(R_1)_{\mathfrak{p}_1}=(R_2)_{\mathfrak{p}_2}$, or $(R_2)_{\mathfrak{p}_2}$ is also a DVR. This implies $\mathfrak{p}_2$ is regular in $R_2$.
(Q.E.D)
As a consequence, we can see that any number ring has only finitely many singular primes. It is interesting to ask let $R_1$ be a number ring with a singular prime $\mathfrak{p}$, then how can we construct a smallest extension of $R_1$ such that primes above $\mathfrak{p}$ are regular? We need to develop more methods to see this.
2. Desingularization in number rings. Let $R$ be a number ring with $K$ is a field of fraction, then the ring of integer in $K$, denote $\mathscr{O}_K$ is of dimension 1, Noetherian and integrally closed. Or equivalently, it is a Dedekind domain. It is a "smallest" integrally closed ring in $K$, in the sense of the normalization $\tilde{R}$ contains $\mathscr{O}_K$, and any subring of $K$ contains $\mathscr{O}_K$ is Dedekind domain. This reasons why the ring of algebraic integers play an important role in algebraic number theory. We begin this section with a following important
Theorem 2.1. Let $R$ be a number ring with $K$ the fraction field of $R$, $\mathscr{O}_K$ be the ring of integers in $K$, $R$ any subring of $K$, then the following statement holds
(i) R is Dedekind iff $R$ contains $\mathscr{O}_K$.
(ii) The normalization $\tilde{R}$ of $R$ is exactly $R\mathscr{O}_K$.
Proof. (i) If $R$ is a number ring, it must contain $\mathbb{Z}$. And hence, if $R$ is integrally closed, it will contain the integral closure of $\mathbb{Z}$ in $K$, which is $\mathscr{O}_K$. Conversely, one can see that both $\mathscr{O}_K$ and $R$ have the same field of fraction, and $\mathscr{O}_K\subset R$. Hence, any singular prime ideal $\mathfrak{p}$ of $R$ lies above a singular prime in $\mathscr{O}_K$, by Proposition 1.3. But one can see that there exists no singular prime in $\mathscr{O}_K$, which does imply there exists no singular prime in $R\mathscr{O}_K$. Or $R$ must be a Dedekind domain.
(ii) It can be seen that the normalization $\tilde{R}$ of $R$ is a Dedekind domain, and it contains both $R$ and $\mathscr{O}_K$. Hence, $\tilde{R}=R\mathscr{O}_K$. (Q.E.D)
The previous theorem tell us that any number can be desingularized, as irreducible algebraic curves, one wants to find its normalization in its field of fraction. The following theorem states that $\tilde{R}$ is not too large, and it is an important argument, that help us step by step build the normalization of a number ring $R$.
Theorem 2.2. Let $R$ be a number ring, $K$ its field of fractions, and $\tilde{R}$ its normalization. Then $\tilde{R}/R$ is finite.
Proof. We can see that both $R$ and $\mathscr{O}_K$ have the same field of fractions $K$, it follows $S=R\cap \mathscr{O}_K$ is also a domain, with fraction field $K$. It can be considered as a $\mathbb{Z}$-submodule of $\mathscr{O}_K$, which is free of rank $[K:\mathbb{Q}]$. These argument implies $S$ is also a free $\mathbb{Z}$-module of rank $[K:\mathbb{Q}]$. Hence, as $\mathbb{Z}$-module, the quotient $\mathscr{O}_K/S$ is finite. This implies there exists an integer $k$ such that $k\mathscr{O}_K\subset S\subset R$. Also, by Theorem 2.1, $\tilde{R}=R\mathscr{O}_K$, and hence $k\tilde{R}=k\mathscr{O}_KR\subset R$. We know that for any ideal of a number ring, the quotient is always finite. This implies $\tilde{R}/R$ is finite. (Q.E.D)
And we give an important characterization for which prime $p$, $R$ is singular above it, that is $p|[\tilde{R}:R]$, without proof. I have to admit that I have no idea to prove it at this time.
Proof. (i) If $R$ is a number ring, it must contain $\mathbb{Z}$. And hence, if $R$ is integrally closed, it will contain the integral closure of $\mathbb{Z}$ in $K$, which is $\mathscr{O}_K$. Conversely, one can see that both $\mathscr{O}_K$ and $R$ have the same field of fraction, and $\mathscr{O}_K\subset R$. Hence, any singular prime ideal $\mathfrak{p}$ of $R$ lies above a singular prime in $\mathscr{O}_K$, by Proposition 1.3. But one can see that there exists no singular prime in $\mathscr{O}_K$, which does imply there exists no singular prime in $R\mathscr{O}_K$. Or $R$ must be a Dedekind domain.
(ii) It can be seen that the normalization $\tilde{R}$ of $R$ is a Dedekind domain, and it contains both $R$ and $\mathscr{O}_K$. Hence, $\tilde{R}=R\mathscr{O}_K$. (Q.E.D)
The previous theorem tell us that any number can be desingularized, as irreducible algebraic curves, one wants to find its normalization in its field of fraction. The following theorem states that $\tilde{R}$ is not too large, and it is an important argument, that help us step by step build the normalization of a number ring $R$.
Theorem 2.2. Let $R$ be a number ring, $K$ its field of fractions, and $\tilde{R}$ its normalization. Then $\tilde{R}/R$ is finite.
Proof. We can see that both $R$ and $\mathscr{O}_K$ have the same field of fractions $K$, it follows $S=R\cap \mathscr{O}_K$ is also a domain, with fraction field $K$. It can be considered as a $\mathbb{Z}$-submodule of $\mathscr{O}_K$, which is free of rank $[K:\mathbb{Q}]$. These argument implies $S$ is also a free $\mathbb{Z}$-module of rank $[K:\mathbb{Q}]$. Hence, as $\mathbb{Z}$-module, the quotient $\mathscr{O}_K/S$ is finite. This implies there exists an integer $k$ such that $k\mathscr{O}_K\subset S\subset R$. Also, by Theorem 2.1, $\tilde{R}=R\mathscr{O}_K$, and hence $k\tilde{R}=k\mathscr{O}_KR\subset R$. We know that for any ideal of a number ring, the quotient is always finite. This implies $\tilde{R}/R$ is finite. (Q.E.D)
And we give an important characterization for which prime $p$, $R$ is singular above it, that is $p|[\tilde{R}:R]$, without proof. I have to admit that I have no idea to prove it at this time.
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