Algebraic Number Theory

If $𝔞$ is a finitely generated $A$-submodule of the quotient field of $A$, then let $d\in A$ denote the product of the denominators of a set of generators. Then $\mathit{𝑑𝔞}\subseteq A$. Conversely, suppose that $𝔞$ is a fractional ideal and $d\in A$ is nonzero and satisfies $\mathit{𝑑𝔞}\subseteq A$. Then $\mathit{𝑑𝔞}$ is an ideal of $A$, hence finitely generated. Moreover, the multiplication-by-$d$ map carries $𝔞$ isomorphically onto $\mathit{𝑑𝔞}$. □

Let $K$ denote the quotient field of $A$. Let $c,d\in A$ be nonzero such that $\mathit{𝑐𝔞}\subseteq A$ and $\mathit{𝑑𝔟}\subseteq A$. Then $c(𝔞\cap 𝔟)\subseteq A$, $\mathit{𝑐𝑑}(𝔞+𝔟)\subseteq A$, and $\mathit{𝑐𝑑𝔞𝔟}\subseteq A$.

Note that ${𝔞}^{-1}$ is an $A$-submodule of $K$ which is nonzero since there exists $d\in A$ with $\mathit{𝑑𝔞}\subset A$ in that $𝔞$ is a fractional ideal. Let $a\in 𝔞$ be nonzero, and let $e\in A$ be its numerator in a representation of $a$ as a fraction, so $e\in 𝔞$ as well. For any $c\in {𝔞}^{-1}$, we have $\mathit{𝑐𝑒}\in A$ by definition, so $e{𝔞}^{-1}\subseteq A$, and therefore ${𝔞}^{-1}$ is a fractional ideal. □

For the nonobvious direction, suppose that $𝔟$ is an ideal of $A$ such that $\mathit{𝔞𝔟}=A$. Then we must have $𝔟\subseteq {𝔞}^{-1}$ by definition of ${𝔞}^{-1}$. On the other hand,

so we must have ${𝔞}^{-1}𝔞=A$. □

There exists $d\in A$ such that $\mathit{𝑑𝔞}=(b)$ for some $b\in A$. Then $\frac{b}{d}\in 𝔞$ and given any $c\in 𝔞$, we have $\mathit{𝑑𝑐}=\mathit{𝑏𝑎}$ for some $a\in A$, so $c=a\frac{b}{d}$. That is, $𝔞=(\frac{b}{d})$. □

If $x\in {(a)}^{-1}$, then $\mathit{𝑥𝑎}=b$ for some $b\in A$, so $x=b{a}^{-1}\in (a^{-1})$. If $x\in (a^{-1})$, then $x=a^{-1}b$ for some $b\in A$. On other hand, any $z\in (a)$ has the form $z=\mathit{𝑦𝑎}$ for some $y\in A$, and we have $\mathit{𝑥𝑧}=a^{-1}\mathit{𝑏𝑦𝑎}=\mathit{𝑏𝑦}\in A$, so $x\in {(a)}^{-1}$. We then have

completing the proof. □

A PID is Noetherian, and it is a UFD, so it is integrally closed. Its nonzero prime ideals are maximal, generated by its irreducible elements. □

That $𝔟\cap A$ is an ideal is clear, so it suffices to show that $𝔟\cap A$ is nonzero. Let $\beta \in 𝔟$ be nonzero and not a zero divisor. Then $\beta$ is a root of some monic polynomial $g\in A[x]$. Write $g=x^{n}f$ for some nonzero $f\in A[x]$ with nonzero constant term. Since $\beta \in 𝔟$, we have $f(\beta )-f(0)\in 𝔟$, and as $f(\beta )=0$ given that $\beta$ is not a zero divisor, we have $f(0)\in 𝔟$. But $f(0)\ne 0$, so $𝔟$ has a nonzero element. □

Let $𝔓$ be a nonzero prime ideal in $B$, and let $𝔭=𝔓\cap A$. Note that $F=A∕𝔭$ is a field as $𝔭$ is a nonzero (prime) ideal of $A$ by Lemma 2.2.4. For $\beta \in B$, let $f\in A[x]$ be a monic polynomial such that $\beta$ is a root of $f$. Let $\overline{f}\in F[x]$ denote the image of $f$ under the natural quotient map $A[x]\to F[x]$. Let $\overline{\beta }$ denote the image of $\beta$ in $B∕𝔓$. Then $\overline{f}(\overline{\beta })$ is the image of $f(\beta )=0$ in $B∕𝔓$ so is itself $0$. In other words, $\overline{\beta }$ is algebraic over $F$. Thus, $B∕𝔓=F[\{\overline{\beta }\mid \beta \in B\}]$ is a field. In other words, $𝔓$ is maximal. □

Note that $B$ is a finitely generated $A$-module by Corollary 1.4.22. If $𝔟$ is an ideal of $B$, then $𝔟$ is an $A$-submodule of $B$, and as $A$ is Noetherian, it is therefore finitely generated. Thus, $B$ is Noetherian. That $B$ is integrally closed is just Proposition 1.2.23. That every nonzero prime ideal in $A$ is maximal is Proposition 2.2.5. □

Given an ideal $𝔟$ of $S^{-1}A$, set $𝔞=A\cap 𝔟$. Then $𝔞$ is an ideal of $A$, and $𝔟=S^{-1}𝔞$. It follows that any set of generators of $𝔞$ as an ideal of $A$ generates $S^{-1}𝔞$ as an ideal of $S^{-1}A$. Hence $S^{-1}A$ is Noetherian. If, moreover, $𝔟$ is a nonzero prime, then clearly $𝔞$ is as well, and $𝔞$ is maximal since $A$ is a Dedekind domain. Then $S^{-1}A∕\mathit{𝔟\cong 𝐴}∕𝔞$ is a field, so $𝔟$ is maximal as well.

Let $K$ be the quotient field of $A$. Any $\alpha \in K$ that is integral over $S^{-1}A$ satisfies a monic polynomial $f$ with coefficients in $S^{-1}A$. Set $n=\mathrm{deg}f$. If $d\in S$ is the product of the denominators of these coefficients, then $d^{n}f(d^{-1}x)\in A[x]$ is monic with $\mathit{𝑑𝛼}\in K$ as a root. Since $A$ is integrally closed, we have $\mathit{𝑑𝛼}\in A$, so $\alpha \in S^{-1}A$. That is, $S^{-1}A$ is integrally closed. □

Consider the set $X$ of nonzero ideals of $A$ for which the statement of the first part of the lemma fails, and order $X$ by inclusion. Suppose by way of contradiction that $X$ is nonempty. Let $C$ be a chain in $X$. Either $C$ has a maximal element or there exist ${𝔞}_i\in C$ for $i\ge 1$ with ${𝔞}_i⊊{𝔞}_{i+1}$ for each $i$. The latter is impossible as $A$ is a Noetherian. By Zorn’s lemma, $X$ contains a maximal element $𝔞$. Now $𝔞$ is not prime since it lies in $X$, so let $a,b\in A-𝔞$ with $\mathit{𝑎𝑏}\in 𝔞$. Then $𝔞+(a)$ and $𝔞+(b)$ both properly contain $𝔞$, so by maximality of $𝔞$, there exist prime ideals ${𝔭}_1,\dots ,{𝔭}_k$ and ${𝔮}_1,\dots ,{𝔮}_l$ of $A$ for some $k,l\ge 0$ such that ${𝔭}_1\cdots {𝔭}_k\subseteq 𝔞+(a)$ and ${𝔮}_1\cdots {𝔮}_l\subseteq 𝔞+(b)$. We then have

a contradiction of $𝔞\in X$. This proves part a.

Now, suppose that $𝔞$ is proper, and let $𝔭$ be a prime ideal containing $𝔞$. Assume that every nonzero prime ideal of $A$ is maximal. If no ${𝔭}_i$ equals $𝔭$, then since ${𝔭}_i$ is maximal, there exist $b_i\in {𝔭}_i$ with $b\notin 𝔭$ for each $1\le i\le k$. We then have $b_1\cdots b_k\notin 𝔭$ as $𝔭$ is prime, so $b_1\cdots b_k\notin 𝔞$, a contradiction. Hence we have part b. □

Let $a\in 𝔭$ be nonzero. Noting Lemma 2.2.9a, we let $k\ge 1$ be minimal such that there exist nonzero prime ideals ${𝔭}_1,\dots ,{𝔭}_k$ of $A$ with ${𝔭}_1\cdots {𝔭}_k\subseteq (a)$. By Lemma 2.2.9b, we may without loss of generality suppose that ${𝔭}_k=𝔭$. Let $b\in {𝔭}_1\cdots {𝔭}_{k-1}$ be such that $b\notin (a)$. Then $a^{-1}b\notin A$, but we have

which implies that $a^{-1}b\in {𝔭}^{-1}$. Moreover, if ${𝔭}^{-1}𝔭=𝔭$, then $a^{-1}\mathit{𝑏𝔭}\subseteq 𝔭$. Since $𝔭$ is finitely generated, Proposition 1.2.4 tells us that $a^{-1}b$ is integral over $A$. But $A$ is integrally closed, so we have a contradiction. That is, we must have $𝔭⊊{𝔭}^{-1}𝔭\subseteq A$, from which it follows that ${𝔭}^{-1}𝔭=A$ by maximality of $𝔭$. □

First suppose that $𝔞$ is a nonzero ideal of $A$. We work by induction on a nonnegative integer $m$ such that there are nonzero prime ideals ${𝔮}_1,\dots ,{𝔮}_m$ of $𝔞$ (not necessarily distinct) with ${𝔮}_1\cdots {𝔮}_m\subseteq 𝔞$, which exists by Lemma 2.2.9a. If $m=0$, then $A\subseteq 𝔞$, so $𝔞=A$. For $m\ge 1$, we may suppose that $𝔞$ is proper, so there exists a nonzero prime ideal $𝔭$ that contains $𝔞$ and $𝔭={𝔮}_i$ for some $i\le m$. Without loss of generality, we take $i=m$. Then

By induction, there exist nonzero prime ideals ${𝔮}_1^{\prime },\dots ,{𝔮}_{\ell }^{\prime }$ of $A$ for some $\ell <m$ such that $𝔞{𝔭}^{-1}={𝔮}_1^{\prime }\cdots {𝔮}_{\ell }^{\prime }$. The desired factorization is given by multiplying by $𝔭$, applying Lemma 2.2.10, and gathering together nondistinct primes.

In general, for a fractional ideal $𝔞$, we let $d\in A$ be such that $\mathit{𝑑𝔞}\subseteq A$. We write $\mathit{𝑑𝔞}={𝔮}_1\cdots {𝔮}_m$ for some $m\ge 0$ and prime ideals ${𝔮}_i$ for $1\le i\le m$. We also write $(d)={𝔩}_1\cdots {𝔩}_n$ for some $n\ge 0$ and prime ideals ${𝔩}_i$ for $1\le i\le n$. By Lemma 2.2.10, we then have

If ${𝔮}_i={𝔩}_j$ for some $i$ and $j$, then we may use Lemma 2.2.10 to remove ${𝔮}_i{𝔩}_j^{-1}$ from the product. Hence we have the desired factorization.

Now suppose that

for some $k,l\ge 0$, distinct primes ${𝔭}_1,\dots ,{𝔭}_k$, distinct primes ${𝔮}_1,\dots ,{𝔮}_l$, nonzero $r_1,\dots ,r_k$, and nonzero $s_1,\dots ,s_l$. If $r_i<0$ (resp., $s_i<0$) for some $i$, we multiply both sides by ${𝔭}_i^{-r_i}$ (resp., ${{𝔮}_{𝔦}}^{-s_i}$) and obtain an equality of two products that involve only integral ideals. So, we assume without loss of generality that all $r_i$ and $s_i$ are positive. We may suppose that $t={\sum }_{i=1}^k{r}_i$ is minimal among all factorizations of $𝔞$. If $t=0$, then $k=0$, and then $l$ must be zero so that ${𝔮}_1^{s_1}\cdots {𝔮}_l^{s_l}$ is non-proper. If $t$ is positive, then ${𝔭}_k$ contains $𝔞$, so Lemma 2.2.9b tells us that ${𝔭}_k={𝔮}_i$ for some $1\le i\le l$. Multiplying both sides by ${𝔭}_k^{-1}$, the quantity $t$ is decreased by one. By induction, we have that the remaining terms are the same up to reordering, hence the result. □

For part a, note that if $𝔭\mid 𝔞$ and $𝔭\mid 𝔟$ for some prime ideal $𝔭$, then $𝔭\mid (𝔞+𝔟)$, so $𝔞+𝔟\ne A$. On the other hand, if there is no such $𝔭$, then $𝔞$ and $𝔟$ are not contained in any common maximal ideal (since $𝔭$ divides $𝔞$ if and only if it occurs in its factorization), so $𝔞+𝔟=A$.

For part b, using Theorem 2.2.11, write $𝔞={𝔭}_1\cdots {𝔭}_k$ (resp., $𝔟={𝔮}_1\cdots {𝔮}_l$) for some nonzero prime ideals ${𝔭}_i$ (resp., ${𝔮}_j$) of $A$. Suppose without loss of generality that ${𝔭}_i={𝔮}_i$ for all $1\le i\le t$ for some nonnegative $t\le \min (k,l)$ and that ${𝔭}_i$ (resp., ${𝔮}_i$) does not occur in the factorization of $𝔟$ (resp., $𝔞$) for $i>t$. Then

the last step using part a. We therefore have $t=l$, so $𝔟$ divides $𝔞$. □

Every element of $I(A)$ has the form $𝔞{𝔟}^{-1}$ where $𝔞$ and $𝔟$ are nonzero ideals of $A$. If $A$ is a PID, then both $𝔞$ and $𝔟$ are principal and, therefore, so is $𝔞{𝔟}^{-1}$. On the other hand, if $𝔞$ is a nonzero ideal of $A$ with $𝔞=(a)$ for some $a\in K$, then clearly $a\in A$, so $\mathrm{Cl}(A)$ being trivial implies that $A$ is a PID. □

We need only show that a Dedekind domain that is a UFD is a PID. Let $A$ be such a Dedekind domain. By Theorem 2.2.11, it suffices to show that each nonzero prime ideal $𝔭$ of $A$ is principal. Since $𝔭$ is prime and $A$ is a UFD, any nonzero element of $𝔭$ is divisible by an irreducible element in $𝔭$. If $\pi$ is such an element, then $(\pi )$ is maximal and contained in $𝔭$, so $𝔭=(\pi )$. □

This is a simple consequence of the fact that in a PID, every nonzero prime ideal is maximal generated by any irreducible element it contains. □

A DVR is a PID, hence a Dedekind domain, and it is local by definition. Conversely, suppose that $A$ is Noetherian, integrally closed, and has a unique nonzero prime ideal $𝔭$. We must show that $A$ is a PID. Since nonzero ideals factor uniquely as products of primes in $A$, every ideal of $A$ has the form ${𝔭}^n$ for some $n$. In particular, $𝔭=(\pi )$ for any $\pi \in 𝔭-{𝔭}^2$, and then ${𝔭}^n=({\pi }^n)$ for all $n$. Therefore, $A$ is a PID and hence a DVR. □

We have seen in Proposition 2.2.8 that $A_{𝔭}$ is a Dedekind domain for all nonzero prime ideals $𝔭$. By Proposition 2.3.4, each such localization is therefore a DVR.

Conversely, suppose $A$ is a Noetherian integral domain such that $A_{𝔭}$ is a DVR for every nonzero prime ideal $𝔭$. We can and do assume that $A$ is not a field and consider the intersection $B={\bigcap }_{𝔭}{A}_{𝔭}$ over all nonzero prime ideals $𝔭$ of $A$, taken inside the quotient field $K$ of $A$. Clearly, $B$ contains $A$, and if $\frac{c}{d}\in B$ for some $c,d\in A$ with $d\ne 0$, then we set

By definition of $B$, we may write $\frac{c}{d}=\frac{r}{s}$ with $r\in A$ and $s\in A-𝔭$, and we see that $\mathit{𝑠𝑐}=\mathit{𝑟𝑑}$, so $s\in 𝔞$. In other words, we have $𝔞⊈𝔭$ for all prime ideals $𝔭$ of $A$, which forces $𝔞=A$. This implies that $c\in (d)$, so $\frac{c}{d}\in A$.

Next, suppose that $𝔮$ is a nonzero prime ideal of $A$, and let $𝔪$ be a maximal ideal containing it. Then $𝔮A_{𝔪}$ is a nonzero prime ideal of $A_{𝔪}$, which is a DVR, so $𝔮A_{𝔪}=𝔪A_{𝔪}$. Since $𝔮$ and $𝔪$ are prime ideals contained in $𝔪$, we therefore have

Thus, every nonzero prime ideal is maximal.

Finally, each $A_{𝔭}$ is integrally closed in $K$ by Corollary 1.2.20, and then the intersection $A$ is as well, since any element of $K$ that is integral over $A$ is integral over each $A_{𝔭}$, hence contained in each $A_{𝔭}$. That is, $A$ satisfies the conditions in the definition of a Dedekind domain. □

Let $a,b\in K$ be nonzero (without loss of generality). Write $(a)={𝔭}^r𝔞$ and $(b)={𝔭}^s𝔟$ for $r=v_{𝔭}(a)$ and $s=v_{𝔭}(b)$ and fractional ideals $𝔞$ and $𝔟$ of $A$. Note that $(\mathit{𝑎𝑏})={𝔭}^{r+s}\mathit{𝔞𝔟}$, so $v_p(\mathit{𝑎𝑏})=r+s$. We have

so

□

Note that $2v(-1)=v(1)=0$, so we have $v(-a)=v(-1)+v(a)=v(a)$. □

If $v(a)<v(b)$, then

so we have $v(a)=\min (v(a+b),v(b))$, which forces $v(a+b)=v(a)$. □

That ${𝒪}_v$ is a ring follows from the fact that if $a,b\in {𝒪}_v$, then $v(\mathit{𝑎𝑏})=v(a)+v(b)\ge 0$, $v(-a)=v(a)\ge 0$, and $v(a+b)\ge \min (v(a),v(b))\ge 0$. For $a\in {𝒪}_v$ and $x,y\in {𝔪}_v$, we have $v(x+y)\ge \min (v(x),v(y))\ge 1$ and $v(\mathit{𝑎𝑥})=v(a)+v(x)\ge 1$, so ${𝔪}_v$ is an ideal. It is also the unique maximal ideal: given $a\in {𝒪}_v-{𝔪}_v$, we have $v(a^{-1})=v(a)+v(a^{-1})=v(1)=0$, so $a\in {𝒪}_v^{\times }$. Given an ideal $𝔞$ of ${𝒪}_v$, let $a\in 𝔞$ be an element of minimal valuation $n$. Let $\pi \in {𝒪}_v$ with $v(\pi )=1$, and write $a={\pi }^{n}u$ for some $u\in {𝒪}_v^{\times }$. Then $v(u)=0$, so $u\in {𝒪}_v^{\times }$. Therefore, $({\pi }^n)\subseteq 𝔞$. On the other hand, since $n$ is minimal, we have $𝔞\subseteq ({\pi }^n)$, and therefore $𝔞$ is a principal. By Lemma 2.3.2, we conclude that ${𝒪}_v$ is a DVR. □

Let $\{a_1,\dots ,a_m\}$ be a set of generators of $A$ as an $R$-module, and for each $i\le m$, let $r_i\in R-\{0\}$ be such that $r_i{a}_i\in R$. Note that $r_i$ exists as $A$ is contained in the quotient field of $R$. Then $r=r_1\cdots r_n$ is a nonzero element of ${𝔣}_R$.

Conversely, let $r\in {𝔣}_R$ be nonzero. Then multiplication by $r$ is an $R$-module isomorphism from $A$ to an ideal of $R$, which is finitely generated as $R$ is noetherian. □

This is a direct consequence of Proposition 1.4.20. □

First, suppose that ${𝔣}_R⊈𝔭$, and let $x\in {𝔣}_R$ with $x\notin 𝔭$. We have that $\mathit{𝑥𝐴}\subseteq R$ and $x\in R_{𝔭}^{\times }$, so $A\subseteq R_{𝔭}$. Let $𝔮=A\cap 𝔭R_{𝔭}$, which is a prime ideal of $A$ containing $𝔭$. We must then have $𝔭=𝔮\cap R$, since $𝔮\cap R$ is a prime ideal of $R$. Note that $R_{𝔭}\subseteq A_{𝔮}$. On the other hand, if $\frac{a}{s}\in A_{𝔮}$ for some $a\in A$ and $s\in A-𝔮$, then note that $\mathit{𝑥𝑎}\in R$ and $\mathit{𝑥𝑠}\notin 𝔭$, so $\frac{a}{s}\in R_{𝔭}$ as well. Thus, $R_{𝔭}=A_{𝔮}$ is a DVR by Proposition 2.3.4.

We claim that $𝔮=\mathit{𝔭𝐴}$. Clearly, $𝔮$ occurs in the factorization of $\mathit{𝔭𝐴}$, and if any other prime ideal ${𝔮}^{\prime }$ occurred in said factorization, then $A_{{𝔮}^{\prime }}$ would contain $R_{𝔭}=A_{𝔮}$, contradicting the fact that ${𝔮}^{\prime }$ is maximal. So $\mathit{𝔭𝐴}={𝔮}^e$ for some $e\ge 1$, but $𝔭R_{𝔭}=𝔭A_{𝔮}={𝔮}^e{A}_{𝔮}$ is the maximal ideal of $R_{𝔭}=A_{𝔮}$, so it equals $𝔮A_{𝔮}$, which forces $e=1$.

Conversely, suppose that $R_{𝔭}$ is a DVR, hence integrally closed. Since $A$ is integral over $R$, every element of $A$ is integral over the larger ring $R_{𝔭}$. In other words, we have that $A\subseteq R_{𝔭}$. Note that this also implies that $𝔭=\mathit{𝔭𝐴}\cap R$ since $𝔭=𝔭R_{𝔭}\cap R\supseteq \mathit{𝔭𝐴}\cap R$, while the other containment is immediate. Let $\{a_1,\dots ,a_n\}$ be a set of generators of $A$ as an $R$-module, and write $a_i=\frac{y_i}{s_i}$ for some $y_i\in R$ and $s_i\in R-𝔭$ for each $1\le i\le n$. Let $s=s_1\cdots s_n$. Then $s{a}_i\in R$ for each $i$, so $s\in {𝔣}_R$. Since $s\in R-𝔭$, we have that ${𝔣}_R⊈𝔭$. □

Let $R$ be a finitely generated $\mathbb{Z}$-submodule of $K$ of rank $[K:\mathbb{Q}]$. That $R$ is an order is an immediate corollary of Lemma 2.4.2. On the other hand, if $R$ is an order in ${𝒪}_K$, then its quotient field is $K$, so its rank as a $\mathbb{Z}$-module is $[K:\mathbb{Q}]$. □

Let $f\ge 1$ be such that $(f)={𝔣}_R\cap \mathbb{Z}$. Since $f\cdot {𝒪}_K\subseteq R$, we have that $f$ is a multiple of the exponent of ${𝒪}_K∕R$. On the other hand, suppose that some prime number $p$ divides $f$ but not $[{𝒪}_K:R]$. Since $p\mid f$, there exists a nonzero prime ideal $𝔭$ of ${𝒪}_K$ with $𝔭\cap {𝒪}_K=\mathit{𝑝\mathbb{Z}}$ that divides ${𝔣}_R$. Let $𝔤={𝔭}^{-1}{𝔣}_R$, which is an ideal of ${𝒪}_K$. Since multiplication by $p$ is invertible on ${𝒪}_K∕R$, we have

In other words, we have $𝔤{𝒪}_K\subseteq R$. But ${𝔣}_R⊊𝔤$, which is a contradiction. □