Structural Applications of the CRT

The next lemma is an application of the Chinese Remainder Theorem. We will use it to prove that every ideal of $\O_K$ can be generated by two elements. Suppose that $I$ is a nonzero integral ideals of $\O_K$. If $a\in I$, then $(a)\subset I$, so $I$ divides $(a)$ and the quotient $(a)I^{-1}$ is an integral ideal. The following lemma asserts that $(a)$ can be chosen so the quotient $(a)I^{-1}$ is coprime to any given ideal.

Lemma 5.2.1   If $I$ and $J$ are nonzero integral ideals in $\O_K$, then there exists an $a\in I$ such that the integral ideal $(a)I^{-1}$ is coprime to $J$.

Before we give the proof in general, note that the lemma is trivial when $I$ is principal, since if $I=(b)$, just take $a=b$, and then $(a)I^{-1} = (a)(a^{-1})= (1)$ is coprime to every ideal.

Proof. Let $\mathfrak{p}_1,\ldots, \mathfrak{p}_r$ be the prime divisors of $J$. For each $n$, let $v_n$ be the largest power of $\mathfrak{p}_n$ that divides $I$. Since $\mathfrak{p}_n^{v_n}\neq \mathfrak{p}_n^{v_n+1}$, we can choose an element $a_n\in \mathfrak{p}_n^{v_n}$ that is not in $\mathfrak{p}_n^{v_n+1}$. By Theorem 5.1.4 applied to the $r+1$ coprime integral ideals

$$\mathfrak{p}_1^{v_1+1}, \ldots, \mathfrak{p}_r^{v_r+1}, I\cdot \left(\prod \mathfrak{p}_n^{v_n}\right)^{-1},$$

there exists $a\in\O_K$ such that

$$a \equiv a_n \pmod{\mathfrak{p}_n^{v_n+1}}$$

for all $n=1,\ldots,r$ and also

$$a \equiv 0 \quad \left(\text{mod} I\cdot \left(\prod \mathfrak{p}_n^{v_n}\right)^{-1}\right).$$

To complete the proof we show that $(a)I^{-1}$ is not divisible by any $\mathfrak{p}_n$, or equivalently, that each $\mathfrak{p}_n^{v_n}$ exactly divides $(a)$. First we show that $\mathfrak{p}_n^{v_n}$ divides $(a)$. Because $a\equiv a_n \pmod{\mathfrak{p}_n^{v_n+1}}$, there exists $b \in \mathfrak{p}_n^{v_n+1}$ such that $a = a_n + b$. Since $a_n\in \mathfrak{p}_n^{v_n}$ and $b \in \mathfrak{p}_n^{v_n + 1} \subset \mathfrak{p}_n^{v_n}$, it follows that $a\in \mathfrak{p}_n^{v_n}$, so $\mathfrak{p}_n^{v_n}$ divides $(a)$. Now assume for the sake of contradiction that $\mathfrak{p}_n^{v_n+1}$ divides $(a$); then $a_n=a-b\in \mathfrak{p}_n^{v_n+1}$, which contradicts that we chose $a_n \not\in\mathfrak{p}_n^{v_n+1}$. Thus $\mathfrak{p}_n^{v_n+1}$ does not divide $(a)$, as claimed. $\qedsymbol$

Suppose $I$ is a nonzero ideal of $\O_K$. As an abelian group $\O_K$ is free of rank equal to the degree $[K:\mathbf{Q}]$ of $K$, and $I$ is of finite index in $\O_K$, so $I$ can be generated as an abelian group, hence as an ideal, by $[K:\mathbf{Q}]$ generators. The following proposition asserts something much better, namely that $I$ can be generated as an ideal in $\O_K$ by at most two elements.

Proposition 5.2.2   Suppose $I$ is a fractional ideal in the ring $\O_K$ of integers of a number field. Then there exist $a,b\in{}K$ such that $I=(a,b)=\{\alpha a + \beta b : \alpha,\beta \in \O_K\}$.

Proof. If $I=(0)$, then $I$ is generated by $1$ element and we are done. If $I$ is not an integral ideal, then there is $x\in K$ such that $xI$ is an integral ideal, and the number of generators of $xI$ is the same as the number of generators of $I$, so we may assume that $I$ is an integral ideal.

Let $a$ be any nonzero element of the integral ideal $I$. We will show that there is some $b\in I$ such that $I=(a,b)$. Let $J=(a)$. By Lemma 5.2.1, there exists $b\in I$ such that $(b)I^{-1}$ is coprime to $(a)$. Since $a,b\in I$, we have $I\mid (a)$ and $I\mid (b)$, so $I\mid (a,b)$. Suppose $\mathfrak{p}^n\mid (a,b)$ with $\mathfrak{p}$ prime and $n\geq 1$. Then $\mathfrak{p}^n\mid (a)$ and $\mathfrak{p}^n\mid (b)$, so $\mathfrak{p}\nmid (b)I^{-1}$, since $(b)I^{-1}$ is coprime to $(a)$. We have $\mathfrak{p}^n\mid(b) = I\cdot (b)I^{-1}$ and $\mathfrak{p}\nmid (b)I^{-1}$, so $\mathfrak{p}^n \mid I$. Thus by unique factorization of ideals in $\O_K$ we have that $(a,b)\mid I$. Sine $I\mid (a,b)$ we conclude that $I=(a,b)$, as claimed. $\qedsymbol$

We can also use Theorem 5.1.4 to determine the $\O_K$-module structure of $\mathfrak{p}^n/\mathfrak{p}^{n+1}$.

Proposition 5.2.3   Let $\mathfrak{p}$ be a nonzero prime ideal of $\O_K$, and let $n\geq 0$ be an integer. Then $\mathfrak{p}^n/\mathfrak{p}^{n+1} \cong \O_K/\mathfrak{p}$ as $\O_K$-modules.

Proof. ^5.1Since $\mathfrak{p}^n\neq \mathfrak{p}^{n+1}$, by unique factorization, there is an element $b\in \mathfrak{p}^n$ such that $b\not\in \mathfrak{p}^{n+1}$. Let $\varphi :\O_K\to\mathfrak{p}^n/\mathfrak{p}^{n+1}$ be the $\O_K$-module morphism defined by $\varphi (a)=ab$. The kernel of $\varphi$ is $\mathfrak{p}$ since clearly $\varphi (\mathfrak{p})=0$ and if $\varphi (a)=0$ then $ab\in\mathfrak{p}^{n+1}$, so $\mathfrak{p}^{n+1}\mid (a)(b)$, so $\mathfrak{p}\mid (a)$, since $\mathfrak{p}^{n+1}$ does not divide $(b)$. Thus $\varphi$ induces an injective $\O_K$-module homomorphism $\O_K/\mathfrak{p}\hookrightarrow \mathfrak{p}^{n}/\mathfrak{p}^{n+1}$.

It remains to show that $\varphi$ is surjective, and this is where we will use Theorem 5.1.4. Suppose $c\in \mathfrak{p}^{n}$. By Theorem 5.1.4 there exists $d\in \O_K$ such that

$$d \equiv c\pmod{\mathfrak{p}^{n+1}}$$   and$$\qquad d \equiv 0\pmod{(b)/\mathfrak{p}^{n}}.$$

We have $\mathfrak{p}^n\mid (d)$ since $d\in\mathfrak{p}^n$ and $(b)/\mathfrak{p}^n\mid (d)$ by the second displayed condition, so since $\mathfrak{p}\nmid(b)/\mathfrak{p}^n$, we have $(b)=\mathfrak{p}^n\cdot(b)/\mathfrak{p}^n\mid (d)$, hence $d/b\in \O_K$. Finally

$$\varphi \left(\frac{d}{b}\right) \quad =\quad \frac{d}{b}\cdot d ... ...{\mathfrak{p}^{n+1}} \quad=\quad b\pmod{p^{n+1}} \quad=\quad c\pmod{p^{n+1}},$$

so $\varphi$ is surjective. $\qedsymbol$