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

$\displaystyle \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

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

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

$\displaystyle 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

$\displaystyle d \equiv c\pmod{\mathfrak{p}^{n+1}}$   and$\displaystyle \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

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

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

William Stein 2012-09-24