# 数学代写|数论作业代写number theory代考|Factorization into Prime Ideals

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## 数学代写|数论作业代写number theory代考|Factorization into Prime Ideals

We now use Theorem 8.2.4 to prove the fundamental property of a Dedekind domain $D$, namely, that every proper integral ideal of $D$ can be expressed uniquely (up to order) as a product of prime ideals.

Theorem 8.3.1 If $D$ is a Dedekind domain every integral ideal $(\neq\langle 0\rangle, D)$ is a product of prime ideals and this factorization is unique in the sense that if
$$P_1 P_2 \cdots P_k=Q_1 Q_2 \cdots Q_l,$$
where the $P_i$ and $Q_j$ are prime ideals, then $k=l$ and after relabeling (ifnecessary)
$$P_i=Q_i, i=1,2, \ldots, k$$
Proof: Suppose there exist integral ideals $(\neq\langle 0\rangle, D)$ of $D$ that are not products of prime ideals. As $D$ is a Dedekind domain, it is Noetherian, and so by the maximal principle (Theorem 3.1.3) there is an ideal $A(\neq\langle 0\rangle, D)$ of $D$ maximal with respect to the property of not being a product of prime ideals. By Theorem 8.2.1 there exist prime ideals $P_1, \ldots, P_k(k \geq 1)$ of $D$ such that
$$P_1 \cdots P_k \subseteq A$$
Let $k$ be the smallest positive integer for which such a product exists. If $k=1$ then $P_1 \subseteq A \subset D$. As $P_1$ is a prime ideal, it is a maximal ideal since $D$ is a Dedekind domain. Thus $A=P_1$. This is impossible as $A$ is not a product of prime ideals. Hence $k \geq 2$. By Theorem 8.2.4 we have $\tilde{P}_1 P_1=D$ so that
$$\tilde{P}_1 P_1 P_2 \cdots P_k=D P_2 \cdots P_k \text {. }$$
Hence
$$\tilde{P}_1 A \supseteq \tilde{P}_1 P_1 \cdots P_k=P_2 \cdots P_k .$$

## 数学代写|数论作业代写number theory代考|Order of an Ideal with Respect to a Prime Ideal

Let $A$ be a nonzero fractional or integral ideal of a Dedekind domain $D$. Then $A$ can be written uniquely in the form
$$A=\prod_{i=1}^n P_i^{a_i}$$

where the $P_i$ are distinct prime ideals and the $a_i$ are integers (positive, negative, or zero).

Definition 8.4.1 (Order of an ideal with respect to a prime ideal) With the preceding notation, the order of the nonzero ideal $A$ of the Dedekind domain $D$ with respect to the prime ideal $P_i(i=1,2, \ldots, n)$, written $\operatorname{ord}{P_i}(A)$, is defined by $$\operatorname{ord}{P_i}(A)=a_i \text {. }$$
For any prime ideal $P \neq P_1, \ldots, P_n$ we define
$$\operatorname{ord}_P(A)=0 .$$
Clearly $\operatorname{ord}_P(\langle 1\rangle)=0$ and $\operatorname{ord}_P\left(P^k\right)=k$ for all prime ideals $P$.
Example 8.4.1 Let $D=\mathbb{Z}+\mathbb{Z} \sqrt{6}$. Let $B$ be the ideal $\langle 12,6 \sqrt{6}\rangle$. Then, with the notation of Example 8.3.3, we have
\begin{aligned} B & =\langle 12,6 \sqrt{6}\rangle=\langle 6\rangle\langle 2, \sqrt{6}\rangle=\langle\sqrt{6}\rangle^2\langle 2, \sqrt{6}\rangle \ & =A^2 P=(P Q)^2 P=P^3 Q^2, \end{aligned}
so that
$$\operatorname{ord}_P(B)=3, \operatorname{ord}_Q(B)=2$$
We now extend the concept of divisibility from integral ideals (Definition 8.3.1) to fractional ideals.

# 数论作业代写

## 数学代写|数论作业代写number theory代考|Factorization into Prime Ideals

$$P_1 P_2 \cdots P_k=Q_1 Q_2 \cdots Q_l,$$

$$P_i=Q_i, i=1,2, \ldots, k$$

$$P_1 \cdots P_k \subseteq A$$

$$\tilde{P}_1 P_1 P_2 \cdots P_k=D P_2 \cdots P_k \text {. }$$

$$\tilde{P}_1 A \supseteq \tilde{P}_1 P_1 \cdots P_k=P_2 \cdots P_k .$$

## 数学代写|数论作业代写number theory代考|Order of an Ideal with Respect to a Prime Ideal

$$A=\prod_{i=1}^n P_i^{a_i}$$

$$\operatorname{ord}_P(A)=0 .$$

\begin{aligned} B & =\langle 12,6 \sqrt{6}\rangle=\langle 6\rangle\langle 2, \sqrt{6}\rangle=\langle\sqrt{6}\rangle^2\langle 2, \sqrt{6}\rangle \ & =A^2 P=(P Q)^2 P=P^3 Q^2, \end{aligned}

$$\operatorname{ord}_P(B)=3, \operatorname{ord}_Q(B)=2$$

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