# 代數導論 Week 3/4 - Unique Factorization Domain

[TOC]

## 定義：UFD

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假定 $R$ 是一個 *integral domain*。若對於 $R$ 中任意 ==非零== 且 ==非 *unit*== 的元素 $r$，$r$ 都滿足下面 2 點：

1. 存在 $R$ 中 *irreducible* 的元素 $p_1, p_1 \dots p_n$，使得 $r$ 是這些 *irreducible* 元素的乘積：

$$
\begin{align}
\exists &p_1,\ p_2 \dots ,p_n \text{ irreducibe}.
\newline
&\text{s.t. } r = p_1 \cdot p_2 \dots p_n
\end{align}
$$

2. 不同的分解在經過某個重排之後，每個元素最多差一個 *unit*：若 $r$ 恰好可以表為另外一組 *irreducible* 元素的乘積：

$$
r = q_1 \cdot q_2 \dots q_m \quad q_i \text{ irreducible}
$$

則兩組分解所需要的 *irreducible* 元素是相同的。即：

$$
m = n
$$

且在經過某個重排之後，對應位置的元素最多只差一個 *unit*。即：

$$
q_i = u_i\cdot p_{\sigma(i)} \quad u_i \text{ unit}
$$

其中，$\sigma$ 是某個 *permutation*，即 $\sigma \in S_n$。

則稱 $R$ 是一個 *Unique Factorization Domain*

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這裡的第二點有時會用「唯一性」來稱呼，因為是 *unique up to relabling and unit*。所以後面講唯一性的時候，指得都是在這種意義下的唯一。

## 引理：PID 滿足 Ascending Chain Condition (L4)

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假定 $R$ 是一個 *PID*，且 $\{I_k\}_{k \in \Bbb N}$ 在 *inclusion* 之下是一個 *chain*。即：

$$
I_1 \subseteq I_2 \subseteq \dots \subseteq I_n \subseteq \dots
$$

則存在 $n \in \Bbb N$，使得在 $k > n$ 之後，每個 $I_k$ 都跟 $I_n$ 一樣。即：

$$
\exists n \in \Bbb N.I_k = I_n \quad \forall k \geq n
$$

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如果這個 *chain* 裡面只有有限個元素，那沒有什麼好證明的，所以假定這個 *chain* 是無限的。考慮：

$$
I = \bigcup_{j \in \Bbb N}I_j
$$

由之前的證明 (證明 *maximal ideal* 時) 的結果：*chain* 中的所有 *ideal* 聯集起來，還會是一個 *ideal*：

$$
I \lhd R
$$

而 $R$ 是一個 *PID*，所以這個由 *chain* 中所有 *ideal* 聯集起來形成的 *ideal* $I$ 也可以被單一元素生成。假定它叫 $a$：

$$
\exists a \in R.(a) = I
$$

但是既然 $a$ 在 $\bigcup I_j$ 中，所以 $a$ 應該會在某個 $I_j$ 裡面。假定他在 $I_n$ 中：

$$
a \in I_n
$$

這時候就會發現，$I_n$ 就會是 $I$。因為 $I$ 是所有 $I_j$ 的聯集，所以也包含了 $I_n$; 但是 $I$ 的生成元 $a$ 又落在 $I_n$ 中，所以他生成的 *ideal* $I$ 也會包含在 $I_n$ 中：

$$
I_n \subseteq \bigcup I_j = (a) \subseteq I_n
$$

同樣的事情也會發生在所有的 $k \geq n$ 之後。因為他們是 *chain*，所以從 $I_n$ 之後，他們都會包含 $I_n$ 包含的所有東西，也包含 $I$; 但另外一方面，$I$ 又是 $\bigcup I_j$，所以這些 $I_k$ 又都要反過來包含在 $I$ 當中：

$$
I_k \subseteq \underbrace{\bigcup I_j}_{I} \subseteq I_k \Rightarrow I = I_k
$$

所以就證明了：

$$
I_k = I = I_n \quad \forall k \geq n
$$

## 定理：PID 都是 UFD

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若 $R$ 是一個 *PID*，則 $R$ 是一個 *UFD*

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### Step 1：非零非可逆元素可以拆出一個 irreducible

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Suppose $r \in R \setminus \{0\}$ and is not a unit. Then there exists an irreducible element $r_1 \in R$, and an element $a_1 \in R$, such that

$$
r = r_1a_1
$$

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If $r$ is irreducible, then choose

$$
r_1 = r, a_1 = 1
$$

Then the proof is done.If othersie $r$ is not irreducible, let

$$
r = b_1b_2
$$

such that $b_1, b_2$ are not unit. Since $r$ is not zero, $b_1, b_2$ can't be zero. Again, if $b_1$ is irreducible, them the proof is done. If $b_1$ is not irreducible, then let

$$
b_{1} = b_{11}b_{12}
$$

where $b_{11}, b_{12}$ are not unit. Observe that $(r)$ is properly contained in $(b_1)$

$$
(r) \subseteq (b_1)
$$

if $(r) = (b_1)$, them $r$ and $b_1$ only differs in a unit, so the proof is also done. Suppose otherwise:

$$
(r) \subset (b_1)
$$

and simliarily, assume $(b_1) \neq (b_{11})$, then:

$$
(r) \subset (b_1) \subset (b_{11})
$$

If the process stop, tehen we find an irreducible $r_1$ such that

$$
r = r_1 a
$$

If the process continues infinitely, then there's an ascending chain of ideal such that:

$$
(r) \subset (b_1) \subset (b_{11}) \subset (b_{111}) \dots
$$

Which contradicts to lemma.

### Step 2：非零非可逆元素可以分解為 Irreducible 的積

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Suppose $r \in R \setminus \{0\}$ and is not a unit. Then there exists an irreducible element $p_1 \dots p_n \in R$, such that

$$
r = p_1 p_2 \dots p_n
$$

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By step 1, exists irreducible $p_1$, such that

$$
r = p_1 u \quad u_1 \in R
$$

If $u$ is an unit, then the proof is done. Suppose otherwise $u_1$ is not a unit, then

$$
(r) \subseteq (u_1)
$$

Then again we can write

$$
u_1 = p_2 u_2 \quad p_2 \text{ irreducible}
$$

By step 1, if $u_2$ is a unit, the proof is done. If not, then we have

$$
(r) \subset (u_1) \subset (u_2)
$$

Either the process stops at some $n$, or we get an ascending chain. In the case of scending chain, by lemma 4 there's some $n$ such that

$$
u_{n - 1} = p_nu_n
$$

where $u_n$ is a unit. Apply backward substitution, then

$$
r = p_1 \dots p_n
$$

where $p_i$ are irreducible.

### Step 3：這樣的分解具有唯一性

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Suppose $r \in R \setminus \{0\}$ and is not a unit. And there exists an irreducible element $p_1 \dots p_n \in R$, and irreducible $q_1 \dots q_m$ such that

$$
\begin{align}
r &= p_1 \dots p_n
\newline
&= q_1 \dots q_m
\end{align}
$$

Then

1. $m = n$; and
2. $q_i = u_ip_{\alpha(i)}$ for some $\alpha(i) \in \{1 \dots n\}$ and unit $u_i$.

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Since $R$ is a PID, $(q_1)$ is a prime ideal. Then

$$
\exists \alpha(1) \in \{1 \dots n\}.(p_{\alpha(1)}) \subseteq (q_1)
$$

Thus

$$
p_{\alpha(1)} = q_1 u_1
$$

Since $p_{\alpha(1)}$ is irreducible, then either $q_1$ or $u_1$ is a unit. Since $q_1$ is also irreducible, then

$$
u_1 \text{ is a unit}
$$

Substitude $p_1$ with $q_1u_1$, then

$$
r = u_1q_1 \left(\prod_{i = 2}^{n}p_i\right) = q_1 \left(\prod_{i = 2}^{m}q_i\right)
$$

Since $R$ is an integral domain, cancellation $q_1$

$$
r = u_1 \underbrace{p_2p_3 \dots p_n} = \underbrace{q_2q_3 \dots q_m}
$$

Continue this process, we have

$$
u_1u_2 \dots u_n = q_{n+1} \dots q_m
$$

If $n \neq m$, then there exists $v$ such that

$$
uv = 1 = q_{n + 1} \dots q_mv
$$

Which leads to the conclusion that $q_{n + 1}$ is an unit, contradicting the fact that $q_{n + 1}$ is prime. Hence

$$
n = m
$$

Thus

$$
q_i = p_{\alpha(i)}u_i \quad u_i \in R\text{ is a unit }
$$

## 觀察：UFD 中的 Prime 都 Irreducible

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Suppose $R$ is an UFD, then a nonzero element is prime if and only if it's irreducible

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Since UFD is also a PID, then every prime is irreducible. Suppose that $r \in R\setminus \{0\}$ is an irreducible. Claim that $(r)$ is a prime ideal.

Suppose that $x, y \in R$ such that

$$
xy \in (r)
$$

Then

$$
xy = rb \quad b \in R
$$

Since $r$ is an irreducible element, by the uniqueness of factorization of $x$ and $y$, we know that

$$
r \mid x \text{ or }r \mid y
$$

Thus

$$
x \in (r) \text{ or }y \in (r)
$$

Which implies $(r)$ is a prime ideal.

## 觀察：整數是 UFD

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$$
\Bbb Z \text{ is a UFD}
$$

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因為他是 *Euclidean Domain* 所以是個 *PID* 所以是個 *UFD*