Unique Factorization Domain - 1

2020-2학기, 대학에서 ‘현대대수1’ 수업을 듣고 공부한 바를 정리한 글입니다. 지적은 언제나 환영입니다 :)

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Associates / Associated Element

Definition 1. Associates

For two elts $a$, $b$ in a (Ring + Unity),

$a$ and $b$ are associates when $a \mid b$ and $b \mid a$. (Divisible by each other)

(cf)

$a \mid b \implies b = ax$

$b \mid a \implies a = by$

$a = by = (ax)y = axy$

따라서 $xy = 1$

Definition 2. Associates

For two elts $a$, $b$ in a (Ring + Unity),

$a$ and $b$ are associates each other,

if one can be obtained from the other by multiplying by some unit.

$\equiv$ $a = bu$ for some unit $u$.

• planetmath.org

Example. from StackExchange

In the Ring of Integers $\mathbb{Z}$, the units are $1$ and $-1$.

If $a, b \in \mathbb{Z}$ are associates, then $a = ub$ for some unit $u$.

Therefore $a = 1 \cdot b = b$ or $a = (-1) \cdot b = -b$.

Two Integers are associates in $\mathbb{Z}$ $\iff$ they are the same up to sign.

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Unique Factorization Domain

Definition 1. Unique Factorization Domain

An Integral Domain in which every non-zero elt has a unique factorization;
i.e. unique decomposition as the product of prime elements (or irreducible elements).

※ Note: In UFD, every irreducible element is a prime element.

• MathWorld/Unique Factorization Domain

Definition 2. Unique Factorization Domain

• Every $a \in D$ ,which is a non-zero & non-unit, can be written as a product of irreducibles.
• This decomposition is unique up to reordering, and up to associates.

Assume that two decompositions $a = p_1 \cdots p_n = q_1 \cdots q_m$ (all $p_i$, $q_i$ are irreducibles)

Then, $n=m$, and there exist a permutation $\sigma \in S_n$ s.t. $p_i = q_{\sigma(i)}$ are associates for all $i=1, \dots, n$.

• herzig’s note

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Theorem. ¹

In UFD, Irreducible elt is also a Prime elt.

일반적으로 Prime element면 Irreducible element다. link
하지만, Irreduciblility가 Primality를 항상 보장하는 것은 아니다.

그런데 UFD에서는 특별히 Irreducibility가 Primality를 보장한다!!

proof.

Let $p \in D$ be a irreducible element.

Let $p \mid ab$ for some $a, b \in D$.

(Goal) show $p \mid a$ or $p \mid b$

$ab = p\cdot c$ for some $c \in D$.

then, since $a$, $b$, $c$ are in UFD, they can be written as below

\[ab = (a_1 ... a_n)(b_1 ... b_m) = p \cdot (c_1 ... c_k)\]

이때 $a$, $b$, $c$의 factorization이 unique 하므로 $p$는 $ab$의 irreducible 중 하나와 associate 해야 한다.

즉, $p \sim a_i$ or $p \sim b_i$.

$\implies$ $p = a_i u$ or $p = b_i u$ for some unit $u$.

Let assume $p \sim a_i \equiv p = a_i u$.

\[a = a_1 \cdots a_n = a_1 \cdots (pu^{-1}) \cdots a_n\]

이것은 $p \mid a$를 의미한다.

마찬가지로 $p \sim b_i$를 가정하면, $p \mid b$를 얻는다.

따라서 $p \mid ab$에 대해 $p \mid a$ or $p \mid b$이므로

Irreducible element인 $p$는 Prime element이기도 하다. $\blacksquare$

Corollary.

$D$: Integral Domain

$D$ is an UFD

$\iff$ Every non-zero elt is a product of finitely many prime elt.

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PID와 UFD 사이의 관계에 대해선 UFD - 2에서 이어집니다.

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1. Dummit & Foote, “Abstract Algebra”, 286p ↩