Prime & Irreducible element
2020-2학기, 대학에서 ‘현대대수1’ 수업을 듣고 공부한 바를 정리한 글입니다. 지적은 언제나 환영입니다 :)
Definition. prime element
Let $R$ be a commutative ring, and $p \in R$.
if
- $p$ is not a zero, and not a unit
- for $a, b \in R$, $p \mid ab$ implies $p \mid a$ or $p \mid b$
then, $p$ is a “prime element”.
Definition. irreducible element
Let $D$ be an integral domain, and $a \in D$.
if $a$ is not a product of two non-units,
then $a$ is a “irreducible element”.
원래는 Irreducible Polynomial에 입각해 Irreducibility를 생각했는데, 그렇게 Irreducibility를 생각하는 것보단 Ring Theory에서 정의하는 Irreducibility 자체를 받아들이는게 좋을 듯!!
An element $a$ is irreducible
$\iff$ possible decompositions of $a$ into the product of two factors are of the form
\[a = u^{-1} \cdot (ua)\]즉, 이것을 다시 말하면,
“If $a = pq$, then $p$ or $q$ is an unit.”
왜냐하면, 만약 $p$가 unit element라면,
\[\begin{aligned} a &= pq\\ p^{-1}a &= p^{-1}(pq) \\ p^{-1}a &= q \\ \end{aligned}\]따라서 $a = pq = p(p^{-1}a) = 1 \cdot a$가 된다!!
Theorem: Prime - Irreducible
$D$: Integral Domain
Prime element in $D$ $\implies$ Irreducible in $D$
일반적으로 역방향은 성립하지 않는다!!
proof.
($\implies$)
Since $p$ is a Prime element,
If $p \mid ab$ for some $a, b \in D$
then, $p \mid a$ or $p \mid b$.
Supp. $p = ab$, (Check) $a$ or $b$ is an unit.
Sine $p$ is a Prime element,
$p \mid ab$ $\implies$ $p \mid a$ or $p \mid b$
Say $p \mid a$, then $a = p \cdot a’$.
Then,
\[\begin{aligned} p &= ab = (p \cdot a') b \\ 1 &= a' b \end{aligned}\]따라서 $b$는 $a’$를 multiplicative inverse로 갖는 unit이다.
반대로 $p \mid b$라면, $a$가 unit이다.
따라서 $p$는 irreducible이다. $\blacksquare$
($\impliedby$의 반례)
Integer Domain(또는 Qudratic Integr Ring)과 Algebraic Norm에 대한 개념을 알아야 반례를 쉽게 찾을 수 있다.
Moreover, while an ideal generated by a prime element is a prime ideal,
it is not true in general that an ideal generated by an irreducible element is an irreducible ideal.