Factorizations in Integral domains

Let $F$ be a field. Then in $F[x]$ the irreducible factors behave a lot like prime numbers. Building on Introductory Ring Theory and needed for Half-Factorial Domains

Definitions

Unit

Multiplicative inverses. We have it that $a$ and $b$ are units if $ab=1$.

Associates

Can be reached with unit multiplication, that is if $a$ and $b$ are associates then $a=ub$ for some unit $u$.

Irreducible Elements

Note elements, not ideals. An element in a integral domain is said to be irreducible if one of its factors is always an unit. I.E, for any $a\in D$, if $a=bc$, either $b$ or $c$ is a unit.

Prime Elements:

For $p\in D$, if $a\mid bc$, then either $a\mid b$ or $a\mid c$.

Unique Factorization Domain (UFD):

Any integral domain D is an UFD iff the following hold:

1. Let $a\in D$ such that $a\ne 0$ and a is not an unit. Then a can be written as a product of irreducible elements in D.
2. Let $a=p_1...p_r=q_1...q_s$ where the $p_i^{\prime }s$ and $q_i^{\prime }s$ are irreducible. Then r = s and there is a $\pi \in S_r$ (permutation in the cycle group of $S_r$) such that $p_i$ and $q_{\pi }(j)$ are associates for $j=1,...,r$. (Just a super fancy way of saying ever p is as associate of some q)

review Examples skipped in the book after this

Principal I

deal Domain (PID) Domain where every ideal is principal.

Integral domain:

In an integral domain:

1. $a\mid b⟺⟨b⟩\subset ⟨a⟩$.
2. $a$ and $b$ are associates $⟺⟨a⟩=⟨b⟩$
3. $a$ is a unit in $D$ $⟺⟨a⟩=D$

Proofs:

1. Lets prove the first theorem:
   1. $\Rightarrow$ Assume $a\mid b$.
      1. Then we know $a$ can generate $b$ and therefore every element within the ideal.
   2. $\Leftarrow$ Assume $⟨b⟩\subset ⟨a⟩$
      1. Then we know that every element of $b$ is generated within $a$. This is only possible if $a$ can generate $b$ which means $a\mid b$
2. Second theorem:
   1. $\Rightarrow$ Assume $a$ and $b$ are associates
      1. Then we have it so $a=ub$ for some unit $u$.
         1. Clearly, $b\mid a$ and $a\mid b$ so they are both subsets of each other.
3. Third Theorem:
   1. $\Rightarrow$ let $a$ be an unit in D
      1. Then we have it so $a{a}^{-1}=1$. Meaning $a$ is an associate of $1$
      2. By the previous logic $⟨a⟩=⟨1⟩$