UFDs

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Introduction

Recall a UFD is an integral domain in which every nonzero, nonunit element has a unique factorization. The uniqueness part is the most important part of the factorization.

The obvious remark

In elementary number theory, we use the Fundamental Theorem of Arithmetic to make a lot of statements about divisibility of integers. Among one of the most important ones is that $a \mid b$ if and only if all of the prime factors of $a$ are prime factors of $b$. A version of this holds in the much more general notion of a UFD.

[theorem] [thmtitle]Theorem (“The obvious remark”).[/thmtitle]

Let $R$ be a UFD and let $a, b, c$ be nonzero elements of $R$. Then

  1. $(a) \subseteq (b)$ if and only if the multiset of irreducible factors of $b$ is contained in the multiset of irreducible factors of $a$.
  2. $a$ and $b$ are associates if and only if the multisets of irreducible factors of $a$ and $b$ are the same.
  3. The irreducible factors of $ab$ are the “union” of the multisets of irreducible factors of $a$ and $b$. [/theorem]

(1) Assume that $(a) \subseteq (b)$. In particular, we know that this implies $b \mid a$. Since $R$ is a UFD, we may uniquely write

\[b = p_1 \cdots p_s\]

where all the factors listed above are irreducibles. Now, since $p_i \mid b$, it follows that $b = p_ir$ for some $r\in R$. Since $b \mid a$, we know that $a = bs$ for some $s\in R$. So $a = p_i (rs)$ and $p_i \mid a$. So $p_i$ is an irreducible factor of $a$. Conversely, if we start with the assumption that $p_i \mid b$ implies $p_i \mid a$, then we will show that it is also the case that $\prod_{i} p_i$ also divides $a$. We proceed by induction on the number of factors. The basis is immediate. If

\[p_1 \cdots p_{k-1} \mid a\]

then we know that

\[a = rp_1\cdots p_{k-1}\]

for some $a \in R$. Now, consider $p_k$. We claim that $p_k \mid r$.