Exercise A0.V.2.01

Throughout the propositions, $R$ is an UFD, and $a,b,c$ are non-zero elements of $R.$ Let $\mathsf{F}(a)$ denote the multiset of irreducible factors of $a$.

We will be using multiset operations and relations as given by wikipedia.

Lemma P2.1

$\mathsf{F}(ab) = \mathsf{F}(a) + \mathsf{F}(b).$

Proof

For convenience, we will ignore units as for any unit $u,$ $\mathsf{F}(ua) = a.$

Write down $ab$ as a product of irreducible factors of $a$ and $b$:

$$ab = a_1^{p_1}a_2^{p_2}\ldots a_n^{p_n}b_1^{q_1}b_2^{q_2}\ldots b_m^{q_m}.$$

Whenever $a_i = b_j$ we add up the powers. When all factors are distinct, we have an irreducible factorization of $ab.$ The multiplicity of each individual irreducible factor is the sum of multiplicities of this factor in $a$ and in $b,$ which matches the definition of the sum of two multisets. By uniqueness this is precisely $\mathsf{F}(ab).$

$\blacksquare$

Proposition A1

$(a) \subseteq (b)$ if and only if $\mathsf{F}(b) \subseteq \mathsf{F}(a).$

Proof

Suppose $(a) \subseteq (b).$ Then there exists an element $w \in R$ such that $a = wb.$ Hence $\mathsf{F}(a) = \mathsf{F}(w) + \mathsf{F}(b).$ Since addition of multisets can only increase the multiplicity of any element, $\mathsf{F}(a)$ must, at the very least, contain all elements of $\mathsf{F}(b).$ Thus $\mathsf{F}(b) \subseteq \mathsf{F}(a)$ as required.

Suppose conversely that $\mathsf{F}(b) \subseteq \mathsf{F}(a).$ In that case, there exists a (possibly empty) multiset of irreducible factors $W$ such that $\mathsf{F}(a) = \mathsf{F}(b) + W.$ Let $w \in R$ denote the product (with multiplicity) of all elements of $W.$ We have $W = \mathsf{F}(w)$ and hence

$$\mathsf{F}(a) = \mathsf{F}(b) + \mathsf{F}(w) = \mathsf{F}(bw).$$

Since $a$ and $bw$ have the same multiset of irreducible factors, there exists a unit $u$ such that $a = bwu.$ Hence $a \in (b)$ and so $(a) \subseteq (b).$

$\blacksquare$

Proposition A2

$(a) = (b) \iff \mathsf{F}(a) = \mathsf{F}(b)$

Proof

Follows from Proposition A1.

$\blacksquare$

Proposition A3

The multiset of irreducible factors of $bc$ is $\mathsf{F}(b) + \mathsf{F}(c).$

Proof

Follows from Lemma P2.1.

$\blacksquare$