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I'm reading a proof from C. Musili's Rings and Modules that every PID is a factorisation domain.

The author defines a factorisation domain as a commutative integral domain $R$ with a unit such that every non-zero $x \in R$ can be written as a unit times a finite product of irreducible elements. I will write down an outline of the proof here and my queries at the end. Those points that I am not sure about I will put in bold.

Let $R$ be a PID, and $\Omega$ the set of all non-zero elements of $R$ that cannot be written as a product of irreducible elements in $R$. We want to show that $\Omega = \emptyset$. So for a contradiction suppose that $\Omega \neq \emptyset$. Consider the non-empty family of principal ideals

$$\mathcal{F} = \{(x) \subseteq R : x \in \Omega\}$$

that is also a poset with respect to set inclusion. Now given any chain in $\mathcal{F}$ we can check that it has an upper bound in $\mathcal{F}$ so that by Zorn's Lemma $\mathcal{F}$ has a maximal element, say $(a)$.

Now $a$ cannot be a unit or irreducible for $a \in \Omega$. So write $a = bc$ for non-units $b$ and $c$. Now $(a) \subsetneqq (b)$ for otherwise $a$ and $b$ would be associates contradicting the fact that $a$ is irreducible. Since $(a)$ is maximal in $\mathcal{F}$, this means that $b$ and $c$ can be factored into irreducibles so that $a \notin \Omega$, a contradiction. Hence $\Omega = \emptyset$.

(1) For the first sentence in bold, should $\Omega$ not be the set of all non-zero elements in $R$ that cannot be written as a unit times an infinite number of irreducibles?

(2) If $\Omega$ is like what I have written above, then $a$ cannot be a unit for then

$a = a \cdot 1 = $ a unit $\times$ an irreducible element.

However if $\Omega$ is as the author has defined it to be, why must $a$ not be unit?

(3) If $a$ is not irreducible, why can we always write $a = bc$ for non-units $b$ and $c$? Does such a decomposition always exist?


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The product of an infinite number of elements is not well-defined in general. – Qiaochu Yuan Jan 8 '12 at 6:30
@QiaochuYuan I don't understand what you mean, I don't see anywhere in the proof the use of the finiteness condition. – user38268 Jan 8 '12 at 6:34
Dear Benjamin: How do you define the product of "an infinite number of irreducibles"? (I'm just following on Qiaochu's comment.) – Pierre-Yves Gaillard Jan 8 '12 at 6:50
"There exists an $x \in R$ such that all finite products of irreducibles $p_1p_2\dotsb p_n \neq x$"? – kahen Jan 8 '12 at 6:58
Because we're using a topology to define such an infinite product. We're using more than the ring structure. – Pierre-Yves Gaillard Jan 8 '12 at 7:05
up vote 3 down vote accepted

For (1), the set $\Omega$ should consists of all non-zero $x \in R$ such that $x$ cannot be written as a unit times a finite number of irreducibles. As Qiaochu pointed out, an infinite product of elements is undefined (so we are using the finiteness of the number of terms in assuming that the product is well-defined).

(2) If $a$ were irreducible, it would be a product of a unit times a finite number of irreducibles, namely 1 times the single irreducible $a$.

(3) This is basically the definition of an element being irreducible.

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For (3), I am sorry I typed the wrong thing in. It has now been changed to (2), and I am wondering why $a$ cannot be a unit (taking the author's definition of $\Omega$). – user38268 Jan 8 '12 at 7:00
Dear @BenjaminLim: The empty product is equal to $1$. – Pierre-Yves Gaillard Jan 8 '12 at 7:10
@Pierre-YvesGaillard Sorry, I don't get what you mean, you are referring to? – user38268 Jan 8 '12 at 7:14
@BenjaminLim: The auther wrote "a product of irreducible elements". I think she/he should have written "a unit times a product of irreducible elements", to cover the case where our element $a$ is a unit and $a\neq1$. – Pierre-Yves Gaillard Jan 8 '12 at 7:22
I think I get it now, $a$ is irreducible iff we cannot write it as the product of two non-units. So (3) above is settled too. Thanks! – user38268 Jan 8 '12 at 7:47

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