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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?

Thanks.

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    The product of an infinite number of elements is not well-defined in general.2012-01-08
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    @QiaochuYuan I don't understand what you mean, I don't see anywhere in the proof the use of the finiteness condition.2012-01-08
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    Dear Benjamin: How do you define the product of "an infinite number of irreducibles"? (I'm just following on Qiaochu's comment.)2012-01-08
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    @Pierre-YvesGaillard I don't know, but then what is the negation of "for all $x \in R$, $x$ is a unit times a finite product of irreducible elements"?2012-01-08
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    "There exists an $x \in R$ such that all finite products of irreducibles $p_1p_2\dotsb p_n \neq x$"?2012-01-08
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    I agree with @kahen: The negation is "There exists an $x$ in $R$ such that for all unit $u$ and all all finite sequence $p_1,..,p_n$ of irreducibles we have $up_1\cdots p_n\neq x$".2012-01-08
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    @Pierre-YvesGaillard I don't get why the product of an infinite number of irreducibles is not defined. Why can't we define it like say $\prod_{n=1}^\infty \big( \frac{4n^2}{4n^2 -1} \big)$2012-01-08
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    Because we're using a **topology** to define such an infinite product. We're using **more** than the ring structure.2012-01-08
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    One more thing I am confused is the definition of an irreducible element. Would not every non-zero non unit $a$ in a ring be irreducible since we can always write it as $a=a \cdot 1$?2012-01-08
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    Because we need convergence requirements on the product? But then we have things like the ring of formal power series where we have infinite sums, so how come there we can take an infinite sum but now we cannot take an infinite product?2012-01-08
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    Good point! Dealing with formal power series, you can either define your infinite sum as a formal symbol, or (**much** better), introduce a topology. For example, the natural topology on $R[X]$ is the so-called "$X$-adic topology". Then you can consider infinite products on $R[X]$. (+1 for your question!)2012-01-08
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    @Pierre-YvesGaillard I don't get it why can't our infinite product be a formal symbol as well? I get (2) now, I think the author should have said that $\Omega$ is the set of all elements $x$ such that (insert what kahen said). (3) I still don't understand.2012-01-08
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    (I wrote $R[X]$ instead of $R[[X]]$ above.) You can define the ring $R[[X]]$ as the set of sequences $(a_n)_{n\ge0}$ in $R$ without ever using the notation $\sum_{n\ge0}a_nx^n$.2012-01-08
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    @BenjaminLim How does writing $a = a \cdot 1$ show irreducibility? You have to show that _every_ way of writing $a = b \cdot c$ involves $b$ or $c$ being a unit.2012-01-08
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    @DylanMoreland Sorry for not seeing your comment earlier. Thanks that helps. Actually now seeing your comment I am starting to wonder if the reason why $(a) \subsetneqq (b)$ is correct.2012-01-12

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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 $\Omega$).2012-01-08
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    Dear @BenjaminLim: The empty product is equal to $1$.2012-01-08
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    @Pierre-YvesGaillard Sorry, I don't get what you mean, you are referring to?2012-01-08
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    @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 $a\neq1$.2012-01-08
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    @Pierre-YvesGaillard Thanks, (2) is settled now but for (3) I still don't get. If $a$ is irreducible then $a= bc$ where $b$ or $c$ is a unit. However if $a$ is not irreducible then how do we know (other than just taking the negation of the above) that $a = bc$ where $b,c$ are not units?2012-01-08
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    @BenjaminLim: Could you please write precisely your definition of "irreducible"? Thanks.2012-01-08
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    @Pierre-YvesGaillard A non-zero non-unit element $a \in R$ is said to be irreducible if it can written as $a = bc$ where $b$ or $c$ is a unit.2012-01-08
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    @BenjaminLim: Then $4$ is irreducible in $\mathbb Z$: you can write it as $1\cdot4$ where $1$ is a unit.2012-01-08
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    @Pierre-YvesGaillard That's the problem I was having, that way every element in the ring would be irreducible. I think it should be if $a = bc$, then either $b$ or $c$ is a unit.2012-01-08
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    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!2012-01-08
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    @BenjaminLim: Wonderful!!! --- You're welcome!2012-01-08
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    @Pierre-YvesGaillard Thanks :D2012-01-08