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Sources for the proof of this claim: If $a$ is non zero non unit in a PID then $a$ has at least one irreducible divisor.

Definition of irreducible element: a non zero non unit element $p$ is irreducible if whenever $p=ab$, then $a$ or $b$ is a unit.

Note: I need this step to prove that every PID is UFD.

One user's answer:

Let R be said PID and suppose a∈R is a nonzero non-unit. Then, the principal ideal aR is nonzero and proper (i.e. aR≠{0} and aR≠R). Therefore, aR is contained in a maximal ideal M, which is principal. Therefore M=pR and a∈aR⊂pR. That means that a=pr for some r∈R∖{0}, so we just need to prove p is irreducible. Well, otherwise, p=xy for some non-units x,y∈R, which would imply pR⊊xR⊊R contradicting the maximality of pR.

What I don't understand in the answer :

Why if p is not irreducible then it implies p=xy, where x and y are non-units? This is the only step that I still can't grasp?

I0_0I
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1 Answers1

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Let $R$ be said PID and suppose $a\in R$ is a nonzero non-unit. Then, the principal ideal $aR$ is nonzero and proper (i.e. $aR\neq\{0\}$ and $aR\neq R$). Therefore, $aR$ is contained in a maximal ideal $M$, which is principal. Therefore $M=pR$ and $a\in aR\subset pR$. That means that $a=pr$ for some $r\in R\backslash\{0\}$, so we just need to prove $p$ is irreducible. Well, otherwise, $p=xy$ for some non-units $x,y\in R$, which would imply $pR\subsetneq xR\subsetneq R$ contradicting the maximality of $pR$.

David Hill
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  • Is there a way to use the fact that any chain of principle ideals must terminate in PID to prove the claim? – I0_0I Apr 20 '21 at 20:54
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    @I0_0I: it implies that $(a)$ is contained in a maximal ideal, by looking at the collection of all proper ideals that contain $(a)$. – Arturo Magidin Apr 20 '21 at 20:57
  • Thank you. That was very helpful. – I0_0I Apr 20 '21 at 20:58
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    @I0_0I Based on your comment (and prior question), likely you seek this form of the proof via ACCP – Bill Dubuque Apr 20 '21 at 21:05
  • Why if $p$ is not irreducible then it implies $p=xy$, where $x$ and $y$ are non-units? This is the only step that I still can't grasp. – I0_0I Apr 21 '21 at 16:15
  • 14 is not irreducible but 14=14*1 – I0_0I Apr 21 '21 at 16:22
  • @I0_0I: It's the definition of irreducible: an element $a$ is irreducible if and only if whenever $a=xy$, either $x$ is a unit or $y$ is a unit. It doesn't mean every expression of $p$ as a product has both factors non-units, it just says there exists an expression $p$ as a product with both factors nonunits. – Arturo Magidin Apr 21 '21 at 19:09
  • I see your point. But I can't see how the definition of irreducible element implies that! If p is irreducible then the definition tells us something about p. But if p is not irreducible then the definition doesn't tell us anything about the element p. Isn't it the 'convers is true falasy' to imply any property on the element p which is not irreducible? – I0_0I Apr 21 '21 at 19:15
  • I'm thinking of irreducible element in context of a ring. Without using any implicit assumptions and properties. How does it follow that there exist x and y non units such that $p=xy$? I'm not stupid I swear! – I0_0I Apr 21 '21 at 19:20
  • I'm stuck on this since the past 4 days – I0_0I Apr 21 '21 at 19:23
  • Just explain to me like you would explain it to Forrest Gump. – I0_0I Apr 21 '21 at 19:25
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    The definition is $$\mbox{"($p$ is irreducible) $\iff$ (for every factorization $p=xy$, either $x$ is a unit or $y$ is a unit)}$$. Now, negate the statements on both sides of $\iff$ to get the logically equivalent statement $$\mbox{($p$ is not irreducible) $\iff$ (there exists a factorization $p=xy$ where both $x$ and $y$ are not units)}$$ – David Hill Apr 21 '21 at 19:27
  • Thank you sir finally. – I0_0I Apr 21 '21 at 19:32
  • We have to prove that there exists some factorization of p though right? Or maybe If p is not factorlizable it implies it's irreducible? – I0_0I Apr 21 '21 at 19:34
  • $p=p\cdot 1$ is a trivial factorization. – David Hill Apr 21 '21 at 19:38
  • OK ok kk I get it now thank you so much – I0_0I Apr 21 '21 at 19:39