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Background:

Lemma 10.9: Let $a$ and $b$ be elements of an integral domain $R.$ Then

  1. $(a)\subset (b)$ if and only if $b\mid a.$

  2. $(a)=(b)$ if and only if $b\mid a$ and $a\mid b.$

  3. $(a)\subsetneq (b)$ if and only if $b\mid a$ and $b$ is not an associate of $a.$

Theorem 10.12: Let $R$ be a principal ideal domain. Every nonzero, nonunit element of $R$ is the product of irreducible elements, and this factorization is unique up to associates that is, if

$$p_1p_2\cdots p_r=q_1q_2\cdots q_s$$

with each $p_i$ and $q_j$ irreducible, then $r=s$ and, after reordering and relabeling if necessary,

$$p_i \text{ is an associate of } q_i \text{ for } i=1,2,\ldots,r.$$

Proof: Let $a$ be a nonzero, nonunit element in $R.$ We must show that $a$ has at least one factorization. Supose, on the contrary, that $a$ is $\textit{not}$ a product of irreducibles. Then $a$ is not itself irreducible. So $a=a_1b_1$ for some nonunits $a_1$ and $b_1$ (otherwise every factorization of $a$ would include a unit and $a$ would be irreducible). If both $a_1$ and $b_1$ are products of irreducibles, then so is $a.$ Thus at least one of them, say $a_1,$ is not a product of irreducibles. Since $b_1$ is not a unit, $a_1$ is not an associate of $a$. Consequently, $(a)\not\subset (a_1)$....

Here is a screenshot of the theorem and proof.

enter image description here

Questions:

In the first part of the proof of the above theorem, when it says that even though $a$ is not itself an irreducible element, but it still has factorization, where $a=a_1b_1$ with both $a_1,b_1$ being both non units. But then $a,$ and $a_1$ both can not be consider as principal ideals, because it would violate the question posted here: prove $(n) \supseteq (m)\iff n\mid m\ $ (contains = divides for principal ideals). But if that is the case, then $a_1\nmid a?$ According to KCd's comment here: Why are irreducible elements non-units?

The whole point is that reducible elements are supposed to have nontrivial factorizations. In an integral domain, the nonzero elements fall into three classes: units, irreducible elements, and reducible elements. For a unit, every factorization into a product of two terms has both factors equal to units (that is to be proved), for an irreducible element every factorization into a product of two elements has exactly one factor equal to a unit. For a reducible element there is some factorization into a product of two elements that are both non-units.

So for elements $a$ in $R,$ an integral domain, does that mean that if $a$ is not an irreducible element, then it can still have factorization but itself and its factors don't satisfy prove $(n) \supseteq (m)\iff n\mid m\ $ (contains = divides for principal ideals) or Lemma 10.9 $(3)$ above, in the sense that $a$ is divisible by its factors but it can not be consider as a principal ideal or an ideal? I am a bit confused here. Basically if the factors of $a$ say $a=bc$ where both $b$ and $c$ are not units, then does it make sense to talk about $b\mid a$ or $(a)\subset (b)$

Thank you in advance

KReiser
  • 65,137
Seth
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  • The proof seems weird to me, maybe I missed something. Do you mind sharing the reference so I could read the whole proof? – Tri Feb 06 '24 at 03:45
  • @Tri I just edit my post with an uploaded screenshot. – Seth Feb 06 '24 at 03:50
  • Okay so I guess your problem is the notation: you wrote $(a)\not\subset (a_1)$ while in the book said $(a) \subsetneq (a_1)$. The former indicates that $(a)$ is not a subset of $(a_1)$ while the latter says that $(a)$ is a subset of $(a_1)$ but not $(a_1)$. – Tri Feb 06 '24 at 03:56
  • @Tri my problem is not exactly notation. When lemma 10.9, it says that $(a)\not\subset (b)$ off $a\mid b$ and $b$ is not an ssociate of $a$. But in the proof of theorem 10.12, $a$ is not an irreducible, but $a=a_1b_1$ so that means $a_1\mid a$ but is $a$ an associate of $a$ and does lemma 10.9 $(3)$ apply. I am getting the imprression that lemma 10.9 $(3)$ is only talking about factoriation of $a$ where it is an irreducible element, but what happen when $a$ is not an irreducible element, and it can still factorize. Then we can't talk about $a$ as a principal ideal element of $R$? – Seth Feb 06 '24 at 04:01
  • @Tri if you have a moment, can I also ask you to take a look at this other post, I have a quick question about finding the corresponding$d$ in the theorem. Is just a quick question. Thank you in advance. – Seth Feb 06 '24 at 04:04
  • your comment is really hard to understand. Once again, I must emphasize using the correct notation. If $(a) \subsetneq (b)$ then $b$ is a divisor of $a$ but not $a$ (think of the similarity in $\mathbb{Z}$, we have $(6) \subsetneq (2)$ means $2 | 6$.) The point of the proof is that if you continue in this manner, you will get an ascending chain of ideal of infinite length. This is not true for PID. For eg, in $\mathbb{Z}$, we have $(6) \subsetneq (2) \subsetneq (1)$ and it stops there. You can't extend it further. – Tri Feb 06 '24 at 04:25
  • @Tri when the proof states: "that $a$ is $\textit{not}$ a product of irreducibles. Then $a$ is not itself irreducible. So $a=a_1b_1$ for some nonunits $a_1$ and $b_1$", can such factorization occur. Because I thought when we have something like $a\mid b$, then $b=ac$ for some unit $c$. What if as in the proof states that if $a$ and $c$ are both nonunits. It still factorizes but it won't satisfy lemma 10.19 $(3)$? – Seth Feb 06 '24 at 04:30
  • Let us continue this discussion in chat. – Tri Feb 06 '24 at 04:32
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    Again, MathJax is not for formatting text. Please use markdown for that. – KReiser Feb 08 '24 at 05:30

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