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Edit: I have determined this initial proof to be incorrect. I have answered with what I believe to be a correct proof.

Definition A Noetherian domain is a domain whose ideals are all finitely generated.

Let $R$ be a commutative, Noetherian domain and suppose for contradiction that $a\in R$ cannot be factored into irreducibles. Then $a$ must be reducible, so there exists $a_1$ and $a_2$ such that $a=a_1a_2$. If both $a_1$ and $a_2$ are irreducible, then $a$ is a product of irreducibles, so it must be that they are equal to products of reducible elements and process continues ad infinitum. Given such an $a$ exists, we can find an endless list of elements satisfying \begin{align*} &a\:\:=a_1b_1 \\ &a_1=a_2b_2 \\ &a_2=a_3b_3 \\ &\;\;\vdots \qquad \; \vdots \\ \end{align*} Where each of the $b_i$'s is assumed to not be a unit. Consider the ideal $$I=\sum\limits_{n=1}^{\infty}Ra_n$$ We first note that since $I$ is an ideal in a Noetherian ring, it must be the case that there is a finite subset $\{a_k\}_{k\in K}\subset\{a_n\}$ such that $$I=\sum\limits_{k\in K}Ra_k$$

(this is the step I'm most nervous about, I'm not 100% convinced that the spanning set should be a subset of the elements whose infinite linear combination is equal to $I$)

Now the construction of $a_n$ is such that $(a_n)\subsetneq (a_{n+1})$, and since $K\subset \mathbb{N}$ is finite and non-empty, it has a greatest element $h$. So then the subset $\{a_h\}$ generates $I$. But since $(a_h)\subsetneq (a_{h+1})$, and since $$R(a_{h+1})=0+0+\dots + R(a_{h+1})+0+\dots \subset \sum\limits_{n=1}^{\infty}Ra_n,$$ we find that $\{a_h\}$ does not span $I$. Therefore $I$ cannot be finitely generated, contradicting $R$ Noetherian. It follows then that for some $a_k$ in the aforementioned list, $a_k$ is irreducible. Therefore every element of $R$ is either irreducible or a product of irreducibles.

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  • $\begingroup$ Rings where multiplication is commutative are named “commutative rings”, not “abelian rings”. Are you sure the statement doesn't require $R$ being a domain? $\endgroup$ – egreg Nov 17 '18 at 23:20
  • $\begingroup$ My bad. I'm not even sure that commutativity is important for this proof, but it is blanketly assumed for all rings encountered in my algebra course. I will modify the beginning. Also my bad, the definition I gave is for a Noetherian domain. $\endgroup$ – Daniel Nov 17 '18 at 23:23
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Okay, I have concluded that the step I was concerned about cannot be justified. I have modified to proof into something that I believe is correct:

Let $R$ be a commutative, Noetherian domain and suppose for contradiction that $a\in R$ cannot be factored into irreducibles. Then $a$ must be reducible, so there exists $a_1$ and $a_2$ such that $a=a_1a_2$. If both $a_1$ and $a_2$ are irreducible, then $a$ is a product of irreducibles, so it must be that they are equal to products of reducible elements and process continues ad infinitum. Given such an $a$ exists, we can find an endless list of elements satisfying \begin{align*} &a\:\:=a_1b_1 \\ &a_1=a_2b_2 \\ &a_2=a_3b_3 \\ &\;\;\vdots \qquad \; \vdots \\ \end{align*} Where each of the $b_i$'s is assumed to not be a unit. Consider the ideal $$I=\sum\limits_{n=1}^{\infty}Ra_n$$ We first note that since $I$ is an ideal in a Noetherian ring, it must be the case that there is a finite set $\{r_k\}_{k\in K}\subset R$ such that $$I=\sum\limits_{k\in K}Rr_k$$

Now, given an arbitrary $\hat{k}\in K$, we consider the fact that there is some set $J\subset \mathbb{N}$ such that $$(r_{\hat{k}})=\sum\limits_{j\in J}Ra_j$$ Since $J\subset \mathbb{N}$, it inherits its ordering from $\mathbb{N}$. Then by the construction of the $a_n$, we know that if $j_a<j_b$, then $(a_{j_a})\subset (a_{j_b})$. This means that $$r_{\hat{k}}=\bigcup\limits_{j\in J}(a_j).$$ Since $r_{\hat{k}}\in (r_{\hat{k}})=\bigcup\limits_{j\in J}(a_j)$, it follows that $r_{\hat{k}}\in (a_{\hat{j}})$ for some $\hat{j}\in J$, so $(r_{\hat{k}})\subset (a_{\hat{j}})$. However, we also know that $a_{\hat{j}}\in (r_{\hat{k}})$, so we conclude that $$(r_{\hat{k}})=(a_{\hat{j}}).$$ It follows then by the construction of the $a_n$ that $$(r_{\hat{k}})\subsetneq (a_{\hat{j}+1}).$$ Thus for each $k\in K$ we can find a $N_k\in \mathbb{N}$ such that $$(r_k)\subsetneq (a_{N_k}).$$ We then find that $$\sum\limits_{k\in K}r_k\subsetneq \sum\limits_{k\in K}(a_{N_k})\subset \sum\limits_{n=1}^{\infty}Ra_n$$ which means that $\{r_k\}$ does not span $I$. We conclude that $I$ has no finite spanning set, contradicting $R$ being a Noetherian domain. Therefore it must be the case that $a$ may be factored into irreducibles.

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