Vrftjkry

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 $ain R$ 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
$$I=sumlimits_{n=1}^{infty}Ra_n$$
We first note that since $I$ is an ideal in a Noetherian ring, it must be the case that there is a finite subset ${a_k}_{kin K}subset{a_n}$ such that
$$I=sumlimits_{kin K}Ra_k$$

(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 $Ksubset mathbb{N}$ 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
$$R(a_{h+1})=0+0+dots + R(a_{h+1})+0+dots subset sumlimits_{n=1}^{infty}Ra_n,$$ 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.

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 $ain R$ 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
$$I=sumlimits_{n=1}^{infty}Ra_n$$
We first note that since $I$ is an ideal in a Noetherian ring, it must be the case that there is a finite set ${r_k}_{kin K}subset R$ such that
$$I=sumlimits_{kin K}Rr_k$$

Now, given an arbitrary $hat{k}in K$, we consider the fact that there is some set $Jsubset mathbb{N}$ such that
$$(r_{hat{k}})=sumlimits_{jin J}Ra_j$$
Since $Jsubset mathbb{N}$, 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
$$r_{hat{k}}=bigcuplimits_{jin J}(a_j).$$ Since $r_{hat{k}}in (r_{hat{k}})=bigcuplimits_{jin J}(a_j)$, 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 $kin K$ we can find a $N_kin mathbb{N}$ such that
$$(r_k)subsetneq (a_{N_k}).$$
We then find that
$$sumlimits_{kin K}r_ksubsetneq sumlimits_{kin K}(a_{N_k})subset sumlimits_{n=1}^{infty}Ra_n$$ 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.