# Proof Checking: Elements of Noetherian Domains may be factored into irreducibles

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.

• Rings where multiplication is commutative are named “commutative rings”, not “abelian rings”. Are you sure the statement doesn't require $R$ being a domain? – egreg Nov 17 '18 at 23:20
• 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. – Daniel Nov 17 '18 at 23:23

## 1 Answer

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 $$j_a, then $$(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.