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I want to prove that in a noetherian ring $R$ which is also an integral domain, every non invertible element can be expressed as product of irreducible elements.

I really do not know where to start. Can someone give me hint how to prove this?

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    $\begingroup$ Here is what I would try first: Take a non-unit $x$. If it is irreducible, we are done. Otherwise, write $x = yz$ where $y, z$ are non-units. If both are irreducible, we are done. Otherwise, write each of them that is reducible as a product of two non-units. Rinse, repeat, and then use Noetherianness of the ring to show that we eventually have to come to a stop. That last part might be easy, it might be hard. I don't know. $\endgroup$ – Arthur Mar 16 '14 at 21:18
  • $\begingroup$ Thank you very much, sounds good. I will try to bring that idea on the paper. $\endgroup$ – Thorben Mar 16 '14 at 21:28
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Let $X$ be your set of nonzero, nonunits which cannot be written as a product of irreducibles. Towards a contradiction, suppose $X\neq\emptyset$, and pick $x_0\in X$. Then $x_0$ itself is not irreducible, so we may write $x_0=xy$, where $x$ and $y$ are nonzero nonunits. If both $x$ and $y$ can be written as a product of irreducibles, then $x_0$ can as well, a contradiction. So at least one of $x$ and $y$ is not a product of irreducibles, say $x$. Call it $x_1\in X$. Then $(x_0)\subset (x_1)$.

Continue this process to yield an ascending chain $$ (x_0)\subset (x_1)\subset\cdots $$ in $R$. Now use the fact that $R$ is Noetherian to find a contradiction.

Since $R$ is Noetherian, $(x_n)=(x_{n+1})$ for some $n$. By our construction, $x_n=yx_{n+1}$ for $y$ a nonunit. But then $x_{n+1}\in (x_n)=(yx_{n+1})$, so for some nonzero $z\in R$, we have $x_{n+1}=zyx_{n+1}$. By cancellation, since we are in a domain, $1=yz$, so $y$ is a unit, a contradiction. So $X=\emptyset$.

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  • $\begingroup$ You might find of interest the remark in my answer, and the papers it links to. $\endgroup$ – Bill Dubuque Mar 16 '14 at 22:33
  • $\begingroup$ @BillDubuque Thanks! $\endgroup$ – BW. Mar 17 '14 at 0:08
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Hint $\ $ Suppose not, and consider the set of principal ideals generated by all counterexamples. By ACC it contains a maximal element $(r),\,$ and $r$ is reducible $\,r = st.\,$ But by maximality, $\,s,t\,$ have irreducible factorizations, thus so too does $\,r = st,\,$ contradiction.

Remark $\ $ Interpreted positively, the proof shows if $S$ is the set of principal ideals disjoint from a monoid $M$ then $(r)$ is maximal in $\,S \iff r\,$ is irreducible. This is a generalization of the well-known case where $\,M = \{1\},\,$ i.e. $\ (r)$ is maximal among principal ideals $\iff r$ is rreducible.

This is a specal case of a wide class of results where elements satisfying such maximality properties are irreducible or prime, e.g. see this answer. One beautiful example along these lines is a famous theorem of Kaplansky that a domain is a UFD $\iff$ every prime ideal $\ne 0$ contains a prime $\ne 0$ (or, equivalently, every prime ideal is generated by primes).

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  • $\begingroup$ May I ask what ACC means? $\endgroup$ – Thorben Mar 16 '14 at 21:38
  • $\begingroup$ Ascending Chain Condition on ideals. Which definition of Noetherian are you using? $\endgroup$ – Bill Dubuque Mar 16 '14 at 21:40
  • $\begingroup$ Ok, so this is the condition that for every chain $(c_i)_{i\in I}$ s.t $c_i\subset c_{i+1}$ there exist some $k\in I$ s.t $\forall j>k$ we have that $c_k=c_j$? We had that this is equivalent to every set of ideal contains some maximal ideal with respect to inclusion and it is equal to that every ideal is finitely generated. $\endgroup$ – Thorben Mar 16 '14 at 21:48
  • $\begingroup$ @user135025 Right. Also equivalent is: every ideal is finitely generated. $\endgroup$ – Bill Dubuque Mar 16 '14 at 21:50
  • $\begingroup$ @user135025 I added some further pedagogical remarks that may shed more light on the essence of the matter. You will meet many more analogous results having this form later if you study number theory or commutative algebra. $\endgroup$ – Bill Dubuque Mar 16 '14 at 22:23

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