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An example of a non Noetherian UFD.

I know an example is $$K[x_1,\ldots,x_n,\dots]$$ with $K$ a field, but I don't know why. Can someone give another example or better an explanation?

Is it not Noetherian because it's not finitely generated? And why do you know it is a UFD?? My question now is how to prove this is a UFD.

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possible duplicate of UFD implies noetherian? – Federico Feb 7 '15 at 13:26

The essential point is that the polynomial ring in infinitely many variables is the ascending union of subrings $K[x_1,\ldots,x_n]$, since any polynomial can involve only finitely-many indeterminates. Each of these rings is a UFD, and it is easy to see that a polynomial in which $x_N$ does not appear has only factorizations in which $x_N$ does not appear, again because everything takes place inside some polynomial ring in finitely-many variables. But the ring is not Noetherian, because the ideal generated by all the indeterminates is certainly not finitely-generated.

Edit: in response to @rschwieb's comment/query about why $x_N$ cannot appear in any factorization of a polynomial $P(x_1,\ldots,x_n)$ not already involving it... If it did, then not only does $P$ have have its factorization in the UFD $K[x_1,\ldots,x_n]$, but also (allegedly) in the UFD $K[x_1,\ldots,x_n,\ldots,x_N]$, with $x_N$ in the latter, not in the former. Impossible.

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I sensed this was the solution, but I could not convince myself about the "factors of things without $x_N$ also do not have $x_N$" step in time. Do you have a handy trick to do that? I know I could find a moderately complicated way, but I wondered if there ws something exceedingly simple that I was overlooking. – rschwieb Jan 18 '13 at 20:54
They key idea is that each successive polynomial ring extension $\,D\subset D[x]\,$ is factorization inert, i.e. the ring extension introduces no new factorizations, i.e. if $\, 0\ne d\ \in D\,$ factors in $\,D[x]\,$ as $\,d = ab\,$ for $\, a,b\in D[x]\,$ then $\,a,b\in D.\,$ From this one easily deduces that the requisite factorization properties extend to $\,R[x_1,x_2,\cdots\,].\,$ The same ideas work for arbitrary inert extensions. See here for more. – Bill Dubuque Feb 2 '15 at 17:11
@rschwieb Suppose that $f\in K[x_1, x_2, ...]$ and $f$ involve indeterminates $x_{j_1}, x_{j_2}, ..., x_{j_n}$. Then $f\in K[x_{j_1}, x_{j_2}, ..., x_{j_n}]$ and $K[x_{j_1}, x_{j_2}, ..., x_{j_n}]$ is a subring of $K[x_1, x_2, ...]$ which is a U.F.D.. Thus, $f$ could be factored as a product of irreducible polynomials. – bfhaha Aug 23 '15 at 6:03

A Noetherian ring will satisfy the Ascending Chain Condition, or equivalently, every proper ideal will be finitely generated. Your example is not Noetherian for both of these reasons, as we can see $$(x_1)\subseteq (x_1,x_2)\subseteq (x_1,x_2,x_3)\subseteq\cdots,$$ and the chain never stabilizes since each indeterminate is distinct from the others. Likewise, $$(x_1,x_2,\ldots)$$ can't be finitely generated since there is no way to generate $x_n$ from $(x_1,\ldots,x_{n-1},x_{n+1},\ldots)$.

Another example of a non-Noetherian ring, for example is $$\prod_{n=1}^\infty \Bbb Z/2\Bbb Z.$$ Can you see why based off of your example and my explanation?

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Now I perfectly understand that is not noetherian, how do you now is UFD, I mean it means an only factorizacion, how do you know that, is the definition? – johnsaa Jan 18 '13 at 21:01
That direct product is not a UFD because it is not a D. – Mariano Suárez-Alvarez Feb 16 at 7:05
@MarianoSuárez-Alvarez: I see your point, but I never claimed it was a domain. I'm not sure if your comment was directed toward me or the OP, hence my response. – Clayton Feb 16 at 20:00
@Clayton, I know. What you wrote is not really an answer to the question at hand. – Mariano Suárez-Alvarez Feb 16 at 20:05
@MarianoSuárez-Alvarez: The second half of his question is: "or better an explanation?" (second half being relative, since he asks more than one question). That was my attempt here; I wasn't trying to give a non-Noetherian UFD example, but to explain the things that were happening in his example. At any rate, it seems neither here nor there for a thread $3$+ years old. :) – Clayton Feb 16 at 20:07

A ring $R$ has a factorization if it's Noetherian. Of course the factorization must not be unique. For the unicity you have to assume that every irreducible is prime.

In your example, $K[x_1, ..]$ is a UFD since $K$ is UFD and each polynomial has a finite number of variables. Furthermore it's not Noetherian because $(x_1, x_2...)$ is not finitely generated, however $K[x_1, ..]$ is finitely generated since $(1) = K[x_1, ...]$

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The notation does not suggest an infinite formal power series. It suggests the polynomial ring in infinitely many indeterminates. – rschwieb Jan 18 '13 at 20:36
Sorry that was my first answer and I always used this notation for formal power series. – user40276 Jan 18 '13 at 20:59
I think the most common notation for power series is $F[[x]]$ and $F[[x_1,x_2,\cdots]]$ – rschwieb Jan 18 '13 at 23:38

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