Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I wonder about how to conclude that $R=K[x_1, x_2,\dots ]$ is a UFD for $K$ a field.

If $f\in R$ then $f$ is a polynomial in only finitely many variables, how do I prove that any factorization of $f$ in $R$ only have factors in these indeterminates, i.e. takes place in the UFD $K[x_1, x_2,\dots , x_n]$ for some $n$?

Somebody argued that $f$ can not have one unique factorization in $K[x_1, x_2,\dots, x_n]$ and another in $K[x_1, x_2,\dots, x_n, \dots, x_m]$ I don't understand this. Do all primes in $K[x_1, x_2,\dots, x_n]$ necessarily stay prime in $R=K[x_1, x_2,\dots ]$? How do we know that the irreducible/prime elements in $K[x_1, x_2,\dots, x_n]$ stays irreducible in $R=K[x_1, x_2,\dots ]$?

share|improve this question
Is there any example of a directed system of rings which are UFD, but whose colimit is not UFD? –  Martin Brandenburg Jun 6 '13 at 20:49
add comment

2 Answers 2

up vote 9 down vote accepted

Note that $K[x_1,x_2,\dots]=\bigcup_{n\in \Bbb N}K[x_1,x_2,\dots,x_n]$. I'm going to denote these rings in the union by $R_n$ to save typing time.

To see that an irreducible $p$ of $R_n$ is irreducible in $R_m$ for $m\geq n$, suppose you have an equation $p=ab$ where $a,b\in R_m$. By evaluating $x_1,\dots, x_n$ all at 1, the variables in $p$ disappear, and you get an equation $\lambda=\overline{ab}\in K[x_{n+1}\dots, x_m]$ where $\lambda\in K$. This is a contradiction unless $a$ and $b$ already fell in $R_n$ in the first place, so that they had no variables above $x_n$. But you already know that in $R_n$, one of $a$ or $b$ must be a unit, so $p$ is irreducible in $R_m$ as well. So between the rings, primes stay prime and irreducibles stay irreducible.

This allows you to conclude that an element has a prime factorization in the first place.

Then you can argue that any two factorizations of a single element into primes must consist of elements in a common $K[x_1,\dots, x_m]$, which will force the factorizations to be equivalent.

share|improve this answer
Very clear and good answer, thank you! –  harajm Jun 6 '13 at 14:49
@harajm Glad you found it useful! –  rschwieb Jun 6 '13 at 16:50
add comment

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 the ascending union $\,K[x_1,x_2,\cdots\,],$ and the same ideas work for arbitrary inert extensions.

Remark $\ $ Paul Cohn introduced the idea of inert extensions when studying Bezout rings. Cohn proved that every gcd domain can be inertly embedded in a Bezout domain, and every UFD can be inertly embedded in a PID. There are a few variations on the notion of inertness that prove useful when studying the relationship between factorizations in base and extension rings, e.g. a weaker form where $\, d = ab\,\Rightarrow\, au, b/u\in D,\,$ for some unit $\,u\,$ in the extension ring.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.