Let $R$ be a UFD. It is well know that $R[x]$ is also a UFD, and so then is $R[x_1,x_2,\cdots,x_n]$ is a UFD for any finite number of variables. Is $R[x_1,x_2,\cdots,x_n,\cdots]$ in countably many variables also a UFD? If not, what about if we take $\tilde{R} = R[x_1,\cdots,x_n,\cdots]$ where each polynomial must be given in finitely many of the variables?
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$\begingroup$ I do not understand what you mean with $\overline{R}$. Every polynomial will always only contain a finite number of terms and thus variables. $\endgroup$– quid ♦May 4, 2015 at 18:06
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$\begingroup$ I wasn't forcing elements of $R[x_1,x_2,.\cdots,x_n,\cdots]$ to have finitely many variables in general, but that would simplify my question. Thanks! $\endgroup$– walkarMay 4, 2015 at 18:35
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$\begingroup$ math.stackexchange.com/questions/412980/kx-1-x-2-dots-is-a-ufd $\endgroup$– user26857May 4, 2015 at 20:27
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$\begingroup$ @user26857 This assumes the base ring is a field, an assumption I'm not making. $\endgroup$– walkarMay 4, 2015 at 20:42
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$\begingroup$ @walkar The proof is almost identical, but need to be read in order to notice it. $\endgroup$– user26857May 4, 2015 at 20:48
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3 Answers
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Key Idea $ $ 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 $\,R[x_1,x_2,\cdots\,].$ The same ideas works 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.
answered May 4, 2015 at 18:03
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We may think of the polynomial ring $R[x_1,x_2,\ldots]$ as a limit of polynomial rings in finitely many variables. There are natural injections $R[x_1,x_2,\ldots,x_n]\hookrightarrow R[x_1,x_2,\ldots]$ and you can show that with these identifications, if $p\in R[x_1,x_2,\ldots,x_n]$, then so is any divisor of $p$. Thus every element of this ring uniquely factors as a product of irreducibles (in the usual sense).
answered May 4, 2015 at 17:58
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$\begingroup$ I did not specify my base ring was a field, just a UFD. $\endgroup$– walkarMay 4, 2015 at 17:59
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$\begingroup$ @walkar Doesn't matter to my argument. I noticed and I edited, thanks. $\endgroup$ May 4, 2015 at 17:59
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Hint: first show that for any $n\in\mathbb N$ and any $f\in R[x_1,x_2,\cdots,x_n]\subset R[x_1,x_2,\cdots]$, $f$ is irreducible in $R[x_1,x_2,\cdots]$ iff it's so in $R[x_1,x_2,\cdots,x_n]$; then notice any element of $R[x_1,x_2,\cdots]$ lies in $R[x_1,x_2,\cdots,x_n]$ for some $n$.
