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It is a known fact that if $R$ is a UFD, then $R[X_1,X_2,\dots]$ is also a UFD, but there is a subtlety that is making me uncomfortable.

The standard approach essentially goes something along the lines of...let $f$ be a polynomial, so it only involves finitely many variables, say up to $X_1,\dots,X_n$. Then since $R[X_1,\dots,X_n]$ is a UFD, $f$ has a factorization into irreducibles. Then $f$ cannot have a factorization involving indeterminates $X_N$ not in $R[X_1,\dots,X_n]$. If it did, then $f$ would have two factorizations in the UFD $R[X_1,\dots,X_n,\dots, X_N]$, a contradiction.

This seems to assume that the prime factorization in $R[X_1,\dots,X_n]$ is still a prime factorization in $R[X_1,\dots,X_n,\dots,X_N]$. But how can you see that irreducibles in $R[X_1,\dots,X_n]$ are still irreducibles in $R[X_1,\dots,X_m]$ for $m>n$? This does not seem obvious.

Let $R_k=R[X_1,\dots,X_k]$. The explanantion I found is this: Suppose $p\in R[X_1,\dots,X_n]$ is irreducible in $R_n$. Suppose $p=ab$ for $a,b\in R_m$. Evaluating $X_1,\dots,X_n$ at $1$, you get $$ \bar{p}=\overline{ab}\in R[X_{n+1},\dots,X_m] $$ But $\bar{p}\in R$, which implies that $a$ and $b$ do not involve any variables $X_{n+1},\dots,X_m$. So $a,b\in R_n$, and thus one is a unit, hence a unit in $R_m$.

The part I don't follow is how evaluation at $1$, implies $a$ and $b$ do not have indeterminates other than $X_1,\dots,X_n$. Isn't it possible that some of the higher indexed indeterminates disappear when you evaluate? Suppose for instance $p\in R_1$, and $a\in R_2$ is $a=-X_1X_2+X_2$. Then evaluting at $1$ gives $\bar{a}=0$ so we may not get the desired contradiction since $\overline{ab}=0\in R$?

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2 Answers 2

up vote 5 down vote accepted

Let $A = R[X_1,\ldots, X_n]$ and $B = R[X_1,\ldots, X_n, \ldots ,X_N]$. Write $B = A[Y_1,\ldots,Y_k]$ to ease notation.

Now $B / pB \cong (A/pA)[Y_1,\ldots, Y_k]$. Since $p$ is irreducible in the UFD $A$, $pA$ is a prime ideal. So $A/ pA$ is an integral domain. So $B / pB$ is also an integral domain. So $p$ is irreducible in $B$.

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(This is also a nice solution...) –  Pete L. Clark Jan 12 at 0:20

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.

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.

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A big +1 for this. Over the years I have been groping for a name for this factorization (or nonfactorization!) property in extensions. Knowing what it is and that it is treated in papers of Cohn (some of whose work I have recently started to read and appreciate) will be helpful. Did you want to indicate why if $R$ is a domain then $R[t]/R$ is an inert extension, or is that left as an exercise to the OP? (I also ask because sometimes you give interesting alternative solutions to things that other people would solve by degree considerations.) –  Pete L. Clark Jan 12 at 0:19

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