If $R$ is a UFD, then $R[x]$ is Noetherian?

Question

If $R$ is Noetherian, then $R[x]$ is Noetherian. However, if $R$ is a UFD, then $R$ might not be Noetherian. $\DeclareMathOperator{\Frac}{Frac}$

I want to show that if $R$ is a UFD, then $R[x]$ is a UFD. I think (hope) I have passed through the gauntlet of proving the uncomfortably diverse range of slightly nontrivial results that are lumped together as the chameleonic "Gauss's lemma" (and cited succinctly in many textbooks/websites/pdfs as "by Gauss's lemma" whenever convenient, even though it is often totally unclear to me which particular result the author is talking about). These include the following results when $R$ is a UFD.

(Say $\sum_i a_ix^i \in R[x]$ is primitive iff $\gcd_R(a_0, a_1, \dots, a_n) = 1$.)

• If $f \in \Frac(R)[x]$, then $f = cf_0$ for $c \in \Frac(R)$ and $f_0 \in R[x]$ primitive. $c$ is unique up to multiplication by a unit in $R^\times$. Additionally, $c \in R$ iff $f \in R[x]$.
• If $f, g \in R[x]$ are primitive, then so is $fg \in R[x]$.
• If $f \in R[x]$ factors into two nonconstant polynomials in $\Frac(R)[x]$, then it factors into two nonconstant polynomials in $R[x]$.
• The constant irreducibles in $R[x]$ are the primes in $R$, and the nonconstant irreducibles in $R[x]$ are nonconstant primitives that are irreducible in $\Frac(R)[x]$.
• If $fg = h$ for $f \in R[x]$ primitive, $g \in \Frac(R)[x]$, and $h \in R[x]$, then $g \in R[x]$.
• In $R[x]$, irreducible implies prime.
• More?

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Say $R$ is an integral domain and $0 \neq r \in R - R^\times$. Pass $r$ into the following algorithm:

1. If $r$ is irreducible, then return this irreducible factorization. If $r$ is reducible, then continue.
2. Write $r = r_1 s_1$ for nonzero $r_1, s_1 \in R - R^\times$.
3. Recursively call this algorithm on $r_1$ and $s_1$.
4. Return the concatenated irreducible factorization of $r_1$ and $s_1$, which is an irreducible factorization of $r$.

Artin's textbook says factoring terminates in an integral domain $R$ to mean that this algorithm is guaranteed to terminate. He proves that "factoring terminates" is equivalent to "$R$ does not contain an infinite strictly increasing chain $(a_1) \subsetneq (a_2) \subsetneq (a_3) \subsetneq \dots $ of principal ideals". If either of these equivalent conditions hold, then an integral domain $R$ is a UFD if and only if "irreducible $\implies$ prime".

Therefore, it suffices to prove that $R[x]$ is Noetherian if $R$ is a UFD. How do we show this? Artin says "It is easy to see that factoring terminates in $\mathbb{Z}[x]$", but this is unclear to me as well.

Answer 1

It's not true that $R[x]$ must be Noetherian (indeed, if $R$ is not Noetherian, then $R[x]$ cannot be Noetherian). To see there can be no infinite strictly increasing chain $(a_1) \subsetneq (a_2) \subsetneq (a_3) \subsetneq \dots$ of principal ideals, note that $\deg a_n$ must be decreasing and so eventually stabilizes. Once the degree stabilizes, you can only get from $a_n$ to $a_{n+1}$ by dividing by a non-unit of $R$. But you can only do this finitely many times, by considering the factorization of the leading coefficients in $R$.

Answer 2

Any quotient ring of a Noetherian ring is Noetherian (if $R$ is Noetherian and $I$ is an ideal, then ideals of $R/I$ are of the form $\bar{J}$, where $J$ is s of $R$ containing $I$. Since $J$ is finitely generated, so is $\bar{J}$). In particular, if $R[X]$ is noetherian, so is $R\simeq R[X]/(X)$. SO any non noetherian UFD will provide you a counterexample (such as $\mathbb{C}[X_n,n\geq 0]$).