Talk about factorisation and primes in a polynomial ring. What is irreducibility? For what rings \(R\) is it true that \(R[x_1, \dots , x_n]\) is a unique factorisation domain? What is wrong with unique factorisation if we don’t have a domain? Now, PIDs are Noetherian, but are there UFDs which are not?
Solution
Let \(R\) be an integral domain. A nonzero nonunit \(a\in R\) is irreducible if \[ a=bc\implies b\in R^\times\text{ or }c\in R^\times, \] and prime if \[ a\mid bc\implies a\mid b\text{ or }a\mid c. \] Every prime element is irreducible. In a UFD, every irreducible is prime.
Gauss’s theorem says that if \(R\) is a UFD, then \(R[x]\) is a UFD. Iterating gives \[ R\text{ UFD}\implies R[x_1,\dots,x_n]\text{ UFD}. \] Conversely, if \(R[x_1,\dots,x_n]\) is a UFD, then every nonzero nonunit \(a\in R\) factors uniquely there. Because the units of the polynomial ring are exactly the units of \(R\), and a factorization of the constant polynomial \(a\) can involve only constant factors, this is already a unique factorization in \(R\). Hence \[ R[x_1,\dots,x_n]\text{ is a UFD}\iff R\text{ is a UFD}. \]
A UFD is by definition a domain. With zero divisors, factorizations into nonunits need not have bounded length: in \(\ZZ/6\ZZ\) the idempotent nonunit \(3\) satisfies \(3=3\cdot3=3^n\) for every \(n\ge1\).
A UFD need not be Noetherian. For example, \[ k[x_1,x_2,x_3,\dots] \] is a UFD. Indeed, every nonzero nonunit belongs to some finite-variable UFD \[ A_m=k[x_1,\dots,x_m] \] and factors there into irreducibles. Each such irreducible \(p\in A_m\) is prime in \(A_m\), and \[ k[x_1,x_2,\dots]/(p) \cong (A_m/(p))[x_{m+1},x_{m+2},\dots] \] is a domain, so \(p\) remains prime in the full ring. Thus these factorizations are factorizations into primes in the full ring, and uniqueness follows. The ring is not Noetherian because \[ (x_1)\subsetneq(x_1,x_2)\subsetneq(x_1,x_2,x_3)\subsetneq\cdots \] does not stabilize.