# Why a certain integral domain is not a UFD.

Let

$$\mathbb{Z}[q]^{\mathbb{N}} = \varprojlim_j \mathbb{Z}[q]/((1-q)\cdots (1- q^j))$$

Why isn't $\mathbb{Z}[q]^{\mathbb{N}}$ a unique factorization domain?

The author proposes a proof whose final step I don't quite understand. We can define an element $f(q) \in \mathbb{Z}[q]^{\mathbb{N}}$ by $f(q) = \prod_{i \geq 1} f_i(q)$ where the $f_i(q)$ are such that $f_i(q) \in (\Phi_{m_i}(q))$ for all $i \geq 1$, where $\Phi_n(q)$ stands for the $n$th cyclotomic polynomial.

This means that $f(q)$ is divisible by $\Phi_{m_1}(q) \cdots \Phi_{m_i}(q)$ for every $i$. Also, we know that $\mathbb{Z}[q]^{\mathbb{N}}$ is an integral domain and I have managed to prove as well that each cyclotomic polynomial is a prime element on the ring $\mathbb{Z}[q]^{\mathbb{N}}$. The author concludes that this means that $\mathbb{Z}[q]^{\mathbb{N}}$ is not a UFD, and this is what I don't get.

For an integral domain to be a UFD, every element $f$ can be written as a product of prime elements and a unit in a unique way. Therefore to prove that $\mathbb{Z}[q]^{\mathbb{N}}$ is not a UFD, we should exhibit an element $f$ with two different factorizations, right? I don't see how this follows from the fact that $f(q)$ is divisible by $\Phi_{m_1}(q) \cdots \Phi_{m_i}(q)$. This proves that $f$ doesn't admit a factorization as a product of cyclotomic polynomials, but there could be some other prime elements in $\mathbb{Z}[q]^{\mathbb{N}}$ such that $f$ does admit a factorization like that.

• What is $q$?... Aug 26 '18 at 6:33
• @user26857 It is a variable, just like $x$ or $t$. Maybe the terminology is not standard, I'm not sure. Aug 26 '18 at 14:02

Therefore to prove that $\mathbb{Z}[q]^\mathbb{N}$ is not a UFD, we should exhibit an element $f$ with two different factorizations, right?
For an element $g$ having a prime factorisation, i.e. that can be written as the product of a unit and finitely many primes, the set of primes dividing $g$ is finite (modulo identification of associated primes). In every divisor, only the primes occurring in the factorisation of $g$ can occur, with at most the power of the prime occurring in $g$.
As was shown, the constructed $f$ is divisible by infinitely many primes, hence it doesn't have a prime factorisation.