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.
Best Answer
There is another option. One can also exhibit an element having no prime factorisation. And prime factorisations are always unique if they exist, it's factorisations into irreducible elements that may be non-unique.
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.