Are multiplicative monoids of different rings isomorphic? I have some algebraic problem and hope some of you can help me!
The problem is about rings and their multiplicative monoids (semi-groups with neutral element $e$). So $M(\mathbb{Z}), M(\mathbb{Z}_2[x])$ and $M(\mathbb{Q}[[x]])$ denote the multiplicative monoids (factorized Modulo association).
The task is to find out whether these monoids are isomorphic or not.
My first idea was to use the Chinese Remainder Theorem, but it only states something for ideals of one and the same ring, which nothing matching to the given problem here.
Furthermore I tried to exclude at least one possible isomorphism by compairing the cardinal number of the monoids, but they seem to be the same. 
What else can I do to find out which of them are isomorphic?
Thank you very much for your help!
 A: Let's find what's $M(\mathbb{Q}[[x]])$. The invertible elements in $\mathbb{Q}[[x]]$ are precisely the power series with nonzero constant term. So, modulo association, we only get
$$
[0],\quad [1],\quad [x],\quad [x^2],\quad \dotsc
$$
which is the free monoid over one generator with the addition of an absorbing element $[0]$. (With $[t]$ I'll denote the equivalence class modulo association.)
The monoid $M(\mathbb{Z})$ instead is the free monoid on a countable basis (the primes) with the addition of an absorbing element (fundamental theorem of arithmetic).
Similarly, $M(\mathbb{Z}_2[x])$ is the free monoid on a countable basis with the addition of an absorbing element, because $\mathbb{Z}_2[x]$ is a principal ideal domain, hence a unique factorization domain, with a countable number of irreducible elements (up to association), because there is at least an irreducible monic polynomial for every positive degree.
Actually, the last mentioned fact is not necessary: what we need is to prove that there are infinitely many irreducible monic polynomials in $\mathbb{Z}[x]$. The proof is essentially the same as Euclid's proof for the integers: if $p_1(x), \dots,p_n(x)$ are monic irreducible polynomials, then
$$
P(x)=p_1(x)p_2(x)\dots p_n(x)+1
$$
is a monic polynomial that has positive degree, so it is divisible by an irreducible (monic) polynomial, which cannot be in the list $p_1(x),\dots,p_n(x)$. Thus no finite list of irreducible polynomials contains all irreducible monic polynomials. Consequently there are infinitely many irreducible monic polynomials. Being monic, they are pairwise not associate. Since the ring $\mathbb{Z}_2[x]$ is countable, we have proved the statement that $M(\mathbb{Z}_2[x])$ is the free monoid on a countable basis with an absorbing element added.
