In one of my assignment, I was told that $(\mathbb Z[x], +)$ and $(\mathbb Q_{>0}, .)$ are isomorphic, with $$\phi (\sum_{k=0}^n a_k x^k) = \prod_{k=0}^n p_k^{a_k}$$ is a one-to-one surjective group homomorphism, where $p_1 = 2$; $p_2 = 3$; $p_3 =5$, and so on.
I know that this proof is based on the decomposition of numbers into a product of primes. But I don't think that it is correct. For instance, for $\frac12 \in \mathbb Q$, I cannot find a corresponding polynomial $f \in \mathbb Z[x]$ such that $\phi(f) = \frac12$.
If this proof is wrong, then are $(\mathbb Z[x], +)$ and $(\mathbb Q_{>0}, \cdot)$ really isomorphic?
I think you have a typo. The isomorphism is $$\phi (\sum_{k=0}^n a_k x^k) = \prod_{k=0}^n p_{k+1}^{a_k}$$
Then, $\frac 12 = p_1^{-1}$ and so the corresponding polynomial is $-1x^0$.
The point is that $(\mathbb Q_{>0}, .) \cong (\Pi_p \mathbb Z, +)$. That is exactly what $(\mathbb Z[x], +)$ is.