For an algebraic curve, I know what the divisors are, they are $\mathbb{Z}$ linear combinations of points of the curve or equivalently sums of the maximal ideals. (If the variety of a higher dimension, then the Weil divisors are linear combinations of codimension one irreducible subvarieties). If one believes the analogy between curves and rings of integers, then the divisors of $spec\mathbb{Z}$ should be a $\mathbb{Z}$ linear combination of the maximal ideals. But one could take this and show it to be isomorphic to $\mathbb{Q}^*_+$. My question is : is $div(spec\mathbb{Z})$ isomorphic to $\mathbb{Q}^*_+$?
Let me talk about my motivation for asking this question. I was looking for a nice proof that the number, $x^{1/n},x\in\mathbb{Q}_+$ is rational if and only if $x=\frac{p_1^{m_1}...p_k^{m_k}}{q_1^{n_1}...q_l^{n_l}}$, in lowest terms, and each $n_i,m_i$ are divisible by $n$. If we write $x$ in terms of what I think the divisors of $spec\mathbb{Z}$ are, then $div(x)=m_1(p_1)+...m_k(p_k)-n_1(q_1)-...-n_l(q_l)$. Thus exponentiating by $1/n$ is the same as multiplying the divisor by $1/n$ which can only be done if each coeffecient is a multiple of $1/n$