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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 $1/n$

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Yes, divisors on $Spec(\mathbb Z)$ are are the group $Div(\mathbb Z)$ of $\mathbb Z$-linear combinations of maximal ideals $p\mathbb Z$ ($p$ prime). As an abstract group $Div(\mathbb Z)\cong \mathbb Q^*_+$ (the positive rationals), not $\mathbb Q^*$. Notice that these groups are not isomorphic because the correct group has no torsion and the other has. –  Georges Elencwajg May 22 '12 at 8:32
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up vote 6 down vote accepted

The maximal ideals of $\mathbb Z$ are those generated by prime numbers, and the free Abelian group on them is isomorphic to $\mathbb Q^\times/\mathbb Z^\times=\mathbb Q^\times/\{1,-1\}$.

You can use this as you indicated to prove your result, but it seems simpler to just apply unique factorization in $\mathbb Z$.

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Unique factorization is what will give the iso between $div\mathbb{Z}$. Thus what I was doing was recasting the proof by applying unique factorization in the language of divisors in hopes of better understanding divisors! I do wonder if this sort of proof is in any way indicative of the way divisors are used in the real world. –  Baby Dragon May 22 '12 at 18:17
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