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We know that for a polynomial $p(x)\in \mathbb Z[x]$, we can write $\frac{1}{p(x)}$ in partial fraction decomposition, i.e. we can write $\frac{1}{p(x)}$ as combinaison of $\frac{1}{(aX+b)^i}$ or $\frac{aX+b}{(cX^2+dX+e)^k}$ where $cX^2+dX+e$ irreducible. Is it the same for number in $\mathbb Z$ ? i.e. if $n\in\mathbb Z$, can we write $\frac{1}{n}$ as sum of $\frac{1}{p^{\alpha _i}}$ ?

user386627
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  • I think you meant "$p(x)\in \mathbb R[x]$" (otherwise you should allow irreducible polynomials of higher degree). See here. – Watson Feb 02 '17 at 14:05

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This follows from the fact that for a collection of integers we can express their greatest common divisor as a linear combination (with coefficients in $\mathbb{Z}$) of these integers. Let $n=\prod_{i=1}^k p_i^{a_i}$. Then the greatest common divisor of the numbers $\frac{n}{p_1^{a_1}},\ldots, \frac{n}{p_k^{a_k}}$ is 1 and hence there exist $x_i\in\mathbb{Z}$ such that $$ 1 = \sum x_i \frac{n}{p_i^{a_i}}. $$ Dividing by $n$ you get the result you look for.

MichalisN
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  • And thus, that's why it also work on $\mathbb Z[X]$ ? (or maxbe I should say $Frac(\mathbb Z[x])$ ? – user386627 Feb 02 '17 at 16:03
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    @user386627: It works over PIDs like $\mathbb{Q}[X]$ but $\mathbb{Z}[X]$ is not a PID (only a unique factorisation domain). In particular $\frac{1}{2X}$ cannot be written as a sum of inverses of powers of irreducible elements. – MichalisN Feb 02 '17 at 19:52