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Is $\mathbb{Z}[\frac{1}{p}]=\{\sum_{i=0}^n \frac{a_i}{p^i} : a_i\in \mathbb{Z}, n \in \mathbb{N}\cup\{0\}\}$ projective for any fixed prime $p$?

I don't know how you would go about proving this. Can we simple say $\{1, p^{-1}, p^{-2}, \dots \}$ is a basis so that $\mathbb{Z}[\frac{1}{p}]$ is free over $\mathbb{Z}$?

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    I would not say that set is a basis: $p \cdot p^{-1} -1= 0$! – 2017-01-17
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    @Andrew: Thanks for pointing this out, I was thinking of the polynomial ring in a single variable and thought I'd get a basis similarly here too. – 2017-01-17
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    @user406564 do you have an injective ring morphism $\mathbb{Z}[X] \to \mathbb{Z}[1/p]$ ? So I'd say this is more related to the [localization](https://en.wikipedia.org/wiki/Localization_of_a_ring) of a ring. – 2017-01-17
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    @user1952009: I don't know. I can't seem to think of any. – 2017-01-17
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    A ring morphism $\phi : \mathbb{Z}[X] \to R$ (sending $1$ to $1$) is determined by $\phi(X)$. Such a morphism isn't injective when $R = \mathbb{Z}[1/p]$. – 2017-01-17
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    Since $\mathbb{Z}$ is a PID, projective would imply free, which is impossible for $\mathbb{Z}[1/p]$. – 2017-01-17
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    You can say it is a basis, but you that's not enough. Did you try proving it is a basis? It isn't of course. – 2017-01-17

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No. If it were projective, it would actually free, and no two elements of $\mathbf Z\biggl[\dfrac1p\biggr]$ are linearly independent.

Indeed two elements can be written as $x=\dfrac a{p^r}$, $y=\dfrac b{p^s}$ and $$bp^r\cdot x-ap^s\cdot y=0.$$ Thus if $\mathbf Z\biggl[\dfrac1p\biggr]$ is free, it has rank $1$, so that $\mathbf Z\biggl[\dfrac1p\biggr]=\mathbf Z$, and $p$ is a unit in $\mathbf Z$. That would be great news.

Edit: The same kind of argument shows no ring of fractions of an integral domain $R$ can be a free $R$-module.

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    I like great news (+1). – 2017-01-17
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A projective $\mathbb Z$-module is in particular a submodule of some free $\mathbb Z$-module $\mathbb Z^{(I)}$.
But a non-zero element of a free $\mathbb Z$-module can only be divided by finitely many integers, while any element of $\mathbb Z[\frac 1p]$ can be divided by the infinitely many $p^r \: (r\in \mathbb N)$.

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    (Infinitely $p$-divisible elements!) – 2017-01-17