Is $\mathbb{Z}[\frac{1}{p}]$ a projective $\mathbb{Z}$-module, where $p$ is a fixed prime?

Question

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 $\mathbb{Z}$?

Answer 1

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.

Answer 2

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 $p^r \: (r\in \mathbb N)$.