R is projective Q-module?

Question

The set of real numbers $\mathbb{R}$ is project $\mathbb{Q}$-module?

I think it is not but I cannot prove it.

How can I prove it?

Answer 1

The answer is more general. If $\mathbb{F}$ is a field, then the category of left (resp. right) modules over $\mathbb{F}$ has global dimension $0$, which means that every object in this category is projective. Note that in the case of a field, modules over it are simply vector spaces over it.

Now in general, if $R$ is any ring, then the projective (right or left) modules over $R$ are exactly the direct summands of free modules over $R$: this means that an $R$-module $M$ is projective if and only if there exists some index-set $I$, as large as it may be, and another $R$-module $N$ such that $\bigoplus_{i\in I}R\cong N\oplus M$, where $$\bigoplus_{i\in I}R:=\{(r_i)_{i\in I}: r_i\in R,\text{ only finitely many of the }r_i\text{ are non-zero}\} $$

Now $\mathbb{R}$ is obviously a module over $\mathbb{Q}$ via the usual number multiplication, i.e. $\mathbb{R}$ is a vector space over $\mathbb{Q}$. To check whether it is a projective module, one should wonder if we can find an index set $I$ and another vector space over $\mathbb{Q}$ such that $V\oplus\mathbb{R}\cong\bigoplus_{i\in I}\mathbb{Q}$. So take a basis of $\mathbb{R}$ as a vector space over $\mathbb{Q}$ (by linear algebra, every vector space has a basis). Use this as an index set and take $V=0$ and you will see that this gives you the desired isomorphism.

(yes, $\mathbb{R}$ is a projective $\mathbb{Q}$-module).