$\mathbb{Q}$ - linear function from $\mathbb{R}$ to $\mathbb{R}$ with kernel $\mathbb{Q}$ I read somewhere that it is possible to construct a $\mathbb{Q} $ - linear function $f:\mathbb{R} \rightarrow \mathbb{R}$ with $\ker f = \mathbb {Q}$ . Can someone enlighten me on that matter? I thought it might be an easy matter of extending the canonic quotient map $\mathbb{R} \rightarrow \mathbb{R}/\mathbb{Q}$ with an isomorphism  $\mathbb{R}/\mathbb{Q}  \rightarrow \mathbb{R}$, but that is probably nothing more than going round in circles.
 A: Well, it depends on what you mean by "construct," but here is one way:
$\mathbb{R}$ is a $\mathbb{Q}$ vector space, and so by the axiom of choice has a $\mathbb{Q}$ basis $V$. By cardinality reasons $V$ must be infinite, so there exists a bijection between $V\setminus \{v\}$ and $V$, for any $v\in V$. Take the map from $\mathbb{R}$ to $\mathbb{R}$ as $\mathbb{Q}$ vector spaces induced by this bijection on the basis, and sends $v$ to zero.
A: Choose a $\mathbb{Q}$ basis $B$ of $\mathbb{R}$ that contains $1$. Then define $f(x)$ by expanding $x$ as a finite linear combination of elements of $B$ and setting the coefficient of $1$ in that expansion to be $0$ (if it isn't already).
Note: $B$ will be uncountable and you need the axiom of choice to construct it.
A: More of the same:
Assume you have $f$ a non-zero $\mathbb{Q}$-linear map from $\mathbb{R}$ to $\mathbb{Q}$. Consider a $\beta$ so that $\alpha\colon=f(\beta)\ne 0$. The map     $$p\colon \mathbb{R} \to \mathbb{Q}\\
x\mapsto \frac{1}{\alpha} f(\beta x)$$  is a $\mathbb{Q}$-linear projection. Then $p'\colon= 1_{\mathbb{R}}-p$ has kernel $\mathbb{Q}$. 
