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Let $V$ and $W$ be vector spaces over a field $\mathbb{F}\neq \mathbb{C}$. Give an example of a non-linear map $T:V\to W$ such that \begin{equation}T(x+y) = T(x)+T(y), \forall x,y\in V.\end{equation}

I asked myself this question when I was resolving an excercise list os Linear Algebra. This example is pretty easy when $\mathbb{F}=\mathbb{C}$. We take $V=W=\mathbb{C}$ and $T:z\mapsto \bar{z}$, and we have that $T(\lambda z) = \bar{\lambda}T(z)$. However, I couldn't find any examples for non-complex vector spaces.

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    Possible duplicate of [Does a nonlinear additive function on R imply a Hamel basis of R?](http://math.stackexchange.com/questions/366010/does-a-nonlinear-additive-function-on-r-imply-a-hamel-basis-of-r)2017-01-17
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    @FlybyNight but that's specifically asking about $\Bbb R$2017-01-17
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    @Omnomnomnom The OP said they "couldn't find any examples for non-complex vector spaces." The reals are such an example. A simple Google search of "additive non linear functions" turns up lots of hits, e.g. on stackexchange, stackoverflow and some other sites.2017-01-17
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    @FlybyNight still, there is no one source that answers the question sufficiently. That seems like enough of a reason to leave it open2017-01-17

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Every such map with $\Bbb F = \Bbb Q$ will be linear. The same will be true for finite fields.

With $\Bbb F = \Bbb R$, this may or may not be true, depending on your axioms (see the wiki page for more).

If $\Bbb F$ is an extension of $\Bbb Q$ with degree $1<[\Bbb F: \Bbb Q] < \infty$, then such maps exist (for example: $\Bbb Q(\sqrt{2})$). This same idea works if $\Bbb F$ satisfies $1 < [\Bbb F: \Bbb F'] < \infty$ for any subfield $\Bbb F'$, for that matter.

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This is just a slight variation on your counterexample. Consider the field $\mathbb{Q}(\sqrt{2}) = \{a+b\sqrt{2}\mid a,b\in\mathbb{Q}\}$. Then $T: \mathbb{Q}(\sqrt{2}) \to \mathbb{Q}(\sqrt{2}),\ a+b\sqrt{2}\mapsto a-b\sqrt{2}$ is such a non-linear map.

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A discontinuous additive function $a\colon\Bbb R\to\Bbb R$ is non-linear map of a linear space $\Bbb R$ over a field $\Bbb R$ to itself. Observe that any additive map is linear, if the field of scalars is $\Bbb Q$.