Map closed under addition but not multiplication I have been helping undergrads in an introduction to linear algebra course. When solving some exercise consisting in showing that a map is linear some get lazy after proving that it is closed under addition and do not prove the closure under scalar multiplication. I wanted to confront them with an example of a map closed under addition but not under the multiplication but could not come up with an example. Do you have any?
 A: If you take any field $F$, and a homomorphism of additive groups $(F,+)\to (F,+)$ which does not preserve multiplication, then this will be just such a map.
This is will never exist when $F$ is the field of rational numbers, or more generally, when it is generated by the unity (such as any ${\bf F}_p$ for a prime $p$), because in those, we can define multiplication in terms of addition.
A similar thing happens if you look at continuous additive maps in a topological field generated topologically by the unity (that is, the smallest subfield is dense), like the reals or $p$-adics -- every continuous additive map is linear in this case.
On the other hand, if you don't care about continuity, it is pretty easy to define such a map when $F=K[a]$ is a finite extension of another field $K$. Then you can just take $f\colon F\to F$ as the map such that $f(a^n)=0$ for $0<n<\deg a$ and $f(k)=k$ for $k\in K$. For example, if $K={\bf R}$ and $a=i$, $f$ takes real part of a complex number. In this case, the map is even continuous.
For fields which are not finite extensions of other fields (such as $F={\bf R}$), the existence of such maps may require a nontrivial application of axiom of choice, or more precisely, basis theorem, and then we can proceed as in the preceding paragraph: if $K\subseteq F$ is a field extension, then the map $F\to F$ which is identity on $K$ and takes a basis complementary to $1$ to zero is $K$-linear, and therefore additive.
A: If $T:\mathbb{C}\to\mathbb{C}$ is defined by $T(z)=\bar{z}$ then $T(z_1+z_2)=T(z_1)+T(z_2)$, but $T(cz)\neq cT(z)$
A: Consider any field homomorphism $L \to L$, and consider $L$ as a vector-space over a subfield, which is not fixed by the map.
From that point of view, the "easiest" example is Frobenius on $\mathbb F_4$.
Note that there are 8 additive maps on the field with 4 elements, which of 4 are also linear.
A: Over the real numbers, this is tough. Any additive map is linear over $\mathbb{Q}$, so will be linear over $\mathbb{R}$ as soon as it's continuous. However, there are non-continuous additive maps, even $\mathbb{R}\to\mathbb{R}$, for instance, $\sqrt{2}\mapsto \pi, \pi\mapsto\sqrt{2}$, extend by $\mathbb{Q}$-linearity and fix everything else. $\pi$ and $\sqrt{2}$ are linearly independent over $\mathbb{Q}$, so this is well defined.
