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 map is not "closed" under addition, sets are. – Carsten S Nov 6 '15 at 18:12
• A more appropriate word than "closed under" would be "commutes with" if I'm interpreting you correctly. – Milo Brandt Nov 6 '15 at 20:55

$T:\mathbb{C}\to\mathbb{C}$ defined by $T(z)=\bar{z}$ then $T(z_1+z_2)=T(z_1)+T(z_2)$ but $T(cz)\neq cT(z)$

• I was thrown for a bit since I'm mostly doing math in $\mathbb{R}$ right now, but $c$ is complex, right? – porglezomp Nov 6 '15 at 17:52
• yes, you are right, $c$ is complex. – R.N Nov 6 '15 at 18:01

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.

• Does "extend by $\mathbb{Q}$-linearity" use axiom of choice? – lisyarus Nov 6 '15 at 16:54
• Given a real $x$, why should checking whether $x=r\pi$ for some rational $r$ depend on choice? – Zach Stone Nov 6 '15 at 17:27
• Yes, it uses choice. When extending the map, you actually use that $\mathbb R$ has a basis as $\mathbb Q$-vectorspace (hence it is free). And this uses choice. – MooS Nov 6 '15 at 18:13
• Original source: zbmath.org/?q=an%3A36.0446.04 :) – Carsten S Nov 6 '15 at 18:17
• Just to be totally clear; there are models of ZF (that is, set theory without choice) in which every additive map from R to R is actually linear. So you will have to look elsewhere (eg. C) for nice answers. – Richard Rast Nov 6 '15 at 21:41

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.

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.