What is $\operatorname{Hom}_{\mathbb{Z}}(\mathbb{R},\mathbb{R})$?

Question

We already know that $\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Q},\mathbb{Q}) = \mathbb{Q}$. Why? Because each homomorphism $f$ is uniquely determined by the value $f(1)$ and we can then calculate for any $a,b\in\mathbb{Z}$ using the $\mathbb{Z}$-linearity of $f$:

\begin{align} a\cdot f(1) &= f(a) =f\left(b\cdot \frac{a}{b}\right) = b\cdot f\left(\frac{a}{b}\right)\\ \Rightarrow f\left(\frac a b\right) &= \frac a b \cdot f(1) \end{align}

However, this trick cannot be used for real numbers. How can we then calculate $\operatorname{Hom}_{\mathbb{Z}}(\mathbb{R},\mathbb{R})$? I would suggest that it is $\neq \mathbb{R}$, but I have no idea how to show or disprove that.

Answer 1

A $\mathbb Z$-linear map $f:\mathbb R\rightarrow\mathbb R$ must also be $\mathbb Q$-linear map due to the argument you provide. Specifically, $\mathbb Z$-linearity gives $$b\cdot f\left(\frac{a}b\cdot x\right) = f(a\cdot x) = a\cdot f(x)$$ which can be solved as $$f\left(\frac{a}b\cdot x\right) = \frac{a}b\cdot f(x).$$ This is a big improvement since vector spaces tend to be better behaved than modules - and we've reduced our question to asking about linear maps $\mathbb R\rightarrow\mathbb R$ as vector spaces over $\mathbb Q$.

These maps can be described using the axiom of choice - which is equivalent to the following statement:

Every vector space $V$ has a basis $B$.

As a particular case of this, $\mathbb R$ must have some basis $B$ as a vector space over $\mathbb Q$ - meaning that every element of $\mathbb R$ is a finite sum of elements of $B$ with weights from $\mathbb Q$. Note that such a basis has the same cardinality as $\mathbb R$ and depends on the axiom of choice to construct.

Then we can use typical vector space facts to finish - in particular, that defining a map from a vector space can be done just by choosing its values on each basis element. Formally:

If $V$ and $W$ are vector spaces and $B$ is a basis for $V$, then for any function $g:B\rightarrow W$, there is a unique linear map $f:V\rightarrow W$ such that $f(b)=g(b)$ whenever $b\in B$.

We then apply that to the spaces in question: the set of $\mathbb Q$-linear maps $\mathbb R\rightarrow\mathbb R$ is in bijection with the set of functions $B\rightarrow\mathbb R$ where $B$ is a $\mathbb Q$-basis of $\mathbb R$. To say the least, this is a very big set and not at all as nice a result as one has with $\mathbb Q\rightarrow \mathbb Q$.

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Here's a nicer, less axiom-of-choice flavor of the same intuition: consider $\mathbb Q$-linear maps $\mathbb Q[\sqrt{2}]\rightarrow \mathbb Q[\sqrt{2}]$ - where $\mathbb Q[\sqrt{2}]$ is the set of expressions of the form $a+b\sqrt{2}$ for rational $a$ and $b$. You could explicitly note $\{1,\sqrt{2}\}$ as a $\mathbb Q$-basis and note that such linear maps are just pairs of values from $\mathbb Q[\sqrt{2}]$ - one specifying the image of $1$ and the other specifying the image of $\sqrt{2}$.

The axiom of choice lets us claim that $\mathbb R$ is somehow like that, except that instead of adding one new element ($\sqrt{2}$) we add uncountably many new elements instead!

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It's probably worth noting that this whole discussion with $\mathbb R$ relies crucially on the axiom of choice - without the axiom of choice, it is not only consistent that there might not be a $\mathbb Q$-basis $B$ for $\mathbb R$, but it's even consistent that every $\mathbb Q$-linear map $\mathbb R\rightarrow \mathbb R$ could also be $\mathbb R$-linear - so without the axiom of choice, it's hard to say much interesting about the set you're asking about.