1

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.

LegNaiB
  • 2,757
  • 4
  • 26
  • Do you believe in the axiom of choice? – Arthur May 28 '21 at 22:09
  • 1
    You've tagged this with [tag:ring-homomorphism] - but you're talking about $\mathbb Z$-linear maps in the question. To be clear, are you asking about $\mathbb Z$-linear maps $\mathbb R\rightarrow\mathbb R$? – Milo Brandt May 28 '21 at 22:12
  • @Arthur In what kind does this matter for this question? I would be happy for any solution - doesn't matter if the axiom of choice is used or not. Anyhow, I believe in this axiom – LegNaiB May 28 '21 at 22:14
  • 1
    @MiloBrandt That tag was maybe not perfect as this question came up working with $\mathbb{Z}$-modules (i.e. rings). Indeed I'm talking about $\mathbb{Z}$-linear maps and I chose ring-homomorphism as there is no tag for module-homomorphism. Question: As rings and $\mathbb{Z}$-modules are equivalent, aren't the corresponding module homomorphisms ring homomorphisms as well? – LegNaiB May 28 '21 at 22:16
  • @Arthur thanks for the link, but this only gives an example of non-linear map between $\mathbb{R}$. But how does the set of all linear maps look like? – LegNaiB May 28 '21 at 22:19
  • @LegNaiB $\mathbb Z$-modules are abelian groups, not rings - perhaps you're confusing that a $\mathbb Z$-algebra is the same as a ring (...at least with the conventions I'm used to). – Milo Brandt May 28 '21 at 22:23
  • Oh yes, all rings are $\mathbb{Z}$-modules, but not the other way round, sorry. Thanks for clarification. But I meant $\mathbb{Z}$-linear maps. I edited the tag – LegNaiB May 28 '21 at 22:24
  • The map constructed (or at least alluded to) in my link above is $\Bbb Z$-linear (called "additive" in that post). It just isn't $\Bbb R$-linear, which was what was asked about there. If you believe in the AoC, then the set of $\Bbb Z$-linear functions $\Bbb R\to\Bbb R$ contains such functions as well. And as with most things that involve AoC, concrete descriptions are difficult. – Arthur May 28 '21 at 22:25
  • So you can definitely say that it's $\neq \mathbb{R}$ and has some complex structure which is difficult to describe? That's a frustrating answer... – LegNaiB May 28 '21 at 22:26

1 Answers1

6

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 $\mathbb Q\rightarrow \mathbb Q$.

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!

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.

Milo Brandt
  • 60,888
  • So one could conclude that at least the cardinality of those linear maps of $\mathbb{R}$ as $\mathbb{Q}$-vector spaces is equal to the cardinality of $\mathbb{R}$ because the cardinality of a basis must have the same cardinality? Or is it more the cardinality $\mathbb{R}^\mathbb{R}$ as we are looking for functions $B\to \mathbb{R}$? – LegNaiB May 28 '21 at 22:41
  • @LegNaiB It's $\mathbb R^{\mathbb R}$ because $|B|=|\mathbb R|$. ($\mathbb R^{\mathbb R}$ is also the cardinality of the powerset of $\mathbb R$, to put it in nicer terms) – Milo Brandt May 28 '21 at 22:43
  • So we have the cardinality but can we say that these sets are isomorphic as well? Is $B\simeq \mathbb{R}$ or do they just have the same cardinality? – LegNaiB May 28 '21 at 22:44
  • 1
    @LegNaiB It depends what you mean by "isomorphic" - $B$ is just a set; it doesn't have an addition operation or anything. Some people certainly describe two sets being in bijection by saying they're isomorphic as sets - though they're certainly not isomorphic as any sort of algebraic structures since $B$ doesn't even have algebraic structure. – Milo Brandt May 28 '21 at 22:46
  • Yeah, I would talk about isomorphie including algebraic structures (e.g. ring isomorphism) and not just as sets. Thanks for clarifying, that was indeed helpful! – LegNaiB May 28 '21 at 22:48
  • Interestingly, as soon as one requires any more topological-ish restrictions on the map $f$ (continuous at a point, the limit at some point exists, it is bounded on some interval, etc), the map must be $\mathbb{R}$-linear, i.e. of the form $f(x) = ax$. The alternative solutions that exist because of the axiom of choice were once explained to me like "if you tried to graph them, it would basically look like someone had thrown paint on your screen". – Joppy May 29 '21 at 02:01