What is $\text{End}_\mathbb{Z}(\mathbb{R})$?

Question

I read from another post "When is $G\cong\operatorname{End}(G)$?", $\text{End}_\mathbb{Z}(\mathbb{R})$ is not $\mathbb{R}$, it may be something "monstrous" due to Axiom of Choice.

I am curious what exactly is $\text{End}_\mathbb{Z}(\mathbb{R})$ isomorphic to? Or if a nice description is not possible, what are some of the differences between $\text{End}_\mathbb{Z}(\mathbb{R})$ and $\text{End}_\mathbb{\mathbb{R}}(\mathbb{R})\cong\mathbb{R}$?

Thanks.

Answer 1

Assume AC.

Then $\Bbb R$ is a vector space of dimension $|\Bbb R|$ over $\Bbb Q$. The $\Bbb Z$-endomorphism ring of $\Bbb R$ is the same as the $\Bbb Q$-endomorphism ring. If $V$ is a $\Bbb Q$-vector space with basis $(e_i)_{i\in I}$ then the endomorphisms of $V$ are given by the row-finite $I\times I$ matrices over $\Bbb Q$, that is matrices $(a_{i,j})_{i,j\in I}$ such that for each $i$, only finitely many $a_{i,j}$ are nonzero.

Answer 2

Since a $\mathbb Z$-linear endomorphism of $\mathbb R$ is automatically also $\mathbb Q$-linear, we are in the realm of linear algebra. Thus, we can view $\mathbb R$ as a vector space over $\mathbb Q$, particularly one of dimension $\mathfrak c$. Thus, we can say that $$\mathrm{End}_{\mathbb Z}(\mathbb R) \cong \mathrm{End}_{\mathbb Q}\left(\mathbb R, \mathbb R\right) \cong \mathrm{Hom}_{\mathbb Q}\left(\bigoplus_{\mathfrak c}\mathbb Q, \mathbb R\right) \cong \prod_{\mathfrak c}\mathrm{Hom}_{\mathbb Q}\left(\mathbb Q, \mathbb R\right) \cong \prod_{\mathfrak c} \mathbb R = \mathbb R^{\mathfrak c}. $$ That is, we have $\mathfrak c$ many basis vectors for $\mathbb R$ as a vector space over $\mathbb Q$. A linear mapping is uniquely determined by where the basis vectors are sent; thus, to each of these $\mathbb c$ basis vectors we have to assign an element of our vector space, which is $\mathbb R$. Hence, we can encode $\mathrm{End}_{\mathbb Z}(\mathbb R)$ as a $\mathfrak c$-sized sequence of real numbers -- one way to think about that is as an (arbitrary!) function $\mathbb R \to \mathbb R$. This should come with the understanding that the two copies of $\mathbb R$ play a different role: one has an element for each real number, while the other has an element for each basis vector of $\mathbb R$ as a vector space over $\mathbb Q$.