Is there a theory for rings which are isomorphic to a proper subring? Which of the following rings have this property?

$$ \mathbb{R} , M_2(\mathbb{R}) , \mathbb{C} \; and \; M_2(\mathbb{Z})$$


I know that $\mathbb{C}$ is isomorphic to a proper subring of $\mathbb{C}$, assuming the axiom of choice. I have no idea about $\mathbb{R}$ and $M_2(\mathbb{R})$. I don't know about any study about such rings either. But here is a proof that $\mathbb{C}$ has a proper subring isomorphic to itself.

Let $\mathcal{B}$ be a transcendental basis of $\mathbb{C}$ over $\mathbb{Q}$. Then, note that $\mathcal{B}$ and $\mathcal{B}\cup\{x\}$ are equinumerous, where $x$ is a transcendental variable. Therefore, a bijection $f:\mathcal{B}\cup\{x\}\to \mathcal{B}$ lifts to a field isomorphism $\varphi:\mathbb{Q}\big(\mathcal{B}\cup\{x\}\big)\to \mathbb{Q}(\mathcal{B})$, which then can be extended to an isomorphism $\Phi:\overline{\mathbb{Q}\big(\mathcal{B}\cup\{x\}\big)}\to \overline{\mathbb{Q}(\mathcal{B})}$. That is, $\Phi:\overline{\mathbb{C}(x)}\to \mathbb{C}$ is an isomorphism of fields. Now, consider the canonical injection $\iota:\mathbb{C}\to\overline{\mathbb{C}(x)}$. Then, we see that the image $S$ of $\mathbb{C}$ under the composition $\Phi\circ \iota$ is a proper subring of $\mathbb{C}$ isomorphic to $\mathbb{C}$.

  • $\begingroup$ And why do you think this wont work for $\mathbb{R}$? $\endgroup$ – Paul K Oct 17 '18 at 7:48
  • $\begingroup$ Because I don't know what kind of extensions I should take to get from $\mathbb{Q}(\mathcal{T})$ to $\mathbb{R}$, if $\mathcal{T}$ is a transcendental basis of $\mathbb{R}$ over $\mathbb{Q}$. For $\mathbb{C}$ it is easy, just take the algebraic closure. Plus, since an algebraic closure is a splitting field, this allows us to extend $\varphi$ to $\Phi$. $\endgroup$ – user593746 Oct 17 '18 at 7:50
  • $\begingroup$ Shouldn't it work if you exchange $\mathbb{Q}$ by its algebraic closure and for $\mathbb{R}$ you intersect it with $\mathbb{R}$? Maybe I'm missing something. $\endgroup$ – Paul K Oct 17 '18 at 7:54
  • $\begingroup$ I don't know the answer to that question. I will think about it. $\endgroup$ – user593746 Oct 17 '18 at 7:56
  • 2
    $\begingroup$ @Zvi I just remembered that the only non-zero function from reals into reals that preserves addition and multiplication is the identity function. So the field of real numbers is not isomorphic to a proper subring. Am I right ?! $\endgroup$ – alex alexeq Oct 17 '18 at 15:05
  1. You should look at A ring isomorphic to a proper subring of itself
  2. If you have enough background in the topics of algebra, then you should read https://www.researchgate.net/publication/321183071_Rings_isomorphic_to_their_nontrivial_subrings

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