Before stating my question, let me recall some preliminaries in rings (especially noncommutative).

Recall that for a noncommutative ring $R$, $‎\textbf{B}‎(R)=\{e\in Z(R): e^2=e\}$.

$\textbf{Definition:}$ Let $R$ and $S$ be two noncommutative rings. A ring homomorphism $\Phi: R\rightarrow S$ is called a $\textbf{conformal}$ map, if $\Phi(\textbf{B}‎(R))\subseteq \textbf{B}‎(S)$. Also, if $\Phi$ is an epimorphism and $\Phi(\textbf{B}‎(R))=\textbf{B}‎(S)$, then $\Phi$ is called a conformal epimorphism.

I should emphasize that if $\Phi$ is an epimorphism, then $\Phi$ is a conformal homomorphism. Now, here is my question:

$\textbf{Question:}$ Can someone help me to find a conformal homomorphism which is an epimorphism but is not a conformal epimorphism?

Many thanks for your notice.

  • $\begingroup$ Could you do the inclusion of $M_n(\mathbb{Z})$ into $M_n(\mathbb{R})$, just out of curiosity? $\endgroup$
    – Matt
    Oct 7, 2018 at 7:43
  • $\begingroup$ Just extend the image ring $S$ to a larger ring. $\endgroup$
    – Wuestenfux
    Oct 7, 2018 at 7:44
  • $\begingroup$ @ Matt abd @Wuestenfux, I have edited the mean problem. I am so sorry for forgeting to emphasize that this map should be epic. $\endgroup$ Oct 7, 2018 at 10:17

1 Answer 1


All you need is an $R$ deficient in idempotents such that $S$ has more idempotents. So, $\mathbb Z\twoheadrightarrow \mathbb Z/6\mathbb Z$ works.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .