# An example of a ring homomorphism

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.

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

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.