I read "Foundations of Module and Ring Theory" of Robert Wisbauer and I got stuck in this problem:
*Show for a ring $R$. The following assertions are equivalent:
(a) $R$ has a unit.
(b) If $R$ is an ideal in ring $S$, then $R$ is a direct summand of $S$.
(c) If $R$ is an ideal in ring $S$, then $R$ is a homomorphic image of $S$.*
Could anyone give some help?
We show the implications (a) $\Rightarrow$ (b) $\Rightarrow$ (c) $\Rightarrow$ (a).
Assume (a) and let $e$ be the unit of $R$. Let $S$ be any ring such that $R\subseteq S$ is an ideal. For any $s\in S$, we have $se, es\in R$ and in fact they are equal: $$ es = e(es) = (es)e = e(se) = (se)e = se, $$ where for the second and fourth equality we have used that $er = re$ for all $r\in R$ and the third equality follows from associativity of multiplication. This shows that $e\in S$ is central. Now, the injection $R\rightarrow S$ has left inverse $S\rightarrow R$, $s\mapsto se=es=ese$, i. e. $R$ is a direct summand in $S$.
Assume (b). Let $R$ be an ideal of $S$ such that $R$ is a direct summand of $S$. Then there is a projection map $S\rightarrow R$, which shows (c).
Assume (c). Take $S = \mathbb Z\times R$ with componentwise addition and multiplication given by $$ (n,r)\cdot (n',r') := (nn', nr'+n'r+rr'). $$ With this, $S$ is a ring with unit and $R$ is an ideal in $S$. By hypothesis, there exists a surjective ring homomorphism $f\colon S\rightarrow R$. Then $f(1)$ is necessarily a unit in $R$.