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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?

Minh Nam
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  • Is $R$ commutative? – Bernard Jul 25 '16 at 10:11
  • @Bernard The ring $R$ is not commutative in general. – Minh Nam Jul 25 '16 at 10:16
  • Then another question: is the ideal a two-sided ideal? Does $S$ have a unit? – Bernard Jul 25 '16 at 10:18
  • It seems eminently clear that "ideal" means "two sided ideal" (as is standard practice) but yes, it would be nice to know if $S$ is supposed to have an identity. I just looked it up in Wisbauer's book (it's freely available online) and the problem seem to be transcribed faithfully. From the way it's written it could hardly be a misprint. I guess we just have to work a little instead of relying on $S$'s identity. – rschwieb Jul 25 '16 at 10:21
  • For (c) $\Rightarrow$ (a) it is certainly necessary that $S$ have a unit; otherwise taking $S=R$ gives a counter example. Or are ideals always understood to be proper? – Claudius Jul 25 '16 at 10:38
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    @user218931 To find a counterexample to the implication you're talking about, you would have to exhibit a ring such that for all $S$ with $R\lhd S$, $R$ is a summand of $S$ and $R$ does not have identity. – rschwieb Jul 25 '16 at 10:44
  • Ah, sorry! This makes it clear now. – Claudius Jul 25 '16 at 10:46

1 Answers1

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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 $R$.

Claudius
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