Condition under which the subring of multiples of an element equal the ring itself? Let $R$ be a commutative ring with identity with $b\in R$. Let $T$ be the subring of all 
multiples of $b$, $T=\{r\cdot b : r \in R\}$. If $u$ is a unit in $R$ with $u \in T$, 
prove that $T=R$.
Could you help me some suggestions?
I really have no clues to do this questions, I can only show $1\cdot R$ belongs to $T$. I even don't know the general way to prove two rings are equal.
 A: The critical thing to realize here is that if $u$ is a unit, and $u = ab$, then $a$ and $b$ are both units.  For if $u$ is a unit, then $uv = 1$ for some $v \in R$, so that $1 = (ab)v = a(bv) = b(av)$, where we have used the commutativity of $R$.  So $u \in T$ a unit implies $b \in T$ is a unit is well, since $u = ab$ for some $a \in R$.  Now since $b \in T = Rb$ is a unit, for any $s \in R$ take $r = sb^{-1} = s(av)$.  Then $s = s1 = sb^{-1}b = rb \in Rb = T$, so in fact $R \subset T$.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
A: I'd like to dispel the notion in the comments that the key idea is a "trick". Instead, I will show that it boils down to the transitivity of "contains" or  "divides" (i.e. $\ b\mid c,\,\ c\mid d\ \Rightarrow\ b\mid d),\ $  using the following simple but fundamental correspondence (ubiquitous in divisibility theory)
Theorem $\ \ bR \supseteq aR\iff b\mid a\,$ in $\,R\quad$ [Contains = Divides for principal ideals]
Proof $\ (\Rightarrow)\,\ a = a\cdot 1\in aR\subset bR\,\Rightarrow\, a = br,\ $ for some $\,r\in R,\,$ so $\,b\mid a\,$ in $R.$
$(\Leftarrow)\ \ b\mid a\,$ in $R\,\Rightarrow\, a = br,\ $ for some $\,r\in R,\ $ so $\ aR = brR\subseteq bR,\,$ by $\,rR\subseteq R.\  $ QED
Corollary $\ $ If $\,c\in bR\,$ and $\,c\mid d\,$ then $\,bR \supset dR$.
Proof $\ $ Translating containment relations to corresponding divisibilty relations we obtain $\ c\in bR\,\Rightarrow\, b\mid c,\,$ so $\, c\mid d\,\Rightarrow\, b\mid d\,\Rightarrow\, bR\supseteq dR,\,$ by transitivity of "divides" and Theorem. $\  $ QED
Your exercise is the special case of the Corollary when $\,c\,$ is a unit, i.e. $\,cr = 1\,$ for some $\,r\in R,\,$ therefore $\,c\mid 1,\,$ so with $\,d=1\,$ in the corollary, we conclude that $\,bR\,\supseteq\, 1\cdot R = R.\,$ Hopefully this clarifies the motivation behind my hint in the comments to your question.
While imposing this conceptual structure costs slightly more effort, the rewards are great, since this structure plays a fundamental conceptual role in divisibility theory and related algebra. Indeed, if you later study algebraic number theory you will learn that ideals satisfying "contains = divides" play a key role in Dedekind's beautiful ideal-theoretic restoration of unique factorization in algebraic number rings (Dedekind domains are those domains whose proper ideals have prime factorizations (necessarily unique); they are the domains whose ideals satisfy "contains = divides", and a suitable finiteness condition, e.g. Noetherian).
