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.

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!!!
