Prove that $a + \langle b \rangle$ is a unit in the quotient ring $R/\langle b \rangle$ if and only if $a$ does not divide $b$ in $R$.

Question

Let $R$ be a commutative ring with $1$. Suppose that $a, b \in R$, $b \notin R^\times$ and $\langle a \rangle$ is a maximal ideal of $R$. Prove that $a + \langle b \rangle$ is a unit in the quotient ring $R/\langle b \rangle$ if and only if $a$ does not divide $b$ in $R$. [Notice that $R^\times = \{ r \in R ~|~ r \text{ is a unit.} \}$].

$\textbf{My Attempt:}$
Prove of "$\implies$":
Which want to prove that if $a + \langle b \rangle$ is a unit in the quotient ring $R/\langle b \rangle$, then $a$ does not divide $b$ in $R$.
By contra-positive, we can prove that if $a$ divide $b$ in $R$, then $a + \langle b \rangle$ is not a unit in the quotient ring $R/\langle b \rangle$.
Assume $a$ divide $b$ in $R$.
Then, there exist $k \in R$ such that $b = ak$.
Since, $b \in R$ and $b \notin R^\times$.
Then, $b = b \cdot 1 \in \langle b \rangle$.
Since, $a \in R$. So, $ak = b \in \langle b \rangle$.
Then, $(a + \langle b \rangle)(k + \langle b \rangle) = ak + \langle b \rangle = \langle b \rangle = 0 + \langle b \rangle \neq 1 + \langle b \rangle$.
So, $a + \langle b \rangle$ is not a unit in the quotient ring $R/\langle b \rangle$.
So, this proves that if $a + \langle b \rangle$ is a unit in the quotient ring $R/\langle b \rangle$, then $a$ does not divide $b$ in $R$.

Prove of "$\Longleftarrow$":
Which want to prove that if $a$ does not divide $b$ in $R$, then $a + \langle b \rangle$ is a unit in the quotient ring $R/\langle b \rangle$.
Assume $a$ does not divide $b$ in $R$.
Then, there exist $k \in R$ such that $b = ak + r$, where $r \in R$ and $r \neq 0$.

Which I am stuck on how to prove that $r = -1$. Which then I can prove that $ak - 1 \in \langle b \rangle$ and conclude that $a + \langle b \rangle$ is a unit in the quotient ring $R/\langle b \rangle$.
Also, is the prove of "$\implies$" correct ?
And, I don't know where to apply the given condition that $\langle a \rangle$ is a maximal ideal of $R$.

Answer 1

Hint: Note that for $ak-1 \in \langle b\rangle$ you don't need to find $k$ with $ak-b=1$ (no such $k$ will exist in general, try $a=7, b=10$ in $R=\mathbb Z$), but only a $k$ with $ak-bl=1$ for some $l\in R$. Maybe now you see where to use the maximality of $\langle a\rangle$?

In your proof for the other direction, you are a bit imprecise at the end: It seems like you want to argue that the existence of some $k$ such that $ak +\langle b\rangle \neq 1 +\langle b\rangle$ shows that $a +\langle b\rangle$ is not a unit. That's not the case. But the fact that there is a $k$ such that $ak +\langle b\rangle = 0 +\langle b\rangle$ indeed shows what we need, namely that there can be no other $k$ with $ak +\langle b\rangle = 1 +\langle b\rangle$. Do you see why?

Answer 2

Hint: it's straightforward using $\rm\color{#c00}{QR}$ = Quotient Reciprocity (Third Isomorphism Theorem)

$$\begin{align} &(a)\!=\!(1)\,\ {\rm in}\,\ R/b\\[.2em] \iff\ & (R/b)/a = (0)\\[.2em] \iff\ & (R/a)/b = (0),\ {\rm by\ \color{#c00}{QR}}\,\\[.2em] \iff\ & (b)\!=\!(1)\ \,{\rm in}\,\ R/a\\[.2em] \iff\ &\ \,b\not \in (a),\ {\rm by}\ R/a\,\ {\rm a\ field,\ by}\ (a)\ {\rm max} \end{align}\qquad$$