Is $R/I$ a field whenever $R$ is a field? is it not a field whenever $R$ is not a field? Let $R$ be a ring $I\subset R$ an ideal.
Are the following true or false conditionals?

*

*If $R$ is a field then so is $R/I$.

*If $R$ is not a field then $R/I$ is not a field as well.

I think both claims are wrong, but I don't have examples for them. (I'll tell why do I think that, because I couldn't find those assertions in my search for proofs, so I guess I can find counterexamples).
Any help?
Thanks!
 A: If $F$ is a field, then its only ideals are $0$ and $F$.
So, $F/I$ is $F$ or $0$, the first is a field while the latter is not.
For the second question, consider $R=\mathbb Z$ and $I=(p)$ for some prime $p$.
Then $R/I=\mathbb Z/(p)$ is a field.
In general, for a commutative ring $R\neq0$ with an ideal $I$, $R/I$ is a field if and only if $I$ is a maximal ideal, i.e., a proper ideal that is not contained in any other proper ideal.
A: Theorem 1 : A commutative ring with unity is a field iff it has only two trivial ideals.
Proof : Let, $R$ be a commutative ring with unity with two trivial ideals.
Claim : $R$ is a field.
We need to show for any $0\neq a \in R$ is unit.
Choose, $I=\langle a \rangle$ principal ideal generated by $a$.
Since $\langle 0 \rangle \neq \langle a \rangle $
Hence, $\langle a \rangle =R$
$1\in R \implies 1\in \langle a \rangle $ and $1= ar$ for some $r\in R$.
Hence, $a$ is unit.
Conversely, suppose $R$ is a field.
Claim : $R$ has only two trivial ideals (zero ideal $\langle 0 \rangle $ and unit ideal $\langle 1 \rangle$)
Choose, $\langle 0 \rangle \neq I $ ideal of $R$.
Then $\exists 0\neq a\in I $.
Since, $R$ is a field, $a$ is unit
i.e  $a^{-1} \in R$
Then, $1=aa^{-1} \in I $ and for any $r\in R $ , $r\cdot 1 =r\in I$
Implies $ I=R$
Theorem 2: $R$ be a commutative ring with unity and $I$ is an ideal of $R$ then $R/I$ is a field iff $I$ is maximal ideal.
Proof : Suppose, $R $ be a commutative ring with unity and $I$ be a maximal ideal of $R$.
Then, $R/I$ is a commutative ring with unity $1+I$.
Choose $I \neq a+I \in R/I$.
Hence, $a\notin I$.
Choose, $J = \langle a, I \rangle$
Then, $J$ is an ideal of $R$  which properly contains $I$ . Then the maximality of $I$ implies $J=R$.
Since, $1\in J $ implies $1= ar +i $
Hence, $1-ar \in I$ and $1+I=ar+I$ .
Implies $(a+I) (r+I) =1+I$
Hence, $a+I$ is unit in $R/ I$ .
Conversely, Suppose $R/I $ is field.
Claim : $I$ is maximal.
Choose, $I\subset J$ an ideal of $R$.
Then, $\exists a \in J \setminus I$
Then, $a+I \neq I $ and since $R / I$ is a field , $\exists r\in R $ such that $(a+I) (r+I) =1+I$
Implies, $ab-1\in I \subset J$
Since,$r\in R $ and $J$ is ideal, $ar\in J$ .
Hence, $(ar)-(ar-1) \in J \implies 1\in J $ .
Hence, $J=R$ i.e $I$ is a maximal ideal of $R$ .

Question 1 : $R$ field implies only two trivial ideals $\langle 0 \rangle$ and $ \langle 1 \rangle$ .
Hence, $R/I \cong \langle 1 \rangle $ or $\langle 0 \rangle $
(That's why we study quotient ring but not quotient field! )
Question 2 : Chhose, any commutative ring with unity and $I$ , a maximal ideal then $R/I$ is a field (doesn't matter whether $U(R)=R\setminus \{0\} $ or not!)
Simplest example : $\Bbb{Z}/ 2\Bbb{Z}\cong \Bbb{Z_2} $ is a field but $\Bbb{Z} $ is not a filed as $U(\Bbb{Z}) =\{\pm {1}\}$
