Shorter proof of $R/I$ is a field if and only if $I$ is maximal Here is a proof I saw somewhere of the fact $R/I$ is a field if and only if $I$ is maximal:
$\implies$ Suppose that $R/I$ is a field and $B$ is an ideal of $R$ that properly contains $I$. Let $b \in B$ but $b \notin I$. Then $b + I$ is a nonzero element of $R/I$ and therefore there exists an element $c + I$ such that $(c + I)(b + I) = 1 + I$. Since $b \in B$ we have $bc \in B$. Because $1 + I = (c + I)(b + I) = bc + I$ we have $1 - bc \in I \subset B$. So $1 = (1-bc) + bc \in B$. Hence $B = R$.
$\Longleftarrow$ Now suppose $I$ is maximal and let $b \in R$ but $b \notin I$. Consider $B = \{br + a \mid r \in R, a \in I \}$. This is an ideal properly containing $I$. Since $I$ is maximal, $B = R$. Thus $1 = bc + a^\prime$ for some $a^\prime \in I$. Then $1 + I = bc + a^\prime + I = bc + I = (b + I)(c + I)$.
I thought this was fairly long so I tried to come up with a shorter proof. Can you tell me if this is right:
$\implies$ Assume that $R/I$ is a field and $I$ is not maximal. Then there exists an $x \in R - I = I^c$ that is not a unit (otherwise $I$ would be maximal). Then $x + I$ does not have an inverse hence $R/I$ is not a field.
$\Longleftarrow$ Assume $I$ is maximal and $R/I$ is not a field. Then there is an $x$ such that $x + I \neq 0 + I$ does not have an inverse. This $x$ is not in $I$ and $x$ is not a unit. Hence $I \subsetneq I + (x) \subsetneq R$. Which contradicts $I$ being maximal.
 A: 
$\mathfrak m$ is a maximal ideal $\Leftrightarrow R/\mathfrak m$ is nonzero and has no proper nonzero ideal.

$\Rightarrow$: Suppose $\{0\}\ne I\subset R/\mathfrak m$, $I=\{a+\mathfrak m:a\in A\}$, and $A$ is as large as possile. $I$ is an ideal, so $a_1,a_2\in A\Rightarrow a_1+\mathfrak m,a_2+\mathfrak m\in I\Rightarrow(a_1+\mathfrak m)+(a_2+\mathfrak m)=(a_1+a_2)+\mathfrak m\in I\Rightarrow a_1+a_2\in A$
and $a\in A\Rightarrow a+\mathfrak m\in I\Rightarrow(r+\mathfrak m)(a+\mathfrak m)=ra+\mathfrak m\in I\Rightarrow ra\in A$ for any $r\in R$. $A$ is an ideal. $I\subset R/\mathfrak m\Rightarrow A\subset R$; $A$ is the largest $\Rightarrow\mathfrak m\subseteq A$; $I\ne\{0\}\Rightarrow A-\mathfrak m\ne\emptyset$.
$\Leftarrow$: $\mathfrak m\subseteq\mathfrak n\subset R\Rightarrow R/\mathfrak n\subseteq R/\mathfrak m\Rightarrow R/\mathfrak n=R/\mathfrak m\Rightarrow\mathfrak m=\mathfrak n$.

$R$ has no proper nonzero ideal $\Leftrightarrow R$ is a field.

$I\ne\{0\}$ and $I\subset R\Rightarrow$ no element in $I$ has an inverse $\Rightarrow R$ is not a field;
$R$ is not a field $\Rightarrow\exists$ nonzero $r\in R$ such that $1\notin rR\Rightarrow rR$ is a proper nonzero ideal.
A: Here is a short proof.
Let $R$ be a ring, $I\triangleleft R$ be an ideal.
We will need two facts:

*

*The correspondence theorem for rings, which says there is a bijection between the ideals of $R$ containing $I$ and the ideals of $R/I$.


*$\{0\}\neq R$ is a field $\iff$ the only ideals of $R$ are $\{0\}$ and $R$. (Short proof below)
Thus:
$R/I$ is a field $\iff$ The  only ideals of $R/I$ are $I$ and $R/I$ (by (2))
$\iff$ The only ideals of $R$ containing $I$ are $\{0\}$ and $R$ (by (1))
$\iff$ $I$ is maximal $\square$

Proof of (2): $\{0\}\neq R$ is a field $\iff$ the only ideals of $R$ are $\{0\}$ and $R$
$\Rightarrow$: Let $\{0\}\neq I\triangleleft R$ be an ideal. Since $R$ is a field, $I$ contains a unit, hence $I$ must contain the identity element $1$. (If $a\in I$ is a unit, $\exists a^{-1}\in R$. For all $r \in R$, $rI\subset I$, so in particular $1 = a^{-1}a \in I$). Hence $I$ contains all the elements of $R$, i.e. $R = I$.
$\Leftarrow$: Let $x \in R - \{0\}$. Then $\{0\}\neq (x) \triangleleft R$, where $(x)$ denotes the ideal generated by $x$. So by the assumption, $(x) = R$, and in particular $\exists y \in R : xy = 1$, i.e. $x$ is a unit. Thus any $x \in R - \{0\}$ is a unit $\Rightarrow R$ is a field. $\square$
A: Both directions of your proof are wrong. If $x$ is not a unit in $R$, it's still possible for $x+I$ to be a unit in $R/I$. If $x$ is not in $I$ and not a unit, it's possible for $I+(x)$ to be $R$. In both cases, you can take $R=\mathbb{Z}$, $I=2\mathbb{Z}$, and $x=3$. 
A: I think m.k.'s comment is right on the money: assuming you can prove that a commutative unitary ring is a field iff it has no non-trivial ideals (when by "trivial ideal" here we understand the whole ring and the zero ideal.):
$R/I\,$ is a field $\,\Longleftrightarrow \nexists\,\,\text{non-trivial}\,\, J/I\leq R/I\Longleftrightarrow \nexists\,\,\text{non-trivial}\,\,I\lneq J\lneq R\Longleftrightarrow I\, $ is a maximal ideal.
A: Let f:R-F be a surjective ring homomorphism. Now F is a field iff there are two ideals, namely (0) and (1), i.e. the zero and unit ideal. The correspondence theorem tells us that there are exactly two ideals in R, containing I= kernel of f.
We know for sure that these two ideals are I itself and R. So I is the maximal ideal of R.
Now F is isomorphic to R/I (from 1st Isomorphic theorem). So, R/I is a field.
Conversely, let h: R- R/I is a surjective ring homomorphism, where R/I is a field. Now there are two ideals in R/I.(same as above). So, by correspondence theorem R has only two ideals that contains the kernel of h. But kernel of h is I as it is the zero element of R/I. So, only two ideals of R (that contains I) are I and R itself. So, I is maximal.
Sorry, if this is not satisfying. Correspondence theorem is the most essential tool to understand this.
