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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.

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For a shorter proof, you could use the correspondence theorem for rings. If $J$ is an ideal such that $I \subseteq J \subseteq R$, then $J/I$ is an ideal of $R/I$. Conversely, any ideal of $R/I$ is of the form $J/I$ with $I \subseteq J \subseteq R$, where $J$ is an ideal of $R$. – Mikko Korhonen May 19 '12 at 14:23
@m.k. Thank you! – Rudy the Reindeer May 19 '12 at 20:27
up vote 10 down vote accepted

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 $x=3$.

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Your example gives a field, but I saw how $I + (x) = R$ meant that after sending $I$ to $0$, if I sent $x$ to $0$, everything collapsed, which meant $x$ was a unit in $R/I$. Further, $I$ needn't be maximal, i.e. this should work when $R/I$ is not a field. I think a more illuminating example is $I = 4\mathbb Z = (4)$. The lattice involves $(1), (2), (4)$, and $(3), (6), (12)$ "below" them. Sending $(4)$ to $(0)$, also sends $(12)$ there, meaning $(6)$ collapses to $(2)$ and $(3)$ collapses to $(1)$. $(2)$ is the only remaining maximal ideal and $3 + I$ isn't in it, so it's a unit. – Travis Bemrose Nov 13 '15 at 8:24

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.

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