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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 $R$.2012-05-19
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    @RudytheReindeer Why in the first proof you take $(1-bc)$ belonging to $I$. This is a bit confusing for me. And then you say that $(1-bc)+bc \in B$ thus $B=R$. Why is this true?2016-09-28
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    @Marion It's explained in the same sentence: because $1 + I = bc + I$.2016-09-28
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    @RudytheReindeer I am missing something very simple then. I agree that $1+I = bc + I \in R/I$. I understand everything except for the last line. The first thing I do not understand is why $1-bc \in I$ and why you (or the author) consider this. Then why $1=1-bc \in B$ means that $B=R$.. sorry for this2016-09-28
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    @Marion Well what happens if you subtract $bc + I$ from $1 + I$?2016-09-28
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    @RudytheReindeer then we get $1-bc$ but why is this in $I$? $bc$ is not in $I$.2016-09-28
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    @Marion No we don't: In $R/I$ the elements are equivalence classes. I cannot tell if you are being sloppy or do not understand this. Please try again and this time try to be very strict with notation.2016-09-28
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    @RudytheReindeer I had the impression that elements in $R/I$ are elements that differ by $I$. This is why I wrote that above. You have to excuse my sloppiness but I am a physicist.2016-09-28
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    @Marion Let's look at a concrete example: $R = \mathbb Z$ and $I = 4\mathbb Z$. What is an example of an element in $R/I$?2016-09-28

3 Answers 3

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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.2015-11-13
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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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    Pedantic comment: $R/I$ needs to be nonzero (which of course is the case if $I\neq R$).2016-08-02
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    This is using rings 4th isomorphism theorem right?2018-11-01
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$\mathfrak m$ 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.