[[Field]]
# Condition for a quotient commutative ring to be a field

Let $R$ be a [[commutative ring]] and $I \trianglelefteq R$ be an [[ideal]].
Then the [[quotient ring]] $R / I$ is a [[field]] iff $I$ is a [[Maximal ideal]]. #m/thm/ring 

> [!check]- Proof
> Assume $R / I$ is a field and $I \subsetneq J \trianglelefteq R$.
> Let $b \in J \setminus I$ so that $b + I \not\equiv 0$,
> whence there exists $a \in R$
> such that $(a + I)(b+I)= 1 + I$.
> Since $ba \in J$, 
> $$
> \begin{align*}
> 1 + I = (b + I)(a + I) = ba + I
> \end{align*}
> $$
> whence $1 - ba \in I \triangleleft J$
> and therefore $1 \in J$, implying $J = R$.
> 
> For the converse, let $I \trianglelefteq R$ be maximal.
> Since $R / I$ is automatically a [[commutative ring]],
> it remains only to show that $R / I$ is a [[division ring]].
> Let $b \in R \setminus I$ and
> $$
> \begin{align*}
> J = \langle b,I \rangle_{\text{ideal}}  = \{ bc + a : a \in I, c \in R \}
> \end{align*}
> $$
> be the ideal generated by $I \cup \{ b \}$.
> Since $I$ is maximal, $J = R$ and in particular $1 \in R$.
> Hence $1 = bc + a$ for some $a \in I$ and $c \in R$, thus
> $$
> \begin{align*}
> 1 + A = bc + a + A = bc + A = (b+A)(c+A)
> \end{align*}
> $$
> as required. <span class="QED"/>

As a corollary, a ring is a field iff it has no nontrivial proper ideals. ^C1

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