Condition for a quotient commutative ring to be a field

Let $𝑅$ be a commutative ring and $𝐼 ⊴ 𝑅$ be an ideal. Then the Quotient ring $𝑅 / 𝐼$ is a field iff $𝐼$ is a Maximal ideal. #m/thm/ring

Proof

Assume $𝑅 / 𝐼$ is a field and $𝐼 ⊊ 𝐽 ⊴ 𝑅$ . Let $𝑏 \in 𝐽 ∖ 𝐼$ so that $𝑏 + 𝐼 ≢ 0$ , whence there exists $𝑎 \in 𝑅$ such that $(𝑎 + 𝐼) (𝑏 + 𝐼) = 1 + 𝐼$ . Since $𝑏 𝑎 \in 𝐽$ ,

1 + 𝐼 = (𝑏 + 𝐼) (𝑎 + 𝐼) = 𝑏 𝑎 + 𝐼

whence $1 - 𝑏 𝑎 \in 𝐼 ◃ 𝐽$ and therefore $1 \in 𝐽$ , implying $𝐽 = 𝑅$ .

For the converse, let $𝐼 ⊴ 𝑅$ be maximal. Since $𝑅 / 𝐼$ is automatically a commutative ring, it remains only to show that $𝑅 / 𝐼$ is a division ring. Let $𝑏 \in 𝑅 ∖ 𝐼$ and

𝐽 = ⟨ 𝑏, 𝐼 ⟩ i d e a l = {𝑏 𝑐 + 𝑎 : 𝑎 \in 𝐼, 𝑐 \in 𝑅}

be the ideal generated by $𝐼 \cup {𝑏}$ . Since $𝐼$ is maximal, $𝐽 = 𝑅$ and in particular $1 \in 𝑅$ . Hence $1 = 𝑏 𝑐 + 𝑎$ for some $𝑎 \in 𝐼$ and $𝑐 \in 𝑅$ , thus

1 + 𝐴 = 𝑏 𝑐 + 𝑎 + 𝐴 = 𝑏 𝑐 + 𝐴 = (𝑏 + 𝐴) (𝑐 + 𝐴)

as required.

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

#state/tidy | #lang/en | #SemBr