Proof that the only ideals of a field are zero and the whole field

Proof

Let $K$ be a field and let $I$ be an ideal of $K$ .
Since $I$ is an additive subgroup of $K$ , $0 \in I$ . So if $I$ has no nonzero element, then $I = \set{0}$ .
Otherwise, let $a \in I$ with $a \neq 0$ . Since $K$ is a field, there is $b \in K$ with $b a = 1$ .
Since $I$ is an ideal, $b a \in I$ . So $1 \in I$ .
Let $r \in K$ . Since $I$ is an ideal and $1 \in I$ , $r = r \cdot 1 \in I$ . So $K \subseteq I$ , and $I = K$ .
Hence $I = \set{0}$ or $I = K$ .

Idea

A nonzero element of an ideal of a field can be multiplied by its inverse. So the ideal contains $1$ , and then it contains every multiple of $1$ , which is every element.

Reading path

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Group concept
Ring definition
Commutative ring definition
Ideal definition
Unit of a ring definition
Field definition
Proof that the only ideals of a field are zero and the whole field proof

Where this leads

70 pages build on this one.

Tag	`HAP2G0H9NC`
URI	`m2://HAP2G0H9NC`
Kind	proof
Writing	W0 stub
MSC	12E99
Proves	The only ideals of a field are zero and the whole field
Uses	Field, Ideal
Changed	4 November 2026

Proof that the only ideals of a field are zero and the whole field

Proof

Idea

Reading path

Where this leads

Further reading

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