The only ideals of a field are zero and the whole field

Statement

Let $K$ be a field and let $I$ be an ideal of $K$. Then $I = \{0\}$ or $I = K$.

Remarks

The proof: if $a \in I$ and $a \neq 0$, then $1 = a^{-1} a \in I$, so $r = r \cdot 1 \in I$ for each $r \in K$. The converse holds: a nonzero commutative ring whose only ideals are $\{0\}$ and the whole ring is a field. For each nonzero $a$, the ideal $(a)$ is the whole ring, so $1 = b a$ for some $b$.

Explanations

L1 human

If an ideal of a field contains a nonzero element, then it contains $1$, because you can divide. Once an ideal contains $1$, it contains every element. So a field has only the two trivial ideals.