Show that $K $ is prime ideal of $K $

Question

Let $K $ be a field. Show that $K$ is a prime ideal of $K$.

The problem I found is $K / K = 0 $ but ring $0$ is not domain. Where is the mistake?

Answer 1

The usual definition of a prime ideal of a ring $A$ is an ideal $\mathfrak p\subsetneq A$ such that for $a,b\in A,$ if $ab\in\mathfrak p,$ $a\in \mathfrak p$ or $b\in \mathfrak p,$ and does not include the case when $\mathfrak p=A.$

Let $\mathfrak a\subseteq K$ be an ideal. If for some $x\in K\setminus \{0\}=K^\times,$ $x\in\mathfrak a,$ then for $y\in K,$ $y=(yx^{-1})x\in\mathfrak a,$ and hence $\mathfrak a =K.$ Otherwise, since $0\in \mathfrak a$ and for any $x\in K^\times,$ $x\not\in \mathfrak a,$ $\mathfrak a=0.$

From the arguments above, $0$ and $K$ are the only ideals of $K.$ As said in the first paragraph, $K$ is not a prime ideal, but since $K$ is an integral domain, $0$ is a prime ideal. Thus, $0\subseteq K$ is the only prime ideal of $K.$