This is example I-22 in Eisenbud's Geometry of Schemes.
Let $K$ be a field and $R=K[x]_{(x)}$ the localization of $K[x]$ at the maximal ideal (x). So $R$ is basically $K(x)$ except the elements where the denominator has a non-zero constant term. In the book it is written that $Spec(R)=\{(0), (x)\}$, but how can $(x)$ be a prime ideal of $R$, since it clearly contains $1$ and hence is equal to $R$, which is by definition not a prime ideal?