Let $\phi:A\to B$ ne a ring homomorphism. Let $X=\text{Spec}(A)$ and $Y=\text{Spec}(B)$. If $q\in Y$, then $\phi^{-1}(q)$ is a prime ideal. Hence, $\phi$ induces a mapping $\phi^*:Y\to X$.
Let $A$ be an integral domain with just one non-zero prime ideal $p$, and let $K$ be the field of fractions of $A$. Let $B=(A/p)\times K$. Define $\phi:A\to B$ by $\phi(x)=(\overline{x},x)$, where $\overline{x}$ is the image of $x$ in $A/p$. Show that $\phi^*$ is bijective.
This is problem 22 (part vii) from pg. 13 of "Introduction to Commutative Algebra" by Atiyah-Macdonald.
I have a doubt regarding the bijectivity of $\phi^*$. I feel that there are more prime ideals in $B$. For instance, $(0,K), ((A/p),0)$ and $(0,0)$ are all prime ideals in $(A/p)\times K$, whose inverses are only $(0)$ and $p$. Isn't this a contradiction?
In general we have that $\operatorname{Spec} (R\times S)$ is the disjoint union $\operatorname{Spec} (R) \sqcup \operatorname{Spec} (S)$.
Hence we have $\operatorname{Spec}(B)=\operatorname{Spec}(A/p) \sqcup \operatorname{Spec}(K)$.
These are both fields, hence $\operatorname{Spec}(B)$ has two points, just like $\operatorname{Spec}(A)$. You just have to check that $\phi^{-1}(0 \times K)=p$ and $\phi^{-1}(A/p \times 0)=0$, which is quite easy.
Note that $\phi$ is only bijective and continuous, but not an homeomorphism: $\operatorname{Spec}(B)$ has two closed points, but $\operatorname{Spec}(A)$ has one closed point and one generic point.