Show that $f^*:Spec(B)\rightarrow Spec(A)$ is a closed mapping. Let $f:A \rightarrow B$ be an integral homomorphism of rings. Show that $f^*:Spec(B)\rightarrow Spec(A)$ is a closed mapping.
My Try:
So, $B$ is integral over $f(A)$. $f^*$ is given by $b\longmapsto b^c$. A closed set of $Spec(B)$ is of the form $V(E)$ where $E\subset B$ . We have to show that $f^*(V(E))=V(F)$ for some $F\subset A$. 
$f(A)$ is a subring of $B$. So, if I take any prime ideal $p$ in $ f(A)$ then by a theorem there is a prime ideal $q$ in $B$ such that $q\cap f(A)=q^c=p$. Now how can I proceed towards the solution from here. Can anybody please help me?
 A: First of all one can reduce to the case that $f$ is injective (hence we can think of $A$ as a subring of $B$): $f$ factors as $A \to A/ker(f) \hookrightarrow B$ and the corersponding morphism of schemes $f^*$ thus factors as $$Spec(B) \to Spec(A/ker(f)) \to Spec(A)$$
Since the second map of this composition is well known to be a closed immersion, we only have to show that $Spec(B) \to Spec(A/ker(f))$ is closed, so we are reduced to the case that $f$ is injective.
We want to show: If $f: A \hookrightarrow B$ is an integral ring extension, we have $f^*(V(I)) = V(f^{-1}(I))$. This shows the result.
We want to reduce this to the case $I=0$. The case $I=0$ states precisely that $f^*$ is surjective, which you already have solved in your question (This is Going Up).
So we are left to show that we can really reduce to this case. We have a commutative diagram
$\require{AMScd}$
\begin{CD}
A @>f>> B\\
@VVV @VVV\\
A/f^{-1}(I) @>>> {B/I} \\
\end{CD}
where the bottom horizontal arrow is also an integral ring extension. This gives rise to a commutative diagram of schemes
$\require{AMScd}$
\begin{CD}
Spec(A) @<f^*<< Spec(B)\\
@AAA @AAA\\
Spec(A/f^{-1}(I)) @<<< Spec(B/I) \\
\end{CD}
where both horizontal arrows are surjective by the treatment of the $I=0$-case. Since the image of $Spec(B/I) \to Spec(B)$ is well known to be $V(I)$, we obtain that $f^*(V(I))$ is the image of $Spec(B/I) \to Spec(B) \to Spec(A)$. By commutativity this is the image of $Spec(B/I) \to Spec(A/f^{-1}(I)) \to Spec(A)$, which is the same as (by surjectivity of the first map) the image of $Spec(A/f^{-1}(I)) \to Spec(A)$, which is precisely $V(f^{-1}(I))$.
